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Existence of Modular Sonar Sequences
of Twin-Prime Product Size
for CDMA Signal Sets
Sung-Jun Yoon
The Graduate School
Yonsei University
Department of Electrical and Electronic
Engineering
2
This certifies that the dissertation of
Sung-Jun Yoon is approved.
_________________________
Thesis supervisor : Hong-Yeop Song
_________________________
Dong-Ku Kim
_________________________
Jong-Moon Chung
The Graduate School
Yonsei University
June 2007
3
Contents
List of Figures
List of Tables
Abstract
1. Introduction
1.1 Motivation
1.2 Overview
2. Interleaved construction of signal set
2.1 Interleaved sequences
2.2 Interleaved constructions for signal sets
2.3 Shift sequence e in the )32,,( 2 vvv signal sets construction
2.3.1 Criterion for shift sequence e
2.3.2 Two known methods to construct the shift sequence e
2.3.3 Comment
4
3. Modular sonar sequence
3.1 Definition of Modular sonar sequence
3.2 Constructions of Modular sonar sequence
3.2.1 Known Modular sonar sequence constructions
3.2.2 Transformation of Modular sonar sequence
3.3. Relation Between vv Modular sonar sequence and shift
sequence e
4. )32,,( 2 vvv signal set with v=p(p+2)
4.1 The case of v=3ⅹ5=15
4.1.1 Result summary
4.1.2 Analysis
4.2 The Case of v=5ⅹ7=35
5. Concluding remarks
Appendix
Bibliography
5
Abstract
Existence of Modular Sonar Sequences
of Twin-Prime Product Size
for CDMA Signal Sets
An exhaustive search for 15ⅹ15 Modular sonar sequences found 9000
examples which are grouped with 5 equivalent classes. Four classes of them
can obtained by the known “Shift Sequence” construction, and the
remaining 1 class turn out to be new, and has some interesting property.
We prove that, for an odd v, any given vv Modular sonar sequence with
self-reciprocal property must satisfy the same interesting property shared by
the new class of size 15 that we have found.
We applied these observation to the case 3575 v for a partial search,
and fail to locate any example. Furthermore, we approached this problem
with some algebraic constructions and transformations of the obtained
Modular sonar sequences, but unfortunately, all failed. This implies the
6
general case of )2()2( pppp Modular sonar sequence seem to be
non-existing.
Index Terms－ Modular sonar sequence, Interleaved construction of
signal set, twin prime
7
Chapter 1
Introduction
1.1 Motivation
A family of pseudorandom sequences with low cross correlation, good
randomness, and large linear span has important application in code-division
multiple-access (CDMA) communications and cryptology [17].
It has long been conjectured that if a balanced binary sequence of period v
has the ideal two-level autocorrelation, then v must be either 12 n for
some positive integer n, a prime p of type 4k+4, or a product of twin-prime.
It is interesting to note that those three types of integers seem to have no
common property except for the above. For the lengths v other than those
listed above, no such binary sequences of period v are currently known,
neither the proof of non-existence of such examples are completed. On the
other hand, for search of these types of lengths, at least one easy
construction for such sequence are well-known [8][12][13][14][15].
8
In [1], Gong introduce a new design for families of binary sequences
with low cross correlation, balance property, and large linear span. The key
idea of this new design is to use short binary periodic sequences with two-
level autocorrelation function and an interleaved structure to construct a set
of long binary sequences with the desired properties. This property also has
significant meaning with the application on signal detection of high-speed
broad-band communication system.
Let vZ be the integers mod v. This Gong’s construction gives a
)32,,( 2 vvv signal set consisting of v binary sequences of period 2v
whose out-of-phase autocorrelation and cross-correlation maximum is
bounded by 2v+3. The construction requires two binary sequences of period
v with the ideal two-level autocorrelation, and a “so-called” shift sequence
e of length v defined over the alphabet vZ . She proved that the
construction works in general if there exists two sequences with the ideal
autocorrelation together with a “shift sequence” e with desired property,
and specifically gave constructions for e as the following two cases : (1)
when 1 npv for a prime p and a positive integer n, and (2) when
pv which is a prime of type 34 k [1].
9
As long as binary sequences are concerned, the above construction uses
two well-known types of balanced binary sequences of period v with the
ideal two-level autocorrelation : (1) 12 nv and (2) pv is a prime of
type 4k+3. These two case cover all types of two-level binary
autocorrelation sequences as “building blocks” except for a class of two-
level autocorrelation sequences of period v where v is a twin prime p(p+2).
What is happening to these twin prime product type missing here? This is
the main motivation of this paper.
We first have recognized that the construction for “shift sequence” e in
[1] is the same at the construction for Modular sonar sequence of length v
mod v [5]. In fact, it is essentially the same at the one given by Games in [4]
for the case 12 nv or 1np , or the exponential-Welch construction in
[6][5] for the case pv of type 4k+3.
To investigate the missing case, that is, the case )2( ppv , we have to
design a construction for Modular sonar sequences over vZ of length v
mod v, which is unfortunately not completed yet. That is, we were not able
to show the non-existence, nor we could find one single example, except for
the one special case v=15. Note that the case v=15 is very special in that it is
10
the only case which is both product of twin-prime and of the form 12 n .
Then the rest of the construction comes easily for families of
)32,,( 2 vvv signal sets for )2( ppv by way of the interleaved
construction in [1], since the existence of balanced binary sequence of
period )2( ppv with the ideal two-level autocorrelation is well-known
[8][14].
1.2 Overview
In Chapter , we first will review interleaved sequences and constructions
for signal set, and shift sequence e construction with criterion of shift
sequence e and two known methods to construct these sequence. In
Chapter 3, we will introduce nn Modular sonar sequence that has same
property with shift sequence e . Next Chapter we will introduce the
)32,,( 2 vvv signal set with )2( ppv from executive search of the
first small case of 1553 , and consider existence of general case of
)2()2( pppp Modular sonar sequence. Finally all those of this
dissertation are summarized and some discussions follow in Chapter 5.
11
Chapter 2
Interleaved Construction of Signal Set
In this Chapter, we review Gong’s new design methods for families of
sequences over )( pGF with low correlation, balance property, and large
linear span. By Gong’s interleaved construction [1], we can obtain
)32,,( 2 vvv signal set that has interleaved sequences of period 2v ,
v cyclically shift distinct sequences in each family, and the maximal
correlation value 32 v which is optimal with respect to the Welch bound.
In particular, for binary case, cross/out-of-phase autocorrelation values
belong to the set {1, - v, v+2, 2v +3, -2v -1} and any sequence where the
short sequences are quadratic residue sequences achieves the maximal linear
span.
The key idea of Gong’s new design is to use short p -ary sequences of
period v with two-level autocorrelation function together with the
interleaved structure to construct a set of long sequences with the desire
12
properties. So, we start with interleaved sequences, and interleaved
construction in the first two sections. In section 3, we describe Gong’s
interleaved construction for )32,,( 2 vvv signal set with two
construction methods of shift sequence e .
2.1 Interleaved sequences
Gong introduced the concept of interleaved sequences over )( pGF in
[7]. Let },{ 1,1,0 stuuu u be a p-ary sequence of period st where 1s
or 1t . We can arrange the element of the sequence u in to a s by t
matrix as follows :
1)1(1)1()1(
12122
11
110
ttststs
tttt
tttt
t
uuu
uuu
uuu
uuu
A
If each column vector of the above matrix is either a phase shift of a p-ary
sequence, say a of period s, or zero sequence, then we say that u is an (s,
t) interleaved sequence over )( pGF and A is the matrix form of u . Let
13
jA called column(componet) sequence of u be the jth column vector of
A, then ),,,( 110 tAAAA . According to the definition, )(aje
j LA ,
10 tj where we denote je if )0,,0,0( jA and )(ajeL
is left shift operator (that is )(ajeL = ),,,,,,( 1011
jejejaaaaa ve
for a ),,,( 110 vaaa ). ),,,( 110 teee e is called a shift
sequence of u . For a given e and a , the interleaved sequence u is
uniquely determined. So u is an (s, t) interleaved sequence associated with
),( ea .
Example 1 : Let
e = (1, 3, 2, 6, 4, 5, 1)
and
a = (1, 1, 1, 0, 1, 0, 0)
be a binary m-sequence of period 7. Then a 7 by 7 matrix A can be given as
follows :
14
1100111
0011110
0110010
1111001
0101100
1001011
1010101
A
From A, we have the following (7, 7) interleaved sequence of period 49.
u = 1010101110100100110101001111010011001111001110011
2.2 Interleaved construction for signal sets
In this section, we will give a procedure to construct a signal set
consisting of u ),( vv interleaved sequences.
Procedure
1) Choose
),,,( 110 vaaa a
and
),,,( 110 vbbb b
15
two sequence over )( pGF of period v with two-level autocorrelation.
2) choose
),,,( 110 veee e ,
an integer sequence whose elements are taken from vZ .
3) Construct
),,,(110 2
v
uuu u ,
a (v, v) interleaved sequence whose jth column sequence is give by
)(ajeL where vj 0 .
4) Set
),,,(1,1,0, 2
vjjjj sss s , vj 0
whose elements are defined by
,1, jiij bus vji ,0 or equivalently )(bus j
j L
5) A signal set ),,( ebaSS is defined as
}1,,1,0:{ vjS j s.
16
We called a and b base sequences of S and e a shift sequence of S.
From Steps 1)-4) in Produce has a matrix from kk BAS where
111
222
111
110
110
110
110
veveve
eee
eee
eee
v
v
v
v
aaa
aaa
aaa
aaa
A
and
11
11
11
11
vkk
vkk
vkk
vkk
k
bbb
bbb
bbb
bbb
B
If we write ),,,(1,1,0,
vkkkk SSSS , then column j of
kS is given by
,)( kj
e
k bLS j
a vj 0 .
Example 2: Let p=2 and v=7
1) Choose a = (1110100), b =(1001011), m-sequences of period 7.
2) Choose e = (1, 3, 2, 6, 4, 5, 1)
3) Construct the interleaved sequence u =
1010101110100100110101001111010011001111001110011 which is the
same example 1 and has the matrix form A.
4) Construct kS : 6,,1,0 k .
17
0001110
1110111
1011011
0010000
1000101
0100010
0111100
1101001
1101001
1101001
1101001
1101001
1101001
1101001
1100111
0011110
0110010
1111001
0101100
1001011
1010101
00 BAS
In this fashion, we obtain seven sequences in S :
1011100011011110110000100110010101000100111001000 s
11110010010001010101101100001111000011010000101111 s
10101110101000001000110000100011101101011110111002 s
00010111111010110000001001111010110001100010011113 s
01100101011111000011111011000000001000111011001014 s
10000000110100100110011110110101111010001001110015 s
00001011100010101100010101110110111111101100010006 s
5) Set },,,{ 610 sss S .
18
2.3 Shift sequence e in the )32,,( 2 vvv signal sets construction
In this section, first we will give a criterion for the shift sequence e
such that S, constructed by Procedure in above section, is )32,,( 2 vvv
signal set. Then we will provide two methods to construct the shift sequence
),,,( 110 veee e satisfying the criterion.
2.3.1 Criterion for shift sequence e
Theorem 2.1 : Let S be constructed by Procedure in section 2. If the shift
sequence ),,,( 110 veee e satisfies the following condition :
.1 allfor }0|{ vssvsvjee sjj (2. 1)
Then S is a )32,,( 2 vvv signal set.
A proof of this Theorem can be found in [1].
If shift sequence ),,,( 110 veee e satisfies the condition of (2.1) in
Theorem 2.1, any pair of sequences in S are shift distinct, each sequence in
19
S has period 2v , and the maximal correlation of S is 32 v .
2.3.2 Two known methods to construct the shift sequence e
In this subsection, we will describe two known methods to construct the
shift sequence e that satisfies the condition of (2.1) in Theorem 2.1.
Construction A : )12,1,)1(( 2 nnn ppp signal set
1) Choose a and b , two sequence over )( pGF of period 1np
with two-level autocorrelation (may not be different).
2) Choose and , primitive elements of )( 2n
pGF and )( 2n
pGF ,
respectively.
3) Computer je ’s satisfying
10 ),(Tr 12 njn
n
epjj (2.2)
where )(Tr 2 xn
n is the trace function from )( 2n
pGF to )( 2n
pGF
and )( 2n
pGF with .0)(Tr 2 n
n
20
Construction B : )32,,( 2 vvv signal set
1) Choose a and b , two-level autocorrelation sequences over )2(GF
with a prime period v .
2) Choose , a primitive element of )(vGF .
3) Computer je ’s satisfying
vje j
j 0 , . (2.3)
Lemma 2.1 : The shift sequence ),,,( 110 veee e , constructed
form Construction A for 1 npv or Construction B for v a prime,
satisfies the condition (2.1) of in Theorem 2.1.
A proof of this Lemma can be found in [7].
2.3.3 Comment
In )32,,( 2 vvv signal sets constructed form Construction A and
Construction B, we choose two-level autocorrelation sequences for binary
sequences a and b of period v . (In construction A, we also choose p-ary
21
sequences of period v with two-level autocorrelation.) There are many
known constructions (that are m-sequences [9], GMW sequences [11],
Quadratic Residue Sequences(Legendre Sequences) [9], Hall’s Sextic
Residue Sequences [10], Jacobi symbol construction [8], and etc.) for binary
sequences of period v with two-level autocorrelation. The binary sequences
of period v for shift sequence ),,,( 110 veee e in Construction A for
12 nv or Construction B for v prime allow to utilize all types of two-
level autocorrelation sequences as “building blocks” except for a class of
two-level autocorrelation sequences of period v where v is a product of twin
prime p(p+2).
Therefore, if we find shift sequence ),,,( 110 veee e for
)2( ppv , we can design new )32,,( 2 vvv signal sets from
construction similar to Construction A or Construction B. In this new case,
we can use Jacobi symbol construction for binary sequences a and b of
period )2( ppv .
A family of pseudorandom sequences with low cross correlation, good
randomness, and large linear span such above signal set has important
application in code-division multiple-access (CDMA) communications and
22
cryptology. The pseudorandom sequences with low cross correlation
employed in CDMA communications can successfully combat interference
from the other users who share a common channel [7].
23
Chapter 3
Modular sonar sequence
In above Chapter, we review shift sequence ),,,( 110 veee e for
the construction of )32,,( 2 vvv signal sets, and if we find new
construction method of shift sequence ),,,( 110 veee e for
)2( ppv , we can design new )32,,( 2 vvv signal sets. In this Chapter
we review vv Modular sonar sequence that has same property with shift
sequence ),,,( 110 veee e . Because of same property of vv
Modular sonar sequence and shift sequence e , we can construct shift
sequence e by construction methods of vv Modular sonar sequence.
In the first section , we describe Modular sonar sequence with definition
and explain their properties. In the next section we introduce and explain
known construction methods of Modular sonar sequence, and we review
similarity of vv Modular sonar sequence and shift sequence e .
24
3.1 Definition of Modular sonar sequence
Sonar sequence were introduced in [2] as examples of two-dimensional
synchronization patterns with minimum ambiguity.
