1

Generalized Bertrand’s postulate and

proposed solution of Legendre’s conjecture

Rosario Turco (1), Maria Colonnese (

2)

Introduction

In this paper we discuss the links between Legendre's conjecture and Bertrand's

postulate, the latter also known as Chebiscev theorem.

Bertrand‟s postulate says that:

1, : 2n p n p n

In other words, for every integer n greater than 1, there exists at least one

prime number between it 2n, namely that:

(2 ) ( )n n

Legendre's conjecture says that there is always a prime between n2 and (n + 1)2,

namely:

2 21, : ( 1)n p n p n

This conjecture is one of the problems of Landau and, to date, has not yet been

demonstrated.

In [3] we have seen evidence of the conjecture, a proposal so, the formulas for

the estimates of the number of primes in a quadratic interval.

In this paper, we emphasize, however, that Legendre's conjecture is a

generalization of Bertrand's postulate, and we will try to leverage this fact to

see if you can get more results.

Generalized Bertrand's postulate (GBP) can be formulated as follows:

1, p : ( 1)n k n kn p k n (1)

The GBP if k = 1 is already shown Bertrand's postulate, in addition Bachraoui

(see [2]) has proven true for k = 2.

If in (1) arises k = n we obtain a conjecture similar to Legendre, a little more

restrictive as the superior border.

An idea, therefore, may be to work at PBG, and use the combinatorial as in [1]

and [4].

1 Rosario Turco is an engineer, who works in a society Information Technology of the group TelecomItalia (NA – Italy)

2 Maria Colonnese is a teacher of mathematics, who works at the High School “De Bottis” of Torre del Greco (NA - Italy)

2

The work is divided, then, in PART ONE Bertrand's postulate, PART TWO

Generalized Bertrand's Postulate and Legendre's conjecture.

Definitions

We define with ( , )a p b

a b p

the product of primes in the interval a,b with the

primes greater than a, but not greater than b; con ( )p a

a p

the product of

primes not greater than a. We define the binomial coefficient with:

!

!( )!

n n

h h n h

that shows the grouping of a subset of h elements compared with a set of n

elements, with 0 h n. We also use:

n n

h n h

.

We remember the Newton’s binomial:

0

( )n

n h n h

h

na b a b

h

.

Then we define with p|c that p is a divisor of c; while with the symbol x we

show the floor of x, the biggest integer less or equal than x. Next with pk show

the k-th prime number.

PART ONE

Betrand’s Postulate

For better readability, we will introduce a series of terms, which are valid for

theorems, following the paper [4].

Although the proof is long enough, it is not difficult to understand. It simply

uses the binomial calculations and considerations on the exponents in the

factorization in the Fundamental Theorem of Arithmetic and this shows even more

the great genius of Paul Erdos. Especially surprising the depth of his global

vision.

Lemma 1 – For each n>0 is always 2

4nn

n

.

Proof

22 2

0

2 24 2 (1 1)

nn n n

h

n n

h n

3

Lemma 2 – If n is odd for each 0 kn is always 12nn

k

.

Proof

N is odd then:

( 1) / 21

0 0

2 (1 1) 2 2 2nn

n n n

h h

n n n n

h h k k

Lemma 3 – If p is a prime number with p>n then allora p isn‟t a divisor of n!.

Proof

The proof is trivial because a prime p cannot be written as a product of primes

smaller than p. So if p> n then even if we have that:

n! = n *(n-1)*(n-2)…3*2*1,

p cannot be its divisor.

Lemma 4 – If a is a real number with a2, then the product of the prime number

less or equal than a is less than 4a: ( ) 4a

p a

a p

Proof

For 2 a 3 it‟s trivial. If it‟s true for each odd integer n 3 it will be

also true for each real number b 3, because for any b 3 always exists a odd

n b n+2, then as b aboves n+2 so the number of prime numbers less than b are equal to those of n:

4 4n b

p b p n

p p

So we have to prove that the Lemma is true for any integer n 3, and we do by

induction.

For n = 3 is true because 6 <64, which represents the starting point of the

induction.

For an odd n 5, let us assume that the theorem is true for every odd k < n

and we choose a sign such that k = (n 1)/2 is odd. In this mode is also k

3.

It‟s obvious that n = 2k 1 