For integers m and n, let }.,2,1{ and },2,1{ nNmM
Definition 3.1 : Regard M as a set of representatives of the integers modulo
m. A function MNf : has the distinct modular differences property if
for all integers h, i, and j, with ,,1 11 hnjiandnh
jimjfhjfifhif i m p l i e s ) ( m o d )()()()( . (3.1)
Definition 3.2 : An nm Modular sonar sequence is a function
MNf : with the distinct modular differences property.
Example 3.1 : The sequence 1 3 6 11 9 4 4 3 11 4 11 is a 11ⅹ11 Modular
sonar sequence (mod 11). (See Figure 3.1)
3.2 Constructions of Modular sonar sequence
In this section we introduce known sonar sequences constructions and
transformations in an attempt to produce better sonar sequences.
25
1 3 6 11 9 4 4 3 11 4 11
2 3 5 9 6 0 10 8 4 7
5 8 3 4 6 10 7 1 0
10 6 9 4 5 7 0 8
8 1 9 3 2 0 7
3 1 8 0 6 7
3 0 5 4 2
2 8 9 0
10 1 5
3 8
10
Figure 3.1 : 11ⅹ11 Modular sonar sequence (Modulo 11, Length 11)
3.2.1 Known Modular sonar sequence constructions
A. Quadratic [3]
Let p be any odd prime and a, b, c be integer constants with a not
congruent to 0 (mod p). Then },,2,1{}1,,2,1{: ppf defined by
) (mod )( 2 pcbiaiif is )1( pp Modular sonar sequence.
Example 3.2 : If p=7, a=1, b=2, c=1 and )7 (mod )1(12)( 22 iiiif ,
then we can construction a 7ⅹ8 Modular sonar sequence.
26
4 2 2 4 1 7(=0) 1 4 (Modulo 7, Length 8)
B. Exponential Welch [2] [6]
Let be a primitive element modular the prime p. Then
},,2,1{}1,,2,1{: ppf defined by iaif )( is a Modular
)1( pp sonar sequence.
Example 3.3 : If p=7, =3 and )7 (mod 3)( iif , then we can
construction a 7ⅹ6 Modular sonar sequence 3 2 6 4 5 1 (Modulo 7, Length 6).
C. Logarithmic Welch [6]
Let be a primitive element modular the prime p. Then
}1,,2,1{}1,,2,1{: ppf defined by iif alog)( is a modular
)1()1( pp sonar sequence.
Example 3.4 : If p=7, =3 and 7) (mod 3 ; log)( )(
3 iiif if , then we
can construction a 6ⅹ6 Modular sonar sequence 6 2 1 4 5 3 (Modulo 6,
Length 6).
27
D. Lempel [2][6]
Let 2 rpq be a prime power, and let be a primitive element of )( rpGF .
Then }1,,2,1{}2,,2,1{: rr ppf defined by 1 iff )( jijif
is a modular )2()1( rr pp sonar sequence.
Example 3.5 : If 32 rpq and be a primitive element of )2( 3GF
whose irreducible polynomial over )2(GF is 1)( 3 xxxh , then
1264554136231 . We can
construction a 7ⅹ6 Modular sonar sequence 3 6 1 5 4 2 (Modulo 7,
Length 6).
E. Golomb [6]
Let 2 rpq be a prime power, and let and be a primitive
element of )( rpGF . Then }1,,2,1{}2,,2,1{: rr ppf defined by
1 iff )( jijif is a modular )2()1( rr pp sonar sequence.
Example 3.6 : If 32 rpq , be a primitive element of )2( 3GF
28
whose irreducible polynomial over )2(GF is 1)( 3 xxxh and be
a primitive element of )2( 3GF whose irreducible polynomial over )2(GF
is 1)( 23 xxxh , then
1266544133251 .
We can construction a 7ⅹ6 Modular sonar sequence 5 3 1 4 6 2 (Modulo 7,
Length 6).
F. Extended exponential Welch [5]
Let be a primitive element modular the prime p and s an integer.
Then },,2,1{}1,,2,1,0{: ppf defined by
siaif )( (3.2)
is a modular pp sonar sequence.
Example 3.7 : If 7p , 3 , s=0 and )7 (mod 3)( iif , then we can
construction a 7ⅹ7 Modular sonar sequence 1 3 2 6 4 5 1 (Modulo 7,
Length 7).
29
Table 3.1 : Summary of the Sonar Sequence Constructions
Constructions Length Range Modulo Reference
Quadratic 1 pn },,2,1{ p
pm [3]
Exponential Welch 1 pn },,2,1{ p
pm [2], [6]
Logarithmic Welch 1 pn }1,,2,1{ p
1 pm [6]
Lempel 2 rpn }1,,2,1{ rp 1 rpm [2], [6]
Golomb 2 rpn }1,,2,1{ rp 1 rpm [6]
Extended
Exponential Welch pn },,2,1{ p
pm [5]
Shift Sequence rpn }1,,2,1{ rp 1 rpm [4]
G. Shift sequence [4]
Let p be a prime, a primitive element of )( 2rpGF , and a
primitive element in )( rpGF .
For p=2, let }1,,2,1{},,2,1{: rr ppf be defined by
)(Tr)( ; ))((log)( 2)( ir
r
ipiifipi rr
βif (3.3)
where )(Tr2 xr
ris the trace function from )( 2rpGF to )( rpGF .
30
For p odd, defined f similarly, except change the domain to
}.2/)1(2/)1(:{ rr pipi Then, f is a rr pp )1( Modular sonar
sequence.
Note : By known (to us) Modular sonar sequence constructions, the modulo
of Modular sonar sequence have three types that is 1 ,1 , rpp-p (see
Table 1).
3.2.2 Transformation of Modular sonar sequence
An nm Modular sonar sequence can be subjected to three
transformation in an attempt to produce better sonar sequences [5].
A. Addition by a modulo m, )(mod )()( maifif a .
This corresponds to cyclically rotation the rows of the sonar array a units.
Contiguous empty rows are rotated to the top of the sonar array, and
dropped to produce a better sonar array.
B. Multiplication by a unit u modulo m, )(mod )()( miufif u .
This corresponds to a permutation of the rows of the sonar array.
31
Permutation are applied in attempt to increase the number of contiguous
empty rows in the sonar array.
C. Shearing by s modulo m, )(mod )()()( msiifif sshear .
This corresponds to shearing the columns of the sonar array by s units and
reducing modulo m. Shears are applied in an attempt to increase the number
of empty rows in the sonar array.
Note : If f is an nm Modular sonar sequence and u is a unit modulo
m, then g defined by )(mod )()( masiiufig is an nm
Modular sonar sequence.
3.3 Relation between vv Modular sonar sequence and
shift sequence e
Now we consider the definitions of distinct differences property and
Modular sonar sequence in above section. We can find that condition of
(3.1) for distinct differences property is similar to condition of (2.1) in
Theorem 2.1 for shift sequence e . If nm in equation (3.1), n in equation
(3.1) regard as v in equation (2.1), and h in equation (3.1) regard as s in
32
equation (2.1), condition of (3.1) and (2.1) has a same property. Therefore
we can find that shift sequence e is same to vv Modular sonar
Sequence.
Two methods to construct the shift sequence e in section 2.3.2 are very
similar with construction method of Modular sonar sequence in section
3.2.1. From nn pp )1( Modular sonar sequence constructed by “Shift
Sequence” method, we can obtain )1()1( nn pp Modular sonar
sequence by eliminating the last column. Because any nm Modular
sonar sequence can be shortened to a )( inm Modular sonar sequence
by eliminating i columns from the ends of the sonar array. This
)1()1( nn pp Modular sonar sequence is same to shift sequence e ,
constructed with method in Construction A by similarity of condition of
(2.2) and (3.3). Also pp Modular sonar sequence constructed by
Extended Exponential Welch method is sane to shift sequence e , constructed
with method in Construction B by similarity of condition of (2.3) and (3.2).
33
Table 3.2 : All Case of binary sequence with two-level autocorrelation and
vv Modular sonar sequence
Cases of binary sequences of period
v with two-level autocorrelation
Cases of vv Modular sonar
sequence (shift sequence e )
pv pv
12 nv 12 nv
)2( ppv unknown
In above Chapter, We knew that the cases of all known constructions for
binary sequences of period v with two-level autocorrelation are pv ,
1 npv , and )2( ppv . We also knew the construction method of
shift sequence e and vv Modular sonar sequence for pv or
1 npv , but the case of )2( ppv is not. (See Table 3.2)
34
Chapter 4
)32,,( 2 vvv Signal Sets with v=p(p+2)
In Chapter 2, we observed that the two construction methods of
)32,,( 2 vvv signal set by two generating methods of shift sequence e .
The one of the case is 1 npv and the other is pv . From Modular
sonar sequence in Chapter 3, we obtain shift sequence e from vv
Modular sonar sequence that has same property with shift sequence e . By
construction method of Modular sonar sequence that is “Shift Sequence”
and “Extended Exponential Welch”, we construct two types of shift
sequence e in above cases. These two case cover all types of two-level
autocorrelation sequences as “building blocks” except for a class of two-
level autocorrelation sequences of period v where v is a twin prime p(p+2).
Therefore we can obtain new construction method of )32,,( 2 vvv
signal set in the case of )2( ppv if we find generating method of shift
35
sequence e by )2()2( pppp Modular sonar sequence.
In this Chapter, we will start the first small case of 1553 v from
the exhaustive full search.
4.1 The case of v=3ⅹ5=15
4.1.1 Result Summary
By an exhaustive search, we found 9000 1515 Modular sonar
sequences. If we consider transformation by Addition by a modulo m which
is )(mod )()( maifif a , the number is reduced to 600(=9000/15).
Because all possible a are 0 to 14 for modulo 15m . Each these 600
sequences have 1 as the first element of sequence (see Appendix A~E). If
we also consider other transformation that is Multiplication by a unit u
modulo m or Shearing by s modulo m, the 600 1515 Modular sonar
sequences are grouped with 5 equivalent classes. Because }14,,2,1,0{ s
and
}14,13,11,8,7,4,2,1{15)} (mod 5 and 3 tocongruent not is ,150|{ iiiu ,
there are )815(120 equivalent sequences having 1 as the first element
of sequence in each class. Therefore, 1800 1515 Modular sonar
36
sequences are grouped with 5 equivalent classes, and in each equivalent
class has )15158(1800 equivalent sequences having equivalence of
)(&)( igif in given by transformation equation asi iufig )()(
)(mod m .
Four out of five equivalent classes turned out to be the ones that can be
constructed by the known method, which in called “Shift Sequence” [4].
The remaining one class is new.
4.1.2 Analysis
As we know from above section, each of the 5 equivalent classes has
1800 equivalent sequences. Representative sequences from each of the 5
equivalent classes are :
Class 1 : e = (1,3,1,7,11,3,14,15,8,8,13,7,4,14,2)
Class 2 : e = (1,1,4,1,9,7,11,1,8,2,12,13,4,6,2)
Class 3 : e = (1,1,2,14,2,13,4,9,13,12,4,2,11,6,8)
Class 4 : e = (1,1,4,9,4,11,10,8,5,9,10,1,9,5,7)
Class 5 : e = (1,6,12,13,10,14,7,9,7,14,10,13,12,6,1)
37
e 1 3 1 7 11 3 14 15 8 8 13 7 4 14 2
15 ●
14 ● ●
13 ●
12
11 ●
10
9
8 ● ●
7 ● ●
6
5
4 ●
3 ● ●
2 ●
1 ● ●
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Array form of 1515 Modular sonar sequence in Class 1
1 3 1 7 11 3 14 15 8 8 13 7 4 14 2
⑬ 2 ⑨ ⑪ 8 ④ ⑭ 7 0 ⑩ 6 3 ⑤ 12
0 ⑪ ⑤ 4 ⑫ ③ 6 7 ⑩ 1 9 ⑧ 2
⑨ ⑦ ⑬ ⑧ ⑪ ⑩ 6 2 1 4 ⑭ 5
⑤ 0 ② ⑦ 3 ⑩ 1 8 4 ⑨ 11
⑬ ④ ① ⑭ 3 ⑤ 7 11 ⑨ 6
② ③ ⑧ ⑭ ⑬ ⑪ 10 1 6
① ⑩ ⑧ ⑨ 4 ⑭ 0 13
⑧ ⑩ ③ 0 7 ④ 12
⑧ ⑤ ⑨ 3 ⑫ 1
③ ⑪ ⑫ 3 9
⑨ ⑭ ② 5
⑫ ⑥ ⑭
② 1
⑭
Modular difference Triangle of 1515 Modular sonar sequence in Class 1
Figure 4.1 : Array form and Modular difference Triangle of sequence in Class 1
38
e 1 1 4 1 9 7 11 1 8 2 12 13 4 6 2
15
14
13 ●
12 ●
11 ●
10
9 ●
8 ●
7 ●
6 ●
5
4 ● ●
3
2 ● ●
1 ● ● ● ●
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Array form of 1515 Modular sonar sequence in Class 2
1 1 4 1 9 7 11 1 8 2 12 13 4 6 2
0 ⑫ 3 ⑦ 2 ⑪ 10 ⑧ 6 ⑤ ⑭ 9 ⑬ 4
⑫ 0 ⑩ ⑨ ⑬ 6 3 ⑭ ⑪ ④ 8 7 2
0 ⑦ ⑫ ⑤ 8 ⑭ 9 ⑤ ⑩ ⑬ 6 11
⑦ ⑨ ⑧ 0 1 5 ⑭ ③ 4 ⑪ 10
⑨ ⑤ 3 ⑧ 7 ⑩ ⑭ ⑫ 2 0
⑤ 0 ⑪ ⑭ ⑫ ⑨ 7 ⑩ 6
0 ⑧ 2 ④ ⑪ 3 5 ⑭
⑧ ⑭ ⑦ ③ 5 1 9
⑭ ④ ⑥ ⑫ 3 5
④ ③ 0 ⑩ 7
③ ⑫ ⑬ ⑭
⑫ ⑩ 2
⑩ ⑭
⑭
Modular difference Triangle of 1515 Modular sonar sequence in Class 2
Figure 4.2 : Array form and Modular difference Triangle of sequence in Class 2
39
e 1 1 2 14 2 13 4 9 13 12 4 2 11 6 8
15
14 ●
13 ● ●
12 ●
11 ●
10
9 ●
8 ●
7
6 ●
5
4 ● ●
3
2 ● ● ●
1 ● ●
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Array form of 1515 Modular sonar sequence in Class 3
1 1 2 14 2 13 4 9 13 12 4 2 11 6 8
0 ⑭ ③ 12 ④ 9 ⑩ ⑪ 1 8 2 ⑥ 5 ⑬
⑭ ② 0 1 ⑬ 4 ⑥ ⑫ 9 10 ⑧ ⑪ 3
② ⑭ ④ 10 ⑧ 0 ⑦ 5 11 1 ⑬ ⑨
⑭ ③ ⑬ 5 ④ 1 0 7 2 6 ⑪
③ ⑫ ⑧ 1 ⑤ 9 2 ⑬ 7 4
⑫ ⑦ ④ 2 ⑬ 11 ⑧ 3 5
⑦ ③ ⑤ 10 0 2 ⑬ 1
③ ④ ⑬ 12 ⑥ 7 ⑪
④ ⑫ 0 3 ⑪ 5
⑫ ⑭ ⑥ 8 ⑨
⑭ ⑤ ⑪ 6
⑤ ⑩ ⑨
⑩ ⑧
⑧
Modular difference Triangle of 1515 Modular sonar sequence in Class 3
Figure 4.3 : Array form and Modular difference Triangle of sequence in Class 3
40
e 1 1 4 9 4 11 10 8 5 9 10 1 9 5 7
15
14
13
12
11 ●
10 ● ●
9 ● ● ●
8 ●
7 ●
6
5 ● ●
4 ● ●
3
2
1 ● ● ●
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Array form of 1515 Modular sonar sequence in Class 4
1 1 4 9 4 11 10 8 5 9 10 1 9 5 7
0 ⑫ ⑩ 5 ⑧ 1 2 3 ⑪ ⑭ 9 ⑦ 4 ⑬
⑫ ⑦ 0 ⑬ ⑨ 3 5 ⑭ ⑩ 8 1 ⑪ 2
⑦ ⑫ ⑧ ⑭ ⑪ 6 1 ⑬ 4 0 5 ⑨
⑫ ⑤ ⑨ 1 ⑭ 2 0 7 ⑪ 4 3
⑤ ⑥ ⑪ 4 ⑩ 1 9 ⑭ 0 2
⑥ ⑧ ⑭ 0 ⑨ 10 1 3 ⑬
⑧ ⑪ ⑩ ⑭ 3 2 5 1
⑪ ⑦ ⑨ 8 ⑩ 6 3
⑦ ⑥ 3 0 ⑭ 4
⑥ 0 ⑩ 5 ⑫
0 ⑦ ⑭ 2
⑦ ⑪ ⑫
⑪ ⑨
⑨
Modular difference Triangle of 1515 Modular sonar sequence in Class 4
Figure 4.4 : Array form and Modular difference Triangle of sequence in Class 4
41
e 1 6 12 13 10 14 7 9 7 14 10 13 12 6 1
15
14 ● ●
13 ● ●
12 ● ●
11
10 ● ●
9 ●
8
7 ● ●
6 ● ●
5
4
3
2
1 ● ●
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Array form of 1515 Modular sonar sequence in Class 5
1 6 12 13 10 14 7 9 7 14 10 13 12 6 1
⑩ ⑨ ⑭ 3 ⑪ 7 ⑬ 2 ⑧ 4 ⑫ 1 6 5
④ ⑧ 2 ⑭ 3 5 0 ⑩ ⑫ 1 ⑬ 7 11
③ ⑪ ⑬ 6 1 7 ⑧ ⑭ ⑨ 2 4 12
⑥ ⑦ 5 4 3 0 ⑫ ⑪ ⑩ 8 9
② ⑭ 3 6 ⑪ 4 ⑨ ⑫ 1 13
⑨ ⑫ 5 ⑭ 0 1 ⑩ 3 6
⑦ ⑭ ⑬ 3 ⑫ 2 1 8
⑨ ⑦ 2 0 ⑬ 8 6
② ⑪ ⑭ 1 4 13
⑥ ⑧ 0 7 9
③ ⑨ 6 12
④ 0 11
⑩ 5
0
Modular difference Triangle of 1515 Modular sonar sequence in Class 5
Figure 4.5 : Array form and Modular difference Triangle of sequence in Class 5
42
Figures 4.1~4.5 show that array form and modular difference triangle of
these representative sequences.
Following are some observation.
Observation 1 : Class 1 and Class 2 come from (known) “Shift Sequence”.
The representative sequences e = (1,3,1,7,11,3,14,15,8,8,13,7,4,14,2) of
1800 equivalent sequence in Class 1 are generated by “Shift Sequence”
method for )12()12( 44 Modular sonar sequence. All 1800 equivalent
sequence in Class 1 are the sequences of transformations of representative
sequence e =(1,3,1,7,11,3,14,15,8,8,13,7,4,14,2) by asi iufig )()(
)(mod m .
From “Shift Sequence” method, that is
}1,,2,1{}1,,2,1{: rr ppf
be defined by
)(Tr)( ; ))((log)( 2)( ir
r
ipiifipi rr
βif
43
where )(Tr2 xr
ris the trace function from )( 2rpGF to )( rpGF ,
p=2, a primitive element of )( 2rpGF , and a primitive element in
)( rpGF , we can calculate )12()12( 44 Modular sonar sequence )(if .
In this case, a primitive element of )2( 8GF , and a primitive
element in )2( 4GF . Each shift sequences )(if were constructed with a
primitive element of )2( 8GF whose irreducible polynomials over
)2(GF are
1)( 2348 xxxxxf , 1)( 358 xxxxxf ,
1)( 2368 xxxxxf , 1)( 2568 xxxxxf ,
1)( 4568 xxxxxf , 1)( 278 xxxxxf ,
1)( 3578 xxxxxf , and 1)( 678 xxxxxf .
For example, if 12348 xxxxa , then we obtain 1515
Modular sonar sequence )(if = e = (2,4,2,8,12,4,15,1,9,9,14,8,5,15,3), that is
equivalent with representative sequence of Class 1, )(ig = e =
(1,3,1,7,11,3,14,15,8,8,13,7,4,14,2) by the equivalence of )(&)( igif given
44
by transformation equation )15(mod 14)()( ifig )14,0,0( isu .
Similarly, the representative sequences e = (1,1,4,1,9,7,11,1,8,2,12,13,4,6,2)
of 1800 equivalent sequence in Class 2 are also generated by “Shift
Sequence” method for )12()12( 44 Modular sonar sequence. In this
case, “Shift Sequences” were constructed with a primitive element of
)2( 8GF whose irreducible polynomials over )2(GF are
1)( 2358 xxxxxf , 1)( 568 xxxxxf ,
1)( 3568 xxxxxf , 1)( 245678 xxxxxxxf ,
1)( 23468 xxxxxxxf , 1)( 23678 xxxxxxxf ,
1)( 25678 xxxxxxxf , and 1)( 2378 xxxxxf .
For example, if 12358 xxxxa , then we obtain 1515
Modular sonar sequence )(if = e = (1,2,6,4,13,12,2,8,1,11,7,9,1,4,1), that is
equivalent with representative sequence of Class 2, )(ig = e =
(1,1,4,1,9,7,11,1,8,2,12,13,4,6,2) by the equivalence of )(&)( igif given by
transformation equation )15(mod 14)()( iifig )0,14,0( isu .
45
Observation 2 : Class 3 and Class 4 are the mirror image of Class 1 and
Class 2, respectively.
The representative sequences e = (1,1,2,14,2,13,4,9,13,12,4,2,11,6,8) of
1800 equivalent sequence in Class 3 are generated by mirror image of Class
1, that is
140 ),14()( iifig . (4.1)
By equation (4.1), we have e = (2,14,4,7,13,8,8,15,14,3,11,7,1,3,1) from
sequence e = (1,3,1,7,11,3,14,15,8,8,13,7,4,14,2) in Class 1. The sequence
e = (2,14,4,7,13,8,8,15,14,3,11,7,1,3,1) is a equivalent sequence of
representative sequence e = (1,1,2,14,2,13,4,9,13,12,4,2,11,6,8) in Class 3 with
transformation equation )15(mod 912)(8)( iifig )9,12,8( isu .
Each 1800 sequence in Class 3 have one to one corresponding pair (have
mirror image pair) with each 1800 sequence in Class 1 satisfying condition
of equation (4.1).
Similarly, the representative sequences e = (1,1,4,9,4,11,10,8,5,9,10,1,9,5,7)
46
of 1800 equivalent sequence in Class 4 are generated by mirror image of
Class 2.
By equation (4.1), we have e = (2,6,4,13,12,2,8,1,11,7,9,1,4,1,1) from
sequence e = (1,1,4,1,9,7,11,1,8,2,12,13,4,6,2) in Class 2. The sequence
e = (2,6,4,13,12,2,8,1,11,7,9,1,4,1,1) is a equivalent sequence of
representative sequence e = (1,1,4,9,4,11,10,8,5,9,10,1,9,5,7) in Class 4 with
transformation equation )15(mod 44)(14)( iifig )4,4,14( isu .
Each 1800 sequence in Class 4 have also one to one corresponding pair
(have mirror image pair) with each 1800 sequence in Class 2 satisfying
condition of equation (4.1).
Observation 3 : Class 5 is “new”.
The rest 1800 equivalent sequences are sequences of transformation by
)()( asiiufig )15(mod with the representative sequence e =
(1,6,12,13,10,14,7,9,7,14,10,13,12,6,1) in Class 5.
47
Why is this one equivalent class having 1800 equivalent sequences new?
1. For all of known (to us at the present time) construction methods for
nm Modular sonar sequences, )12()12( 44 Modular sonar
sequence are constructed by only one known method, which in called
“Shift Sequence”. As show in Table 3.1, the known construction
methods of Modular sonar sequence having 1 npm for their
modular are “Lempel”, “Golomb”, and “Shift Sequence”. Consider
length of these 3 methods, the “Shift Sequence” method can
mpn n 1 by eliminating the last column, but “Lempel” and
“Golomb” is not. We can construct 1515 Modular sonar sequence by
“Shift Sequence” method, but can not construct by “Lempel” or “Golomb
method. From “Lempel” or “Golomb” method, we only obtain
1415 Modular sonar sequence
2. The sequences of Class 1 and Class 2 cover all possible 1515 Modular
sonar sequences constructed by “Shift Sequence”. Because all possible
number of primitive element of )2( 8GF whose irreducible
polynomials over )2(GF for )12()12( 44 Modular sonar
sequences constructed by “Shift Sequence” method is 16, and 8 primitive
element are used for sequences in Class 1, and the other 8
48
primitive element are used for sequences in Case 2.
3. As we know, Case 3 is obtained from Case 1 by (4.1) and Case 4 is
obtained from Case 2 by (4.1). Is Case 5 also obtained from Case 1 or
Case 2? It is not! Because all of the each 1800 sequence in Case 1
(Case 2 ) has one to one corresponding pair with sequence in Case 3
(Case 4) satisfying condition (4.1). All of the each 1800 sequence in
Case 5 has one to one corresponding with sequence in Class 5. In detail,
each )180015
7(840 sequence in Class 5 has one to one corresponding
with one of the other )180015
7(840 sequences in Class 5, and each
)180015
1(120 sequence has one to one corresponding with itself.
Therefore, Case 5 is a “new” class.
Observation 4: Property of Class 5
In this special case, for each u of Modular sonar sequence transformation
which is ) (mod )()( masiiufig , there is only one sequence having
reciprocal property that is
viivfif 0 ),()( for sequence )(if of period v. (4.2)
49
For each u, there are 15 sequences, and 1 sequence having reciprocal
property, and the other 14 sequences have one to one corresponding with
each other. For example, the sequence obtained by transformation where
u=1,s=0,a=0 is corresponding with the sequence obtained by transformation
where u=1,s=12,a=0. All possible number of u is 8 and a is 15, so there are
120 sequences having reciprocal property.
From sequences having reciprocal property in Class 5, for example
e =(1,6,12,13,10,14,7,9,7,14,10,13,12,6,1), we also find some fact that is
first 8 elements of sequence e satisfies the following condition :
.81 allfor 8}80|0][),({ sssjeejsd sjj (4.3)
where
714 where)(15
0 7 where||
8 0 where
15 8 where)(15
][
sjjsjj
sjjsjj
sjjsjj
sjjsjj
sjj
eeee
eeee
eeee
eeee
ee
and 8][0 sjj ee .
This condition (4.3) means that half sequence (period 8) of Modular sonar
sequence of period 15 has the property similar to distinct difference property
(See Figure 4.4).
50
1 6 12 13 10 14 7 9 7 14 10 13 12 6 1
⑤ ⑥ ① 3 ④ 7 ② 2 ⑦ 4 ③ 1 6 5
④ ⑦ 2 ① 3 5 ⑤ ③ 1 ② 7 ④
③ ④ ② 6 1 ① ⑥ 2 4 ③
⑥ ⑦ 5 4 ④ ⑤ ⑦ ⑥
② ① 3 ③ 1 ②
⑥ ③ 3 6
⑦ ⑦
Figure 4.4 : Difference Triangle for condition (4.2)
General condition (4.3) for sequence of odd period v is that,
.2
11 allfor
2
1}
2
10|0][),({
vss
vs
vjeejsd sjj
(4.4)
where
2
1)1( where)(
02
1 where||
2
10 where
2
1 where)(
][
veeveev
eev
ee
veeee
veev
eev
ee
sjjsjj
sjjsjj
sjjsjj
sjjsjj
sjj
51
and 2
1][0
vee sjj
Theorem 1 : For any sequence ),,,( 110 veee e of odd period v that
has self-reciprocal property of condition (4.2), if the sequence e is a
vv Modular sonar sequence, then a sequence of period 2
1v that has first
2
1v elements of e satisfies condition (4.4).
Proof : If the sequence of period 2
1v that has first
2
1v elements of e
does not satisfy condition (4.4). There exist ),,()',(' jsdjsd
sv
jj
2
1'0 for some
2
11
vs , or .0),( jsd
If sjj eea is a original value of ),( jsd and sjj eea ''' is a
original value of )',(' jsd in condition (4.4),
)},(),,(),,(),,({ jsdvjsdjsdjsdva ,
)}',(),',(),',(),',({' jsdvjsdjsdjsdva .
Because a modular v is ),( jsdv or ),( jsd and 'a modular v is
)',(' jsdv or )',(' jsd , there exist 4 cases.
1. If ),()(mod jsdvva and )',()(mod' jsdvva ,
then )(mod')(mod vava .
52
2. If ),()(mod jsdva and )',()(mod' jsdva ,
then )(mod')(mod vava .
3. If ),()(mod jsdvva and )',()(mod' jsdva ,
then )(mod')(mod vavva .
4. If ),()(mod jsdva and )',()(mod' jsdvva ,
then )(mod')(mod vavva .
In case 1 and 2,
)(mod)(mod '' veevee sjjsjj for sv
jj
2
1'0 (4.5)
In case 3 and 4, ')1()'()1()(mod' jvsjv eevav by self-reciprocal
property of condition (4.2), therefore
)(mod)(mod )'1()'1( veevee ssjvsjvsjj (4.6)
for '12
1
2
10 vsjvs
vs
vj
.
In the case of ,0),( jsd by self-reciprocal property of condition (4.2),
)(mod)(mod )1()1( veevee ssjvsjvsjj =0 (4.7)
for 12
1
2
10 vsjvs
vs
vj
.
53
From (4.5), (4.6) and (4.7), the sequence ),,,( 110 veee e of period v
does not satisfy the distinct modular differences property. Therefore, for
any sequence ),,,( 110 veee e of odd period v having self-reciprocal
property, if the sequence of period 2
1v that has first
2
1v elements of
e does not satisfy condition (4.4), the sequence e is not a vv Modular
sonar sequence. □
4.2 The case of v=5ⅹ7=35
The second short case of )2( ppv is a v=5ⅹ7=35. When we
consider present computer hardware speed, exhaustive search of v=5ⅹ7=35
is impossible. Because hardware complexity is exponentially increased by
adding more length of the sequence. So we approached without exhaustive
search by using Theorem 4.1 and some algebraic idea.
Approach 1 : Using Theorem 4.1
If we use a Theorem 4.1, we can find 35ⅹ35 Modular sonar sequence by
54
considering only first 18 elements of sequence.
If there exist vv Modular sonar sequence of period 3575 v , the
sequence has the same singularity property of 1515 Modular sonar
sequence (property from “New” self-reciprocal sequences of Class 5 in
above section except “known” sequences by “Shift Sequence” method).
Therefore the 3535 Modular sonar sequence will have self-reciprocal
property, too. For all sequence ),,,( 3410 eee e of period 35 having
self-reciprocal property, if there does not exist the sequence that has first 18
elements of e and satisfies condition (4.4), there dose not exist the
vv Modular sonar sequence of period 3575 v
We have exhaustive search for all sequence e of period 35 that has self-
reciprocal property by following algorithm.
1. Choose value 1 for first element of sequence e . We does not have to
consider other 2 to 35 value for first element of sequence, because of
addition transformation by a modulo m.
2. Choose value from 1 to 35 for second element of sequence e . If
second element satisfies condition (4.4) for 1,0 sj with first
element of sequence e , go Stage 3. Else if, return Stage 1.
3. Choose value from 1 to 35 for third element of sequence e . If third
element satisfies condition (4.4) for 31,20 sj with first
55
element and second element of sequence e , go Stage 4. Else if,
return Stage 1.
4. Choose value from 1 to 35 for fourth element of sequence e . If
fourth element satisfies condition (4.4) for 41,30 sj with
first element, second element, and third element of sequence e , go
Stage 5. Else if, return Stage 1.
5. For 5th
~18th
elements of sequence e , computing satisfaction of
condition (4.4) by method of Stage 4.
6. If there are all of 1th
~18th
elements of sequence e are satisfy Stage
1-5, we can obtain 18th
~35th
elements of sequence e by
3518 ),35()( iifif .
If there exists the self-reciprocal sequence e satisfying all stage of
above algorithm, the sequence is or is not 3535 Modular sonar sequence.
Some of self-reciprocal sequence e is a 3535 Modular sonar sequence.
But if there does not exist the self-reciprocal sequence e satisfying all
stage of above algorithm, there does not exist 3535 Modular sonar
sequence.
From several repeated computing simulation (running time of
computing is about 7 days per one simulation), we dose not obtain the self-
reciprocal sequence e satisfying all stage of above algorithm.
56
Approach 2 : Algebraic construction
We may be able to come up with an algebraic construction using the
following observation. There exists n such that )2( ppv is a divisor of
12 n . Consider the finite field nF2
and an element of order v in it.
Successive powers of will produce a sequence of length v over nF2
.
The final step is to see if there exists a transformation that sends this
sequence into those over vZ .
In the case of 3575)2( ppv , 35 is a divisor of 1212 . For
exam, if is a primitive element of )2( 12GF whose irreducible
polynomials over )2(GF is 1)( 4612 xxxxxf , then we can
consider an element 117 of order 35 in )2( 12GF . As see Figure 4.5,
successive powers of will produce a sequence of length 35 over 122F . If
we consider following transformation that is
),,,( 3521 xxxi x ,
),,,( 3521 yyyi
y ,
jx of
5
1
),(2n
n
i jnix (mod 35) and
57
Figure 4.5: Example of Approach 2 for 35 by 35 Modular sonar sequence
jy of
12
1
),(2n
n
ijniy (mod 35)
where ),( jni is a value of ))12(mod(12 ni th-dimensional vector
of 12-dimensional vectors representation for j ,
then we can obtain a sequence of length 35 over 35Z . Finally, we observer
the sequence is 35 by 35 Modular sonar sequence or not.
In spite of several initial attempts, all for these approach failed.
Approach 3: Using Transformation of Modular sonar sequence
A main idea of Approach 3 is same as Approach 2 without using
58
Figure 4.6: Example of Approach 3 for 35 by 35 Modular sonar sequence
transformation method of Modular sonar sequence.
Steps for Approach 3 are that
Step 1 : There exists n such that )2( ppv is a divisor of 12 n .
Step 2 : Consider the finite field nF2
and an element of order v in it.
Step 3 : Successive powers of will produce a sequence of length v
over nF2
.
Step 4 : Find vn 2 Modular sonar sequence by transformation that
sends this sequence into those over nZ2
.
Step 5 : Otain vv Modular sonar sequence by transformation of above
vn 2 Modular sonar sequence.
59
Step 1~3 is same as Approach 2, and Figure 4.6 is example of these steps
in the case of v=35.
From Step 1~4, we can obtain some 35212 Modular sonar sequences.
A example of 35212 Modular sonar sequence is that
e =(3417,1107,2707,1682,2516,413,1607,3489,1591,599,3075,2675,2390,
3517,468,3268,532,1842,165,2947,3486,3124,1271,2954,899,199,2151,
3684,3352,2647,346,3616,965,2863,2048).
But fail to obtain any 3535 Modular sonar sequence by all
transformation method of above Modular sonar sequence.
60
Chapter 5
Concluding Remarks
We review two construction methods of )32,,( 2 vvv signal set by
two generating methods of shift sequence e . The one of the case is
12 nv and the other is pv . From vv Modular sonar sequence, we
can obtain shift sequence e . Because they have same property with each
other. By construction method of Modular sonar sequence that is “Shift
Sequence” and “Extended Exponential Welch”, we construct two types of
shift sequence e in above cases. These two case cover all types of two-
level autocorrelation sequences except for a class of two-level
autocorrelation sequences of period v =p(p+2).
By an exhaustive search of first small case of 1553)2( ppv , we
find 9000 1515 Modular sonar sequences. If we consider transformation
by asi iufig )()( )(mod m , 9000 1515 Modular sonar sequences
are grouped with 5 equivalent classes, and in each equivalent class has
61
)15158(1800 equivalent sequences.
Two out of five equivalent classes (Class 1 and 2) turned out to be the
ones that can be constructed by the known method, which in called “Shift
Sequence”. Each 1800 sequence in other two equivalent Classes 3 and 4 has
one to one corresponding pair (have mirror image pair) with each 1800
sequence in Class 1 and 2. One remain Class 5 is new. Because 1515
Modular sonar sequence is only constructed by “Shift Sequence” method,
Class 1 and 2 cover all sequences constructed by “Shift Sequence” method,
and Class 5 does not come from Class 1~4. Each sequence in Class 5 has
one to one corresponding with each other or itself.
Class 5 that has unknown “new” constructed method satisfies some
singular condition. For any sequence ),,,( 110 veee e of odd period v
that has self-reciprocal property, if the sequence e is a vv Modular sonar
sequence, then a sequence of period 2
1v that has first
2
1v elements of
e satisfies these singular condition in Class 5. Using this theorem, we can
62
find 35ⅹ35 modular sonar sequence by considering only first 18 elements
of sequence. From several repeated computing simulation, we dose not find
35ⅹ35 modular sonar sequence.
Furthermore we approached this problem with some algebraic
constructions and transformations of the obtained Modular sonar sequences,
but unfortunately, all failed.
General case of )2()2( pppp Modular sonar sequence seems to
be nonexistent. Because if general case of )2()2( pppp Modular
sonar sequence is exist, the 35ⅹ35 Modular sonar sequence has to been
constructed by method from same property of unknown “new” case of
15ⅹ15 Modular sonar sequence.
63
Appendix
A. 120 sequences in Class 1 have “1” as the first element of sequence
u= 1, s= 0, a= 0 : 1 3 1 7 11 3 14 15 8 8 13 7 4 14 2
u= 1, s= 1, a= 0 : 1 4 3 10 15 8 5 7 1 2 8 3 1 12 1
u= 1, s= 2, a= 0 : 1 5 5 13 4 13 11 14 9 11 3 14 13 10 15
u= 1, s= 3, a= 0 : 1 6 7 1 8 3 2 6 2 5 13 10 10 8 14
u= 1, s= 4, a= 0 : 1 7 9 4 12 8 8 13 10 14 8 6 7 6 13
u= 1, s= 5, a= 0 : 1 8 11 7 1 13 14 5 3 8 3 2 4 4 12
u= 1, s= 6, a= 0 : 1 9 13 10 5 3 5 12 11 2 13 13 1 2 11
u= 1, s= 7, a= 0 : 1 10 15 13 9 8 11 4 4 11 8 9 13 15 10
u= 1, s= 8, a= 0 : 1 11 2 1 13 13 2 11 12 5 3 5 10 13 9
u= 1, s= 9, a= 0 : 1 12 4 4 2 3 8 3 5 14 13 1 7 11 8
u= 1, s=10, a= 0 : 1 13 6 7 6 8 14 10 13 8 8 12 4 9 7
u= 1, s=11, a= 0 : 1 14 8 10 10 13 5 2 6 2 3 8 1 7 6
u= 1, s=12, a= 0 : 1 15 10 13 14 3 11 9 14 11 13 4 13 5 5
u= 1, s=13, a= 0 : 1 1 12 1 3 8 2 1 7 5 8 15 10 3 4
u= 1, s=14, a= 0 : 1 2 14 4 7 13 8 8 15 14 3 11 7 1 3
u= 2, s= 0, a=14 : 1 5 1 13 6 5 12 14 15 15 10 13 7 12 3
u= 2, s= 1, a=14 : 1 6 3 1 10 10 3 6 8 9 5 9 4 10 2
u= 2, s= 2, a=14 : 1 7 5 4 14 15 9 13 1 3 15 5 1 8 1
u= 2, s= 3, a=14 : 1 8 7 7 3 5 15 5 9 12 10 1 13 6 15
u= 2, s= 4, a=14 : 1 9 9 10 7 10 6 12 2 6 5 12 10 4 14
u= 2, s= 5, a=14 : 1 10 11 13 11 15 12 4 10 15 15 8 7 2 13
u= 2, s= 6, a=14 : 1 11 13 1 15 5 3 11 3 9 10 4 4 15 12
u= 2, s= 7, a=14 : 1 12 15 4 4 10 9 3 11 3 5 15 1 13 11
u= 2, s= 8, a=14 : 1 13 2 7 8 15 15 10 4 12 15 11 13 11 10
u= 2, s= 9, a=14 : 1 14 4 10 12 5 6 2 12 6 10 7 10 9 9
u= 2, s=10, a=14 : 1 15 6 13 1 10 12 9 5 15 5 3 7 7 8
u= 2, s=11, a=14 : 1 1 8 1 5 15 3 1 13 9 15 14 4 5 7
u= 2, s=12, a=14 : 1 2 10 4 9 5 9 8 6 3 10 10 1 3 6
u= 2, s=13, a=14 : 1 3 12 7 13 10 15 15 14 12 5 6 13 1 5
u= 2, s=14, a=14 : 1 4 14 10 2 15 6 7 7 6 15 2 10 14 4
u= 4, s= 0, a=12 : 1 9 1 10 11 9 8 12 14 14 4 10 13 8 5
u= 4, s= 1, a=12 : 1 10 3 13 15 14 14 4 7 8 14 6 10 6 4
64
u= 4, s= 2, a=12 : 1 11 5 1 4 4 5 11 15 2 9 2 7 4 3
u= 4, s= 3, a=12 : 1 12 7 4 8 9 11 3 8 11 4 13 4 2 2
u= 4, s= 4, a=12 : 1 13 9 7 12 14 2 10 1 5 14 9 1 15 1
u= 4, s= 5, a=12 : 1 14 11 10 1 4 8 2 9 14 9 5 13 13 15
u= 4, s= 6, a=12 : 1 15 13 13 5 9 14 9 2 8 4 1 10 11 14
u= 4, s= 7, a=12 : 1 1 15 1 9 14 5 1 10 2 14 12 7 9 13
u= 4, s= 8, a=12 : 1 2 2 4 13 4 11 8 3 11 9 8 4 7 12
u= 4, s= 9, a=12 : 1 3 4 7 2 9 2 15 11 5 4 4 1 5 11
u= 4, s=10, a=12 : 1 4 6 10 6 14 8 7 4 14 14 15 13 3 10
u= 4, s=11, a=12 : 1 5 8 13 10 4 14 14 12 8 9 11 10 1 9
u= 4, s=12, a=12 : 1 6 10 1 14 9 5 6 5 2 4 7 7 14 8
u= 4, s=13, a=12 : 1 7 12 4 3 14 11 13 13 11 14 3 4 12 7
u= 4, s=14, a=12 : 1 8 14 7 7 4 2 5 6 5 9 14 1 10 6
u= 7, s= 0, a= 9 : 1 15 1 13 11 15 2 9 5 5 10 13 7 2 8
u= 7, s= 1, a= 9 : 1 1 3 1 15 5 8 1 13 14 5 9 4 15 7
u= 7, s= 2, a= 9 : 1 2 5 4 4 10 14 8 6 8 15 5 1 13 6
u= 7, s= 3, a= 9 : 1 3 7 7 8 15 5 15 14 2 10 1 13 11 5
u= 7, s= 4, a= 9 : 1 4 9 10 12 5 11 7 7 11 5 12 10 9 4
u= 7, s= 5, a= 9 : 1 5 11 13 1 10 2 14 15 5 15 8 7 7 3
u= 7, s= 6, a= 9 : 1 6 13 1 5 15 8 6 8 14 10 4 4 5 2
u= 7, s= 7, a= 9 : 1 7 15 4 9 5 14 13 1 8 5 15 1 3 1
u= 7, s= 8, a= 9 : 1 8 2 7 13 10 5 5 9 2 15 11 13 1 15
u= 7, s= 9, a= 9 : 1 9 4 10 2 15 11 12 2 11 10 7 10 14 14
u= 7, s=10, a= 9 : 1 10 6 13 6 5 2 4 10 5 5 3 7 12 13
u= 7, s=11, a= 9 : 1 11 8 1 10 10 8 11 3 14 15 14 4 10 12
u= 7, s=12, a= 9 : 1 12 10 4 14 15 14 3 11 8 10 10 1 8 11
u= 7, s=13, a= 9 : 1 13 12 7 3 5 5 10 4 2 5 6 13 6 10
u= 7, s=14, a= 9 : 1 14 14 10 7 10 11 2 12 11 15 2 10 4 9
u= 8, s= 0, a= 8 : 1 2 1 4 6 2 15 8 12 12 7 4 10 15 9
u= 8, s= 1, a= 8 : 1 3 3 7 10 7 6 15 5 6 2 15 7 13 8
u= 8, s= 2, a= 8 : 1 4 5 10 14 12 12 7 13 15 12 11 4 11 7
u= 8, s= 3, a= 8 : 1 5 7 13 3 2 3 14 6 9 7 7 1 9 6
u= 8, s= 4, a= 8 : 1 6 9 1 7 7 9 6 14 3 2 3 13 7 5
u= 8, s= 5, a= 8 : 1 7 11 4 11 12 15 13 7 12 12 14 10 5 4
u= 8, s= 6, a= 8 : 1 8 13 7 15 2 6 5 15 6 7 10 7 3 3
u= 8, s= 7, a= 8 : 1 9 15 10 4 7 12 12 8 15 2 6 4 1 2
u= 8, s= 8, a= 8 : 1 10 2 13 8 12 3 4 1 9 12 2 1 14 1
u= 8, s= 9, a= 8 : 1 11 4 1 12 2 9 11 9 3 7 13 13 12 15
u= 8, s=10, a= 8 : 1 12 6 4 1 7 15 3 2 12 2 9 10 10 14
u= 8, s=11, a= 8 : 1 13 8 7 5 12 6 10 10 6 12 5 7 8 13
65
u= 8, s=12, a= 8 : 1 14 10 10 9 2 12 2 3 15 7 1 4 6 12
u= 8, s=13, a= 8 : 1 15 12 13 13 7 3 9 11 9 2 12 1 4 11
u= 8, s=14, a= 8 : 1 1 14 1 2 12 9 1 4 3 12 8 13 2 10
u=11, s= 0, a= 5 : 1 8 1 7 6 8 9 5 3 3 13 7 4 9 12
u=11, s= 1, a= 5 : 1 9 3 10 10 13 15 12 11 12 8 3 1 7 11
u=11, s= 2, a= 5 : 1 10 5 13 14 3 6 4 4 6 3 14 13 5 10
u=11, s= 3, a= 5 : 1 11 7 1 3 8 12 11 12 15 13 10 10 3 9
u=11, s= 4, a= 5 : 1 12 9 4 7 13 3 3 5 9 8 6 7 1 8
u=11, s= 5, a= 5 : 1 13 11 7 11 3 9 10 13 3 3 2 4 14 7
u=11, s= 6, a= 5 : 1 14 13 10 15 8 15 2 6 12 13 13 1 12 6
u=11, s= 7, a= 5 : 1 15 15 13 4 13 6 9 14 6 8 9 13 10 5
u=11, s= 8, a= 5 : 1 1 2 1 8 3 12 1 7 15 3 5 10 8 4
u=11, s= 9, a= 5 : 1 2 4 4 12 8 3 8 15 9 13 1 7 6 3
u=11, s=10, a= 5 : 1 3 6 7 1 13 9 15 8 3 8 12 4 4 2
u=11, s=11, a= 5 : 1 4 8 10 5 3 15 7 1 12 3 8 1 2 1
u=11, s=12, a= 5 : 1 5 10 13 9 8 6 14 9 6 13 4 13 15 15
u=11, s=13, a= 5 : 1 6 12 1 13 13 12 6 2 15 8 15 10 13 14
u=11, s=14, a= 5 : 1 7 14 4 2 3 3 13 10 9 3 11 7 11 13
u=13, s= 0, a= 3 : 1 12 1 4 11 12 5 3 2 2 7 4 10 5 14
u=13, s= 1, a= 3 : 1 13 3 7 15 2 11 10 10 11 2 15 7 3 13
u=13, s= 2, a= 3 : 1 14 5 10 4 7 2 2 3 5 12 11 4 1 12
u=13, s= 3, a= 3 : 1 15 7 13 8 12 8 9 11 14 7 7 1 14 11
u=13, s= 4, a= 3 : 1 1 9 1 12 2 14 1 4 8 2 3 13 12 10
u=13, s= 5, a= 3 : 1 2 11 4 1 7 5 8 12 2 12 14 10 10 9
u=13, s= 6, a= 3 : 1 3 13 7 5 12 11 15 5 11 7 10 7 8 8
u=13, s= 7, a= 3 : 1 4 15 10 9 2 2 7 13 5 2 6 4 6 7
u=13, s= 8, a= 3 : 1 5 2 13 13 7 8 14 6 14 12 2 1 4 6
u=13, s= 9, a= 3 : 1 6 4 1 2 12 14 6 14 8 7 13 13 2 5
u=13, s=10, a= 3 : 1 7 6 4 6 2 5 13 7 2 2 9 10 15 4
u=13, s=11, a= 3 : 1 8 8 7 10 7 11 5 15 11 12 5 7 13 3
u=13, s=12, a= 3 : 1 9 10 10 14 12 2 12 8 5 7 1 4 11 2
u=13, s=13, a= 3 : 1 10 12 13 3 2 8 4 1 14 2 12 1 9 1
u=13, s=14, a= 3 : 1 11 14 1 7 7 14 11 9 8 12 8 13 7 15
u=14, s= 0, a= 2 : 1 14 1 10 6 14 3 2 9 9 4 10 13 3 15
u=14, s= 1, a= 2 : 1 15 3 13 10 4 9 9 2 3 14 6 10 1 14
u=14, s= 2, a= 2 : 1 1 5 1 14 9 15 1 10 12 9 2 7 14 13
u=14, s= 3, a= 2 : 1 2 7 4 3 14 6 8 3 6 4 13 4 12 12
u=14, s= 4, a= 2 : 1 3 9 7 7 4 12 15 11 15 14 9 1 10 11
u=14, s= 5, a= 2 : 1 4 11 10 11 9 3 7 4 9 9 5 13 8 10
u=14, s= 6, a= 2 : 1 5 13 13 15 14 9 14 12 3 4 1 10 6 9
66
u=14, s= 7, a= 2 : 1 6 15 1 4 4 15 6 5 12 14 12 7 4 8
u=14, s= 8, a= 2 : 1 7 2 4 8 9 6 13 13 6 9 8 4 2 7
u=14, s= 9, a= 2 : 1 8 4 7 12 14 12 5 6 15 4 4 1 15 6
u=14, s=10, a= 2 : 1 9 6 10 1 4 3 12 14 9 14 15 13 13 5
u=14, s=11, a= 2 : 1 10 8 13 5 9 9 4 7 3 9 11 10 11 4
u=14, s=12, a= 2 : 1 11 10 1 9 14 15 11 15 12 4 7 7 9 3
u=14, s=13, a= 2 : 1 12 12 4 13 4 6 3 8 6 14 3 4 7 2
u=14, s=14, a= 2 : 1 13 14 7 2 9 12 10 1 15 9 14 1 5 1
B. 120 sequences in Class 2 have “1” as the first element of sequence
u= 1, s= 0, a= 0 : 1 1 4 1 9 7 11 1 8 2 12 13 4 6 2
u= 1, s= 1, a= 0 : 1 2 6 4 13 12 2 8 1 11 7 9 1 4 1
u= 1, s= 2, a= 0 : 1 3 8 7 2 2 8 15 9 5 2 5 13 2 15
u= 1, s= 3, a= 0 : 1 4 10 10 6 7 14 7 2 14 12 1 10 15 14
u= 1, s= 4, a= 0 : 1 5 12 13 10 12 5 14 10 8 7 12 7 13 13
u= 1, s= 5, a= 0 : 1 6 14 1 14 2 11 6 3 2 2 8 4 11 12
u= 1, s= 6, a= 0 : 1 7 1 4 3 7 2 13 11 11 12 4 1 9 11
u= 1, s= 7, a= 0 : 1 8 3 7 7 12 8 5 4 5 7 15 13 7 10
u= 1, s= 8, a= 0 : 1 9 5 10 11 2 14 12 12 14 2 11 10 5 9
u= 1, s= 9, a= 0 : 1 10 7 13 15 7 5 4 5 8 12 7 7 3 8
u= 1, s=10, a= 0 : 1 11 9 1 4 12 11 11 13 2 7 3 4 1 7
u= 1, s=11, a= 0 : 1 12 11 4 8 2 2 3 6 11 2 14 1 14 6
u= 1, s=12, a= 0 : 1 13 13 7 12 7 8 10 14 5 12 10 13 12 5
u= 1, s=13, a= 0 : 1 14 15 10 1 12 14 2 7 14 7 6 10 10 4
u= 1, s=14, a= 0 : 1 15 2 13 5 2 5 9 15 8 2 2 7 8 3
u= 2, s= 0, a=14 : 1 1 7 1 2 13 6 1 15 3 8 10 7 11 3
u= 2, s= 1, a=14 : 1 2 9 4 6 3 12 8 8 12 3 6 4 9 2
u= 2, s= 2, a=14 : 1 3 11 7 10 8 3 15 1 6 13 2 1 7 1
u= 2, s= 3, a=14 : 1 4 13 10 14 13 9 7 9 15 8 13 13 5 15
u= 2, s= 4, a=14 : 1 5 15 13 3 3 15 14 2 9 3 9 10 3 14
u= 2, s= 5, a=14 : 1 6 2 1 7 8 6 6 10 3 13 5 7 1 13
u= 2, s= 6, a=14 : 1 7 4 4 11 13 12 13 3 12 8 1 4 14 12
u= 2, s= 7, a=14 : 1 8 6 7 15 3 3 5 11 6 3 12 1 12 11
u= 2, s= 8, a=14 : 1 9 8 10 4 8 9 12 4 15 13 8 13 10 10
u= 2, s= 9, a=14 : 1 10 10 13 8 13 15 4 12 9 8 4 10 8 9
u= 2, s=10, a=14 : 1 11 12 1 12 3 6 11 5 3 3 15 7 6 8
u= 2, s=11, a=14 : 1 12 14 4 1 8 12 3 13 12 13 11 4 4 7
u= 2, s=12, a=14 : 1 13 1 7 5 13 3 10 6 6 8 7 1 2 6
67
u= 2, s=13, a=14 : 1 14 3 10 9 3 9 2 14 15 3 3 13 15 5
u= 2, s=14, a=14 : 1 15 5 13 13 8 15 9 7 9 13 14 10 13 4
u= 4, s= 0, a=12 : 1 1 13 1 3 10 11 1 14 5 15 4 13 6 5
u= 4, s= 1, a=12 : 1 2 15 4 7 15 2 8 7 14 10 15 10 4 4
u= 4, s= 2, a=12 : 1 3 2 7 11 5 8 15 15 8 5 11 7 2 3
u= 4, s= 3, a=12 : 1 4 4 10 15 10 14 7 8 2 15 7 4 15 2
u= 4, s= 4, a=12 : 1 5 6 13 4 15 5 14 1 11 10 3 1 13 1
u= 4, s= 5, a=12 : 1 6 8 1 8 5 11 6 9 5 5 14 13 11 15
u= 4, s= 6, a=12 : 1 7 10 4 12 10 2 13 2 14 15 10 10 9 14
u= 4, s= 7, a=12 : 1 8 12 7 1 15 8 5 10 8 10 6 7 7 13
u= 4, s= 8, a=12 : 1 9 14 10 5 5 14 12 3 2 5 2 4 5 12
u= 4, s= 9, a=12 : 1 10 1 13 9 10 5 4 11 11 15 13 1 3 11
u= 4, s=10, a=12 : 1 11 3 1 13 15 11 11 4 5 10 9 13 1 10
u= 4, s=11, a=12 : 1 12 5 4 2 5 2 3 12 14 5 5 10 14 9
u= 4, s=12, a=12 : 1 13 7 7 6 10 8 10 5 8 15 1 7 12 8
u= 4, s=13, a=12 : 1 14 9 10 10 15 14 2 13 2 10 12 4 10 7
u= 4, s=14, a=12 : 1 15 11 13 14 5 5 9 6 11 5 8 1 8 6
u= 7, s= 0, a= 9 : 1 1 7 1 12 13 11 1 5 8 3 10 7 6 8
u= 7, s= 1, a= 9 : 1 2 9 4 1 3 2 8 13 2 13 6 4 4 7
u= 7, s= 2, a= 9 : 1 3 11 7 5 8 8 15 6 11 8 2 1 2 6
u= 7, s= 3, a= 9 : 1 4 13 10 9 13 14 7 14 5 3 13 13 15 5
u= 7, s= 4, a= 9 : 1 5 15 13 13 3 5 14 7 14 13 9 10 13 4
u= 7, s= 5, a= 9 : 1 6 2 1 2 8 11 6 15 8 8 5 7 11 3
u= 7, s= 6, a= 9 : 1 7 4 4 6 13 2 13 8 2 3 1 4 9 2
u= 7, s= 7, a= 9 : 1 8 6 7 10 3 8 5 1 11 13 12 1 7 1
u= 7, s= 8, a= 9 : 1 9 8 10 14 8 14 12 9 5 8 8 13 5 15
u= 7, s= 9, a= 9 : 1 10 10 13 3 13 5 4 2 14 3 4 10 3 14
u= 7, s=10, a= 9 : 1 11 12 1 7 3 11 11 10 8 13 15 7 1 13
u= 7, s=11, a= 9 : 1 12 14 4 11 8 2 3 3 2 8 11 4 14 12
u= 7, s=12, a= 9 : 1 13 1 7 15 13 8 10 11 11 3 7 1 12 11
u= 7, s=13, a= 9 : 1 14 3 10 4 3 14 2 4 5 13 3 13 10 10
u= 7, s=14, a= 9 : 1 15 5 13 8 8 5 9 12 14 8 14 10 8 9
u= 8, s= 0, a= 8 : 1 1 10 1 5 4 6 1 12 9 14 7 10 11 9
u= 8, s= 1, a= 8 : 1 2 12 4 9 9 12 8 5 3 9 3 7 9 8
u= 8, s= 2, a= 8 : 1 3 14 7 13 14 3 15 13 12 4 14 4 7 7
u= 8, s= 3, a= 8 : 1 4 1 10 2 4 9 7 6 6 14 10 1 5 6
u= 8, s= 4, a= 8 : 1 5 3 13 6 9 15 14 14 15 9 6 13 3 5
u= 8, s= 5, a= 8 : 1 6 5 1 10 14 6 6 7 9 4 2 10 1 4
u= 8, s= 6, a= 8 : 1 7 7 4 14 4 12 13 15 3 14 13 7 14 3
u= 8, s= 7, a= 8 : 1 8 9 7 3 9 3 5 8 12 9 9 4 12 2
68
u= 8, s= 8, a= 8 : 1 9 11 10 7 14 9 12 1 6 4 5 1 10 1
u= 8, s= 9, a= 8 : 1 10 13 13 11 4 15 4 9 15 14 1 13 8 15
u= 8, s=10, a= 8 : 1 11 15 1 15 9 6 11 2 9 9 12 10 6 14
u= 8, s=11, a= 8 : 1 12 2 4 4 14 12 3 10 3 4 8 7 4 13
u= 8, s=12, a= 8 : 1 13 4 7 8 4 3 10 3 12 14 4 4 2 12
u= 8, s=13, a= 8 : 1 14 6 10 12 9 9 2 11 6 9 15 1 15 11
u= 8, s=14, a= 8 : 1 15 8 13 1 14 15 9 4 15 4 11 13 13 10
u=11, s= 0, a= 5 : 1 1 4 1 14 7 6 1 3 12 2 13 4 11 12
u=11, s= 1, a= 5 : 1 2 6 4 3 12 12 8 11 6 12 9 1 9 11
u=11, s= 2, a= 5 : 1 3 8 7 7 2 3 15 4 15 7 5 13 7 10
u=11, s= 3, a= 5 : 1 4 10 10 11 7 9 7 12 9 2 1 10 5 9
u=11, s= 4, a= 5 : 1 5 12 13 15 12 15 14 5 3 12 12 7 3 8
u=11, s= 5, a= 5 : 1 6 14 1 4 2 6 6 13 12 7 8 4 1 7
u=11, s= 6, a= 5 : 1 7 1 4 8 7 12 13 6 6 2 4 1 14 6
u=11, s= 7, a= 5 : 1 8 3 7 12 12 3 5 14 15 12 15 13 12 5
u=11, s= 8, a= 5 : 1 9 5 10 1 2 9 12 7 9 7 11 10 10 4
u=11, s= 9, a= 5 : 1 10 7 13 5 7 15 4 15 3 2 7 7 8 3
u=11, s=10, a= 5 : 1 11 9 1 9 12 6 11 8 12 12 3 4 6 2
u=11, s=11, a= 5 : 1 12 11 4 13 2 12 3 1 6 7 14 1 4 1
u=11, s=12, a= 5 : 1 13 13 7 2 7 3 10 9 15 2 10 13 2 15
u=11, s=13, a= 5 : 1 14 15 10 6 12 9 2 2 9 12 6 10 15 14
u=11, s=14, a= 5 : 1 15 2 13 10 2 15 9 10 3 7 2 7 13 13
u=13, s= 0, a= 3 : 1 1 10 1 15 4 11 1 2 14 9 7 10 6 14
u=13, s= 1, a= 3 : 1 2 12 4 4 9 2 8 10 8 4 3 7 4 13
u=13, s= 2, a= 3 : 1 3 14 7 8 14 8 15 3 2 14 14 4 2 12
u=13, s= 3, a= 3 : 1 4 1 10 12 4 14 7 11 11 9 10 1 15 11
u=13, s= 4, a= 3 : 1 5 3 13 1 9 5 14 4 5 4 6 13 13 10
u=13, s= 5, a= 3 : 1 6 5 1 5 14 11 6 12 14 14 2 10 11 9
u=13, s= 6, a= 3 : 1 7 7 4 9 4 2 13 5 8 9 13 7 9 8
u=13, s= 7, a= 3 : 1 8 9 7 13 9 8 5 13 2 4 9 4 7 7
u=13, s= 8, a= 3 : 1 9 11 10 2 14 14 12 6 11 14 5 1 5 6
u=13, s= 9, a= 3 : 1 10 13 13 6 4 5 4 14 5 9 1 13 3 5
u=13, s=10, a= 3 : 1 11 15 1 10 9 11 11 7 14 4 12 10 1 4
u=13, s=11, a= 3 : 1 12 2 4 14 14 2 3 15 8 14 8 7 14 3
u=13, s=12, a= 3 : 1 13 4 7 3 4 8 10 8 2 9 4 4 12 2
u=13, s=13, a= 3 : 1 14 6 10 7 9 14 2 1 11 4 15 1 10 1
u=13, s=14, a= 3 : 1 15 8 13 11 14 5 9 9 5 14 11 13 8 15
u=14, s= 0, a= 2 : 1 1 13 1 8 10 6 1 9 15 5 4 13 11 15
u=14, s= 1, a= 2 : 1 2 15 4 12 15 12 8 2 9 15 15 10 9 14
u=14, s= 2, a= 2 : 1 3 2 7 1 5 3 15 10 3 10 11 7 7 13
69
u=14, s= 3, a= 2 : 1 4 4 10 5 10 9 7 3 12 5 7 4 5 12
u=14, s= 4, a= 2 : 1 5 6 13 9 15 15 14 11 6 15 3 1 3 11
u=14, s= 5, a= 2 : 1 6 8 1 13 5 6 6 4 15 10 14 13 1 10
u=14, s= 6, a= 2 : 1 7 10 4 2 10 12 13 12 9 5 10 10 14 9
u=14, s= 7, a= 2 : 1 8 12 7 6 15 3 5 5 3 15 6 7 12 8
u=14, s= 8, a= 2 : 1 9 14 10 10 5 9 12 13 12 10 2 4 10 7
u=14, s= 9, a= 2 : 1 10 1 13 14 10 15 4 6 6 5 13 1 8 6
u=14, s=10, a= 2 : 1 11 3 1 3 15 6 11 14 15 15 9 13 6 5
u=14, s=11, a= 2 : 1 12 5 4 7 5 12 3 7 9 10 5 10 4 4
u=14, s=12, a= 2 : 1 13 7 7 11 10 3 10 15 3 5 1 7 2 3
u=14, s=13, a= 2 : 1 14 9 10 15 15 9 2 8 12 15 12 4 15 2
u=14, s=14, a= 2 : 1 15 11 13 4 5 15 9 1 6 10 8 1 13 1
C. 120 sequences in Class 3 have “1” as the first element of sequence u= 1, s= 0, a= 0 : 1 1 2 14 2 13 4 9 13 12 4 2 11 6 8
u= 1, s= 1, a= 0 : 1 2 4 2 6 3 10 1 6 6 14 13 8 4 7
u= 1, s= 2, a= 0 : 1 3 6 5 10 8 1 8 14 15 9 9 5 2 6
u= 1, s= 3, a= 0 : 1 4 8 8 14 13 7 15 7 9 4 5 2 15 5
u= 1, s= 4, a= 0 : 1 5 10 11 3 3 13 7 15 3 14 1 14 13 4
u= 1, s= 5, a= 0 : 1 6 12 14 7 8 4 14 8 12 9 12 11 11 3
u= 1, s= 6, a= 0 : 1 7 14 2 11 13 10 6 1 6 4 8 8 9 2
u= 1, s= 7, a= 0 : 1 8 1 5 15 3 1 13 9 15 14 4 5 7 1
u= 1, s= 8, a= 0 : 1 9 3 8 4 8 7 5 2 9 9 15 2 5 15
u= 1, s= 9, a= 0 : 1 10 5 11 8 13 13 12 10 3 4 11 14 3 14
u= 1, s=10, a= 0 : 1 11 7 14 12 3 4 4 3 12 14 7 11 1 13
u= 1, s=11, a= 0 : 1 12 9 2 1 8 10 11 11 6 9 3 8 14 12
u= 1, s=12, a= 0 : 1 13 11 5 5 13 1 3 4 15 4 14 5 12 11
u= 1, s=13, a= 0 : 1 14 13 8 9 3 7 10 12 9 14 10 2 10 10
u= 1, s=14, a= 0 : 1 15 15 11 13 8 13 2 5 3 9 6 14 8 9
u= 2, s= 0, a=14 : 1 1 3 12 3 10 7 2 10 8 7 3 6 11 15
u= 2, s= 1, a=14 : 1 2 5 15 7 15 13 9 3 2 2 14 3 9 14
u= 2, s= 2, a=14 : 1 3 7 3 11 5 4 1 11 11 12 10 15 7 13
u= 2, s= 3, a=14 : 1 4 9 6 15 10 10 8 4 5 7 6 12 5 12
u= 2, s= 4, a=14 : 1 5 11 9 4 15 1 15 12 14 2 2 9 3 11
u= 2, s= 5, a=14 : 1 6 13 12 8 5 7 7 5 8 12 13 6 1 10
u= 2, s= 6, a=14 : 1 7 15 15 12 10 13 14 13 2 7 9 3 14 9
u= 2, s= 7, a=14 : 1 8 2 3 1 15 4 6 6 11 2 5 15 12 8
u= 2, s= 8, a=14 : 1 9 4 6 5 5 10 13 14 5 12 1 12 10 7
u= 2, s= 9, a=14 : 1 10 6 9 9 10 1 5 7 14 7 12 9 8 6
70
u= 2, s=10, a=14 : 1 11 8 12 13 15 7 12 15 8 2 8 6 6 5
u= 2, s=11, a=14 : 1 12 10 15 2 5 13 4 8 2 12 4 3 4 4
u= 2, s=12, a=14 : 1 13 12 3 6 10 4 11 1 11 7 15 15 2 3
u= 2, s=13, a=14 : 1 14 14 6 10 15 10 3 9 5 2 11 12 15 2
u= 2, s=14, a=14 : 1 15 1 9 14 5 1 10 2 14 12 7 9 13 1
u= 4, s= 0, a=12 : 1 1 5 8 5 4 13 3 4 15 13 5 11 6 14
u= 4, s= 1, a=12 : 1 2 7 11 9 9 4 10 12 9 8 1 8 4 13
u= 4, s= 2, a=12 : 1 3 9 14 13 14 10 2 5 3 3 12 5 2 12
u= 4, s= 3, a=12 : 1 4 11 2 2 4 1 9 13 12 13 8 2 15 11
u= 4, s= 4, a=12 : 1 5 13 5 6 9 7 1 6 6 8 4 14 13 10
u= 4, s= 5, a=12 : 1 6 15 8 10 14 13 8 14 15 3 15 11 11 9
u= 4, s= 6, a=12 : 1 7 2 11 14 4 4 15 7 9 13 11 8 9 8
u= 4, s= 7, a=12 : 1 8 4 14 3 9 10 7 15 3 8 7 5 7 7
u= 4, s= 8, a=12 : 1 9 6 2 7 14 1 14 8 12 3 3 2 5 6
u= 4, s= 9, a=12 : 1 10 8 5 11 4 7 6 1 6 13 14 14 3 5
u= 4, s=10, a=12 : 1 11 10 8 15 9 13 13 9 15 8 10 11 1 4
u= 4, s=11, a=12 : 1 12 12 11 4 14 4 5 2 9 3 6 8 14 3
u= 4, s=12, a=12 : 1 13 14 14 8 4 10 12 10 3 13 2 5 12 2
u= 4, s=13, a=12 : 1 14 1 2 12 9 1 4 3 12 8 13 2 10 1
u= 4, s=14, a=12 : 1 15 3 5 1 14 7 11 11 6 3 9 14 8 15
u= 7, s= 0, a= 9 : 1 1 8 2 8 10 7 12 10 3 7 8 11 6 5
u= 7, s= 1, a= 9 : 1 2 10 5 12 15 13 4 3 12 2 4 8 4 4
u= 7, s= 2, a= 9 : 1 3 12 8 1 5 4 11 11 6 12 15 5 2 3
u= 7, s= 3, a= 9 : 1 4 14 11 5 10 10 3 4 15 7 11 2 15 2
u= 7, s= 4, a= 9 : 1 5 1 14 9 15 1 10 12 9 2 7 14 13 1
u= 7, s= 5, a= 9 : 1 6 3 2 13 5 7 2 5 3 12 3 11 11 15
u= 7, s= 6, a= 9 : 1 7 5 5 2 10 13 9 13 12 7 14 8 9 14
u= 7, s= 7, a= 9 : 1 8 7 8 6 15 4 1 6 6 2 10 5 7 13
u= 7, s= 8, a= 9 : 1 9 9 11 10 5 10 8 14 15 12 6 2 5 12
u= 7, s= 9, a= 9 : 1 10 11 14 14 10 1 15 7 9 7 2 14 3 11
u= 7, s=10, a= 9 : 1 11 13 2 3 15 7 7 15 3 2 13 11 1 10
u= 7, s=11, a= 9 : 1 12 15 5 7 5 13 14 8 12 12 9 8 14 9
u= 7, s=12, a= 9 : 1 13 2 8 11 10 4 6 1 6 7 5 5 12 8
u= 7, s=13, a= 9 : 1 14 4 11 15 15 10 13 9 15 2 1 2 10 7
u= 7, s=14, a= 9 : 1 15 6 14 4 5 1 5 2 9 12 12 14 8 6
u= 8, s= 0, a= 8 : 1 1 9 15 9 7 10 5 7 14 10 9 6 11 12
u= 8, s= 1, a= 8 : 1 2 11 3 13 12 1 12 15 8 5 5 3 9 11
u= 8, s= 2, a= 8 : 1 3 13 6 2 2 7 4 8 2 15 1 15 7 10
u= 8, s= 3, a= 8 : 1 4 15 9 6 7 13 11 1 11 10 12 12 5 9
u= 8, s= 4, a= 8 : 1 5 2 12 10 12 4 3 9 5 5 8 9 3 8
71
u= 8, s= 5, a= 8 : 1 6 4 15 14 2 10 10 2 14 15 4 6 1 7
u= 8, s= 6, a= 8 : 1 7 6 3 3 7 1 2 10 8 10 15 3 14 6
u= 8, s= 7, a= 8 : 1 8 8 6 7 12 7 9 3 2 5 11 15 12 5
u= 8, s= 8, a= 8 : 1 9 10 9 11 2 13 1 11 11 15 7 12 10 4
u= 8, s= 9, a= 8 : 1 10 12 12 15 7 4 8 4 5 10 3 9 8 3
u= 8, s=10, a= 8 : 1 11 14 15 4 12 10 15 12 14 5 14 6 6 2
u= 8, s=11, a= 8 : 1 12 1 3 8 2 1 7 5 8 15 10 3 4 1
u= 8, s=12, a= 8 : 1 13 3 6 12 7 7 14 13 2 10 6 15 2 15
u= 8, s=13, a= 8 : 1 14 5 9 1 12 13 6 6 11 5 2 12 15 14
u= 8, s=14, a= 8 : 1 15 7 12 5 2 4 13 14 5 15 13 9 13 13
u=11, s= 0, a= 5 : 1 1 12 9 12 13 4 14 13 2 4 12 6 11 3
u=11, s= 1, a= 5 : 1 2 14 12 1 3 10 6 6 11 14 8 3 9 2
u=11, s= 2, a= 5 : 1 3 1 15 5 8 1 13 14 5 9 4 15 7 1
u=11, s= 3, a= 5 : 1 4 3 3 9 13 7 5 7 14 4 15 12 5 15
u=11, s= 4, a= 5 : 1 5 5 6 13 3 13 12 15 8 14 11 9 3 14
u=11, s= 5, a= 5 : 1 6 7 9 2 8 4 4 8 2 9 7 6 1 13
u=11, s= 6, a= 5 : 1 7 9 12 6 13 10 11 1 11 4 3 3 14 12
u=11, s= 7, a= 5 : 1 8 11 15 10 3 1 3 9 5 14 14 15 12 11
u=11, s= 8, a= 5 : 1 9 13 3 14 8 7 10 2 14 9 10 12 10 10
u=11, s= 9, a= 5 : 1 10 15 6 3 13 13 2 10 8 4 6 9 8 9
u=11, s=10, a= 5 : 1 11 2 9 7 3 4 9 3 2 14 2 6 6 8
u=11, s=11, a= 5 : 1 12 4 12 11 8 10 1 11 11 9 13 3 4 7
u=11, s=12, a= 5 : 1 13 6 15 15 13 1 8 4 5 4 9 15 2 6
u=11, s=13, a= 5 : 1 14 8 3 4 3 7 15 12 14 14 5 12 15 5
u=11, s=14, a= 5 : 1 15 10 6 8 8 13 7 5 8 9 1 9 13 4
u=13, s= 0, a= 3 : 1 1 14 5 14 7 10 15 7 9 10 14 11 6 2
u=13, s= 1, a= 3 : 1 2 1 8 3 12 1 7 15 3 5 10 8 4 1
u=13, s= 2, a= 3 : 1 3 3 11 7 2 7 14 8 12 15 6 5 2 15
u=13, s= 3, a= 3 : 1 4 5 14 11 7 13 6 1 6 10 2 2 15 14
u=13, s= 4, a= 3 : 1 5 7 2 15 12 4 13 9 15 5 13 14 13 13
u=13, s= 5, a= 3 : 1 6 9 5 4 2 10 5 2 9 15 9 11 11 12
u=13, s= 6, a= 3 : 1 7 11 8 8 7 1 12 10 3 10 5 8 9 11
u=13, s= 7, a= 3 : 1 8 13 11 12 12 7 4 3 12 5 1 5 7 10
u=13, s= 8, a= 3 : 1 9 15 14 1 2 13 11 11 6 15 12 2 5 9
u=13, s= 9, a= 3 : 1 10 2 2 5 7 4 3 4 15 10 8 14 3 8
u=13, s=10, a= 3 : 1 11 4 5 9 12 10 10 12 9 5 4 11 1 7
u=13, s=11, a= 3 : 1 12 6 8 13 2 1 2 5 3 15 15 8 14 6
u=13, s=12, a= 3 : 1 13 8 11 2 7 7 9 13 12 10 11 5 12 5
u=13, s=13, a= 3 : 1 14 10 14 6 12 13 1 6 6 5 7 2 10 4
u=13, s=14, a= 3 : 1 15 12 2 10 2 4 8 14 15 15 3 14 8 3
72
u=14, s= 0, a= 2 : 1 1 15 3 15 4 13 8 4 5 13 15 6 11 9
u=14, s= 1, a= 2 : 1 2 2 6 4 9 4 15 12 14 8 11 3 9 8
u=14, s= 2, a= 2 : 1 3 4 9 8 14 10 7 5 8 3 7 15 7 7
u=14, s= 3, a= 2 : 1 4 6 12 12 4 1 14 13 2 13 3 12 5 6
u=14, s= 4, a= 2 : 1 5 8 15 1 9 7 6 6 11 8 14 9 3 5
u=14, s= 5, a= 2 : 1 6 10 3 5 14 13 13 14 5 3 10 6 1 4
u=14, s= 6, a= 2 : 1 7 12 6 9 4 4 5 7 14 13 6 3 14 3
u=14, s= 7, a= 2 : 1 8 14 9 13 9 10 12 15 8 8 2 15 12 2
u=14, s= 8, a= 2 : 1 9 1 12 2 14 1 4 8 2 3 13 12 10 1
u=14, s= 9, a= 2 : 1 10 3 15 6 4 7 11 1 11 13 9 9 8 15
u=14, s=10, a= 2 : 1 11 5 3 10 9 13 3 9 5 8 5 6 6 14
u=14, s=11, a= 2 : 1 12 7 6 14 14 4 10 2 14 3 1 3 4 13
u=14, s=12, a= 2 : 1 13 9 9 3 4 10 2 10 8 13 12 15 2 12
u=14, s=13, a= 2 : 1 14 11 12 7 9 1 9 3 2 8 8 12 15 11
u=14, s=14, a= 2 : 1 15 13 15 11 14 7 1 11 11 3 4 9 13 10
D. 120 sequences in Class 4 have “1” as the first element of sequence
u= 1, s= 0, a= 0 : 1 1 4 9 4 11 10 8 5 9 10 1 9 5 7
u= 1, s= 1, a= 0 : 1 2 6 12 8 1 1 15 13 3 5 12 6 3 6
u= 1, s= 2, a= 0 : 1 3 8 15 12 6 7 7 6 12 15 8 3 1 5
u= 1, s= 3, a= 0 : 1 4 10 3 1 11 13 14 14 6 10 4 15 14 4
u= 1, s= 4, a= 0 : 1 5 12 6 5 1 4 6 7 15 5 15 12 12 3
u= 1, s= 5, a= 0 : 1 6 14 9 9 6 10 13 15 9 15 11 9 10 2
u= 1, s= 6, a= 0 : 1 7 1 12 13 11 1 5 8 3 10 7 6 8 1
u= 1, s= 7, a= 0 : 1 8 3 15 2 1 7 12 1 12 5 3 3 6 15
u= 1, s= 8, a= 0 : 1 9 5 3 6 6 13 4 9 6 15 14 15 4 14
u= 1, s= 9, a= 0 : 1 10 7 6 10 11 4 11 2 15 10 10 12 2 13
u= 1, s=10, a= 0 : 1 11 9 9 14 1 10 3 10 9 5 6 9 15 12
u= 1, s=11, a= 0 : 1 12 11 12 3 6 1 10 3 3 15 2 6 13 11
u= 1, s=12, a= 0 : 1 13 13 15 7 11 7 2 11 12 10 13 3 11 10
u= 1, s=13, a= 0 : 1 14 15 3 11 1 13 9 4 6 5 9 15 9 9
u= 1, s=14, a= 0 : 1 15 2 6 15 6 4 1 12 15 15 5 12 7 8
u= 2, s= 0, a=14 : 1 1 7 2 7 6 4 15 9 2 4 1 2 9 13
u= 2, s= 1, a=14 : 1 2 9 5 11 11 10 7 2 11 14 12 14 7 12
u= 2, s= 2, a=14 : 1 3 11 8 15 1 1 14 10 5 9 8 11 5 11
u= 2, s= 3, a=14 : 1 4 13 11 4 6 7 6 3 14 4 4 8 3 10
u= 2, s= 4, a=14 : 1 5 15 14 8 11 13 13 11 8 14 15 5 1 9
u= 2, s= 5, a=14 : 1 6 2 2 12 1 4 5 4 2 9 11 2 14 8
73
u= 2, s= 6, a=14 : 1 7 4 5 1 6 10 12 12 11 4 7 14 12 7
u= 2, s= 7, a=14 : 1 8 6 8 5 11 1 4 5 5 14 3 11 10 6
u= 2, s= 8, a=14 : 1 9 8 11 9 1 7 11 13 14 9 14 8 8 5
u= 2, s= 9, a=14 : 1 10 10 14 13 6 13 3 6 8 4 10 5 6 4
u= 2, s=10, a=14 : 1 11 12 2 2 11 4 10 14 2 14 6 2 4 3
u= 2, s=11, a=14 : 1 12 14 5 6 1 10 2 7 11 9 2 14 2 2
u= 2, s=12, a=14 : 1 13 1 8 10 6 1 9 15 5 4 13 11 15 1
u= 2, s=13, a=14 : 1 14 3 11 14 11 7 1 8 14 14 9 8 13 15
u= 2, s=14, a=14 : 1 15 5 14 3 1 13 8 1 8 9 5 5 11 14
u= 4, s= 0, a=12 : 1 1 13 3 13 11 7 14 2 3 7 1 3 2 10
u= 4, s= 1, a=12 : 1 2 15 6 2 1 13 6 10 12 2 12 15 15 9
u= 4, s= 2, a=12 : 1 3 2 9 6 6 4 13 3 6 12 8 12 13 8
u= 4, s= 3, a=12 : 1 4 4 12 10 11 10 5 11 15 7 4 9 11 7
u= 4, s= 4, a=12 : 1 5 6 15 14 1 1 12 4 9 2 15 6 9 6
u= 4, s= 5, a=12 : 1 6 8 3 3 6 7 4 12 3 12 11 3 7 5
u= 4, s= 6, a=12 : 1 7 10 6 7 11 13 11 5 12 7 7 15 5 4
u= 4, s= 7, a=12 : 1 8 12 9 11 1 4 3 13 6 2 3 12 3 3
u= 4, s= 8, a=12 : 1 9 14 12 15 6 10 10 6 15 12 14 9 1 2
u= 4, s= 9, a=12 : 1 10 1 15 4 11 1 2 14 9 7 10 6 14 1
u= 4, s=10, a=12 : 1 11 3 3 8 1 7 9 7 3 2 6 3 12 15
u= 4, s=11, a=12 : 1 12 5 6 12 6 13 1 15 12 12 2 15 10 14
u= 4, s=12, a=12 : 1 13 7 9 1 11 4 8 8 6 7 13 12 8 13
u= 4, s=13, a=12 : 1 14 9 12 5 1 10 15 1 15 2 9 9 6 12
u= 4, s=14, a=12 : 1 15 11 15 9 6 1 7 9 9 12 5 6 4 11
u= 7, s= 0, a= 9 : 1 1 7 12 7 11 4 5 14 12 4 1 12 14 13
u= 7, s= 1, a= 9 : 1 2 9 15 11 1 10 12 7 6 14 12 9 12 12
u= 7, s= 2, a= 9 : 1 3 11 3 15 6 1 4 15 15 9 8 6 10 11
u= 7, s= 3, a= 9 : 1 4 13 6 4 11 7 11 8 9 4 4 3 8 10
u= 7, s= 4, a= 9 : 1 5 15 9 8 1 13 3 1 3 14 15 15 6 9
u= 7, s= 5, a= 9 : 1 6 2 12 12 6 4 10 9 12 9 11 12 4 8
u= 7, s= 6, a= 9 : 1 7 4 15 1 11 10 2 2 6 4 7 9 2 7
u= 7, s= 7, a= 9 : 1 8 6 3 5 1 1 9 10 15 14 3 6 15 6
u= 7, s= 8, a= 9 : 1 9 8 6 9 6 7 1 3 9 9 14 3 13 5
u= 7, s= 9, a= 9 : 1 10 10 9 13 11 13 8 11 3 4 10 15 11 4
u= 7, s=10, a= 9 : 1 11 12 12 2 1 4 15 4 12 14 6 12 9 3
u= 7, s=11, a= 9 : 1 12 14 15 6 6 10 7 12 6 9 2 9 7 2
u= 7, s=12, a= 9 : 1 13 1 3 10 11 1 14 5 15 4 13 6 5 1
u= 7, s=13, a= 9 : 1 14 3 6 14 1 7 6 13 9 14 9 3 3 15
u= 7, s=14, a= 9 : 1 15 5 9 3 6 13 13 6 3 9 5 15 1 14
u= 8, s= 0, a= 8 : 1 1 10 5 10 6 13 12 3 5 13 1 5 3 4
74
u= 8, s= 1, a= 8 : 1 2 12 8 14 11 4 4 11 14 8 12 2 1 3
u= 8, s= 2, a= 8 : 1 3 14 11 3 1 10 11 4 8 3 8 14 14 2
u= 8, s= 3, a= 8 : 1 4 1 14 7 6 1 3 12 2 13 4 11 12 1
u= 8, s= 4, a= 8 : 1 5 3 2 11 11 7 10 5 11 8 15 8 10 15
u= 8, s= 5, a= 8 : 1 6 5 5 15 1 13 2 13 5 3 11 5 8 14
u= 8, s= 6, a= 8 : 1 7 7 8 4 6 4 9 6 14 13 7 2 6 13
u= 8, s= 7, a= 8 : 1 8 9 11 8 11 10 1 14 8 8 3 14 4 12
u= 8, s= 8, a= 8 : 1 9 11 14 12 1 1 8 7 2 3 14 11 2 11
u= 8, s= 9, a= 8 : 1 10 13 2 1 6 7 15 15 11 13 10 8 15 10
u= 8, s=10, a= 8 : 1 11 15 5 5 11 13 7 8 5 8 6 5 13 9
u= 8, s=11, a= 8 : 1 12 2 8 9 1 4 14 1 14 3 2 2 11 8
u= 8, s=12, a= 8 : 1 13 4 11 13 6 10 6 9 8 13 13 14 9 7
u= 8, s=13, a= 8 : 1 14 6 14 2 11 1 13 2 2 8 9 11 7 6
u= 8, s=14, a= 8 : 1 15 8 2 6 1 7 5 10 11 3 5 8 5 5
u=11, s= 0, a= 5 : 1 1 4 14 4 6 10 3 15 14 10 1 14 15 7
u=11, s= 1, a= 5 : 1 2 6 2 8 11 1 10 8 8 5 12 11 13 6
u=11, s= 2, a= 5 : 1 3 8 5 12 1 7 2 1 2 15 8 8 11 5
u=11, s= 3, a= 5 : 1 4 10 8 1 6 13 9 9 11 10 4 5 9 4
u=11, s= 4, a= 5 : 1 5 12 11 5 11 4 1 2 5 5 15 2 7 3
u=11, s= 5, a= 5 : 1 6 14 14 9 1 10 8 10 14 15 11 14 5 2
u=11, s= 6, a= 5 : 1 7 1 2 13 6 1 15 3 8 10 7 11 3 1
u=11, s= 7, a= 5 : 1 8 3 5 2 11 7 7 11 2 5 3 8 1 15
u=11, s= 8, a= 5 : 1 9 5 8 6 1 13 14 4 11 15 14 5 14 14
u=11, s= 9, a= 5 : 1 10 7 11 10 6 4 6 12 5 10 10 2 12 13
u=11, s=10, a= 5 : 1 11 9 14 14 11 10 13 5 14 5 6 14 10 12
u=11, s=11, a= 5 : 1 12 11 2 3 1 1 5 13 8 15 2 11 8 11
u=11, s=12, a= 5 : 1 13 13 5 7 6 7 12 6 2 10 13 8 6 10
u=11, s=13, a= 5 : 1 14 15 8 11 11 13 4 14 11 5 9 5 4 9
u=11, s=14, a= 5 : 1 15 2 11 15 1 4 11 7 5 15 5 2 2 8
u=13, s= 0, a= 3 : 1 1 10 15 10 11 13 2 8 15 13 1 15 8 4
u=13, s= 1, a= 3 : 1 2 12 3 14 1 4 9 1 9 8 12 12 6 3
u=13, s= 2, a= 3 : 1 3 14 6 3 6 10 1 9 3 3 8 9 4 2
u=13, s= 3, a= 3 : 1 4 1 9 7 11 1 8 2 12 13 4 6 2 1
u=13, s= 4, a= 3 : 1 5 3 12 11 1 7 15 10 6 8 15 3 15 15
u=13, s= 5, a= 3 : 1 6 5 15 15 6 13 7 3 15 3 11 15 13 14
u=13, s= 6, a= 3 : 1 7 7 3 4 11 4 14 11 9 13 7 12 11 13
u=13, s= 7, a= 3 : 1 8 9 6 8 1 10 6 4 3 8 3 9 9 12
u=13, s= 8, a= 3 : 1 9 11 9 12 6 1 13 12 12 3 14 6 7 11
u=13, s= 9, a= 3 : 1 10 13 12 1 11 7 5 5 6 13 10 3 5 10
u=13, s=10, a= 3 : 1 11 15 15 5 1 13 12 13 15 8 6 15 3 9
75
u=13, s=11, a= 3 : 1 12 2 3 9 6 4 4 6 9 3 2 12 1 8
u=13, s=12, a= 3 : 1 13 4 6 13 11 10 11 14 3 13 13 9 14 7
u=13, s=13, a= 3 : 1 14 6 9 2 1 1 3 7 12 8 9 6 12 6
u=13, s=14, a= 3 : 1 15 8 12 6 6 7 10 15 6 3 5 3 10 5
u=14, s= 0, a= 2 : 1 1 13 8 13 6 7 9 12 8 7 1 8 12 10
u=14, s= 1, a= 2 : 1 2 15 11 2 11 13 1 5 2 2 12 5 10 9
u=14, s= 2, a= 2 : 1 3 2 14 6 1 4 8 13 11 12 8 2 8 8
u=14, s= 3, a= 2 : 1 4 4 2 10 6 10 15 6 5 7 4 14 6 7
u=14, s= 4, a= 2 : 1 5 6 5 14 11 1 7 14 14 2 15 11 4 6
u=14, s= 5, a= 2 : 1 6 8 8 3 1 7 14 7 8 12 11 8 2 5
u=14, s= 6, a= 2 : 1 7 10 11 7 6 13 6 15 2 7 7 5 15 4
u=14, s= 7, a= 2 : 1 8 12 14 11 11 4 13 8 11 2 3 2 13 3
u=14, s= 8, a= 2 : 1 9 14 2 15 1 10 5 1 5 12 14 14 11 2
u=14, s= 9, a= 2 : 1 10 1 5 4 6 1 12 9 14 7 10 11 9 1
u=14, s=10, a= 2 : 1 11 3 8 8 11 7 4 2 8 2 6 8 7 15
u=14, s=11, a= 2 : 1 12 5 11 12 1 13 11 10 2 12 2 5 5 14
u=14, s=12, a= 2 : 1 13 7 14 1 6 4 3 3 11 7 13 2 3 13
u=14, s=13, a= 2 : 1 14 9 2 5 11 10 10 11 5 2 9 14 1 12
u=14, s=14, a= 2 : 1 15 11 5 9 1 1 2 4 14 12 5 11 14 11
E. 120 sequences in Class 5 have “1” as the first element of sequence
u= 1, s= 0, a= 0 : 1 1 2 13 5 4 7 4 12 14 5 3 12 1 6
u= 1, s= 1, a= 0 : 1 2 4 1 9 9 13 11 5 8 15 14 9 14 5
u= 1, s= 2, a= 0 : 1 3 6 4 13 14 4 3 13 2 10 10 6 12 4
u= 1, s= 3, a= 0 : 1 4 8 7 2 4 10 10 6 11 5 6 3 10 3
u= 1, s= 4, a= 0 : 1 5 10 10 6 9 1 2 14 5 15 2 15 8 2
u= 1, s= 5, a= 0 : 1 6 12 13 10 14 7 9 7 14 10 13 12 6 1
u= 1, s= 6, a= 0 : 1 7 14 1 14 4 13 1 15 8 5 9 9 4 15
u= 1, s= 7, a= 0 : 1 8 1 4 3 9 4 8 8 2 15 5 6 2 14
u= 1, s= 8, a= 0 : 1 9 3 7 7 14 10 15 1 11 10 1 3 15 13
u= 1, s= 9, a= 0 : 1 10 5 10 11 4 1 7 9 5 5 12 15 13 12
u= 1, s=10, a= 0 : 1 11 7 13 15 9 7 14 2 14 15 8 12 11 11
u= 1, s=11, a= 0 : 1 12 9 1 4 14 13 6 10 8 10 4 9 9 10
u= 1, s=12, a= 0 : 1 13 11 4 8 4 4 13 3 2 5 15 6 7 9
u= 1, s=13, a= 0 : 1 14 13 7 12 9 10 5 11 11 15 11 3 5 8
u= 1, s=14, a= 0 : 1 15 15 10 1 14 1 12 4 5 10 7 15 3 7
u= 2, s= 0, a=14 : 1 1 3 10 9 7 13 7 8 12 9 5 8 1 11
u= 2, s= 1, a=14 : 1 2 5 13 13 12 4 14 1 6 4 1 5 14 10
76
u= 2, s= 2, a=14 : 1 3 7 1 2 2 10 6 9 15 14 12 2 12 9
u= 2, s= 3, a=14 : 1 4 9 4 6 7 1 13 2 9 9 8 14 10 8
u= 2, s= 4, a=14 : 1 5 11 7 10 12 7 5 10 3 4 4 11 8 7
u= 2, s= 5, a=14 : 1 6 13 10 14 2 13 12 3 12 14 15 8 6 6
u= 2, s= 6, a=14 : 1 7 15 13 3 7 4 4 11 6 9 11 5 4 5
u= 2, s= 7, a=14 : 1 8 2 1 7 12 10 11 4 15 4 7 2 2 4
u= 2, s= 8, a=14 : 1 9 4 4 11 2 1 3 12 9 14 3 14 15 3
u= 2, s= 9, a=14 : 1 10 6 7 15 7 7 10 5 3 9 14 11 13 2
u= 2, s=10, a=14 : 1 11 8 10 4 12 13 2 13 12 4 10 8 11 1
u= 2, s=11, a=14 : 1 12 10 13 8 2 4 9 6 6 14 6 5 9 15
u= 2, s=12, a=14 : 1 13 12 1 12 7 10 1 14 15 9 2 2 7 14
u= 2, s=13, a=14 : 1 14 14 4 1 12 1 8 7 9 4 13 14 5 13
u= 2, s=14, a=14 : 1 15 1 7 5 2 7 15 15 3 14 9 11 3 12
u= 4, s= 0, a=12 : 1 1 5 4 2 13 10 13 15 8 2 9 15 1 6
u= 4, s= 1, a=12 : 1 2 7 7 6 3 1 5 8 2 12 5 12 14 5
u= 4, s= 2, a=12 : 1 3 9 10 10 8 7 12 1 11 7 1 9 12 4
u= 4, s= 3, a=12 : 1 4 11 13 14 13 13 4 9 5 2 12 6 10 3
u= 4, s= 4, a=12 : 1 5 13 1 3 3 4 11 2 14 12 8 3 8 2
u= 4, s= 5, a=12 : 1 6 15 4 7 8 10 3 10 8 7 4 15 6 1
u= 4, s= 6, a=12 : 1 7 2 7 11 13 1 10 3 2 2 15 12 4 15
u= 4, s= 7, a=12 : 1 8 4 10 15 3 7 2 11 11 12 11 9 2 14
u= 4, s= 8, a=12 : 1 9 6 13 4 8 13 9 4 5 7 7 6 15 13
u= 4, s= 9, a=12 : 1 10 8 1 8 13 4 1 12 14 2 3 3 13 12
u= 4, s=10, a=12 : 1 11 10 4 12 3 10 8 5 8 12 14 15 11 11
u= 4, s=11, a=12 : 1 12 12 7 1 8 1 15 13 2 7 10 12 9 10
u= 4, s=12, a=12 : 1 13 14 10 5 13 7 7 6 11 2 6 9 7 9
u= 4, s=13, a=12 : 1 14 1 13 9 3 13 14 14 5 12 2 6 5 8
u= 4, s=14, a=12 : 1 15 3 1 13 8 4 6 7 14 7 13 3 3 7
u= 7, s= 0, a= 9 : 1 1 8 10 14 7 13 7 3 2 14 15 3 1 6
u= 7, s= 1, a= 9 : 1 2 10 13 3 12 4 14 11 11 9 11 15 14 5
u= 7, s= 2, a= 9 : 1 3 12 1 7 2 10 6 4 5 4 7 12 12 4
u= 7, s= 3, a= 9 : 1 4 14 4 11 7 1 13 12 14 14 3 9 10 3
u= 7, s= 4, a= 9 : 1 5 1 7 15 12 7 5 5 8 9 14 6 8 2
u= 7, s= 5, a= 9 : 1 6 3 10 4 2 13 12 13 2 4 10 3 6 1
u= 7, s= 6, a= 9 : 1 7 5 13 8 7 4 4 6 11 14 6 15 4 15
u= 7, s= 7, a= 9 : 1 8 7 1 12 12 10 11 14 5 9 2 12 2 14
u= 7, s= 8, a= 9 : 1 9 9 4 1 2 1 3 7 14 4 13 9 15 13
u= 7, s= 9, a= 9 : 1 10 11 7 5 7 7 10 15 8 14 9 6 13 12
u= 7, s=10, a= 9 : 1 11 13 10 9 12 13 2 8 2 9 5 3 11 11
u= 7, s=11, a= 9 : 1 12 15 13 13 2 4 9 1 11 4 1 15 9 10
77
u= 7, s=12, a= 9 : 1 13 2 1 2 7 10 1 9 5 14 12 12 7 9
u= 7, s=13, a= 9 : 1 14 4 4 6 12 1 8 2 14 9 8 9 5 8
u= 7, s=14, a= 9 : 1 15 6 7 10 2 7 15 10 8 4 4 6 3 7
u= 8, s= 0, a= 8 : 1 1 9 7 3 10 4 10 14 15 3 2 14 1 11
u= 8, s= 1, a= 8 : 1 2 11 10 7 15 10 2 7 9 13 13 11 14 10
u= 8, s= 2, a= 8 : 1 3 13 13 11 5 1 9 15 3 8 9 8 12 9
u= 8, s= 3, a= 8 : 1 4 15 1 15 10 7 1 8 12 3 5 5 10 8
u= 8, s= 4, a= 8 : 1 5 2 4 4 15 13 8 1 6 13 1 2 8 7
u= 8, s= 5, a= 8 : 1 6 4 7 8 5 4 15 9 15 8 12 14 6 6
u= 8, s= 6, a= 8 : 1 7 6 10 12 10 10 7 2 9 3 8 11 4 5
u= 8, s= 7, a= 8 : 1 8 8 13 1 15 1 14 10 3 13 4 8 2 4
u= 8, s= 8, a= 8 : 1 9 10 1 5 5 7 6 3 12 8 15 5 15 3
u= 8, s= 9, a= 8 : 1 10 12 4 9 10 13 13 11 6 3 11 2 13 2
u= 8, s=10, a= 8 : 1 11 14 7 13 15 4 5 4 15 13 7 14 11 1
u= 8, s=11, a= 8 : 1 12 1 10 2 5 10 12 12 9 8 3 11 9 15
u= 8, s=12, a= 8 : 1 13 3 13 6 10 1 4 5 3 3 14 8 7 14
u= 8, s=13, a= 8 : 1 14 5 1 10 15 7 11 13 12 13 10 5 5 13
u= 8, s=14, a= 8 : 1 15 7 4 14 5 13 3 6 6 8 6 2 3 12
u=11, s= 0, a= 5 : 1 1 12 13 15 4 7 4 2 9 15 8 2 1 11
u=11, s= 1, a= 5 : 1 2 14 1 4 9 13 11 10 3 10 4 14 14 10
u=11, s= 2, a= 5 : 1 3 1 4 8 14 4 3 3 12 5 15 11 12 9
u=11, s= 3, a= 5 : 1 4 3 7 12 4 10 10 11 6 15 11 8 10 8
u=11, s= 4, a= 5 : 1 5 5 10 1 9 1 2 4 15 10 7 5 8 7
u=11, s= 5, a= 5 : 1 6 7 13 5 14 7 9 12 9 5 3 2 6 6
u=11, s= 6, a= 5 : 1 7 9 1 9 4 13 1 5 3 15 14 14 4 5
u=11, s= 7, a= 5 : 1 8 11 4 13 9 4 8 13 12 10 10 11 2 4
u=11, s= 8, a= 5 : 1 9 13 7 2 14 10 15 6 6 5 6 8 15 3
u=11, s= 9, a= 5 : 1 10 15 10 6 4 1 7 14 15 15 2 5 13 2
u=11, s=10, a= 5 : 1 11 2 13 10 9 7 14 7 9 10 13 2 11 1
u=11, s=11, a= 5 : 1 12 4 1 14 14 13 6 15 3 5 9 14 9 15
u=11, s=12, a= 5 : 1 13 6 4 3 4 4 13 8 12 15 5 11 7 14
u=11, s=13, a= 5 : 1 14 8 7 7 9 10 5 1 6 10 1 8 5 13
u=11, s=14, a= 5 : 1 15 10 10 11 14 1 12 9 15 5 12 5 3 12
u=13, s= 0, a= 3 : 1 1 14 7 8 10 4 10 9 5 8 12 9 1 6
u=13, s= 1, a= 3 : 1 2 1 10 12 15 10 2 2 14 3 8 6 14 5
u=13, s= 2, a= 3 : 1 3 3 13 1 5 1 9 10 8 13 4 3 12 4
u=13, s= 3, a= 3 : 1 4 5 1 5 10 7 1 3 2 8 15 15 10 3
u=13, s= 4, a= 3 : 1 5 7 4 9 15 13 8 11 11 3 11 12 8 2
u=13, s= 5, a= 3 : 1 6 9 7 13 5 4 15 4 5 13 7 9 6 1
u=13, s= 6, a= 3 : 1 7 11 10 2 10 10 7 12 14 8 3 6 4 15
78
u=13, s= 7, a= 3 : 1 8 13 13 6 15 1 14 5 8 3 14 3 2 14
u=13, s= 8, a= 3 : 1 9 15 1 10 5 7 6 13 2 13 10 15 15 13
u=13, s= 9, a= 3 : 1 10 2 4 14 10 13 13 6 11 8 6 12 13 12
u=13, s=10, a= 3 : 1 11 4 7 3 15 4 5 14 5 3 2 9 11 11
u=13, s=11, a= 3 : 1 12 6 10 7 5 10 12 7 14 13 13 6 9 10
u=13, s=12, a= 3 : 1 13 8 13 11 10 1 4 15 8 8 9 3 7 9
u=13, s=13, a= 3 : 1 14 10 1 15 15 7 11 8 2 3 5 15 5 8
u=13, s=14, a= 3 : 1 15 12 4 4 5 13 3 1 11 13 1 12 3 7
u=14, s= 0, a= 2 : 1 1 15 4 12 13 10 13 5 3 12 14 5 1 11
u=14, s= 1, a= 2 : 1 2 2 7 1 3 1 5 13 12 7 10 2 14 10
u=14, s= 2, a= 2 : 1 3 4 10 5 8 7 12 6 6 2 6 14 12 9
u=14, s= 3, a= 2 : 1 4 6 13 9 13 13 4 14 15 12 2 11 10 8
u=14, s= 4, a= 2 : 1 5 8 1 13 3 4 11 7 9 7 13 8 8 7
u=14, s= 5, a= 2 : 1 6 10 4 2 8 10 3 15 3 2 9 5 6 6
u=14, s= 6, a= 2 : 1 7 12 7 6 13 1 10 8 12 12 5 2 4 5
u=14, s= 7, a= 2 : 1 8 14 10 10 3 7 2 1 6 7 1 14 2 4
u=14, s= 8, a= 2 : 1 9 1 13 14 8 13 9 9 15 2 12 11 15 3
u=14, s= 9, a= 2 : 1 10 3 1 3 13 4 1 2 9 12 8 8 13 2
u=14, s=10, a= 2 : 1 11 5 4 7 3 10 8 10 3 7 4 5 11 1
u=14, s=11, a= 2 : 1 12 7 7 11 8 1 15 3 12 2 15 2 9 15
u=14, s=12, a= 2 : 1 13 9 10 15 13 7 7 11 6 12 11 14 7 14
u=14, s=13, a= 2 : 1 14 11 13 4 3 13 14 4 15 7 7 11 5 13
u=14, s=14, a= 2 : 1 15 13 1 8 8 4 6 12 9 2 3 8 3 12=
79
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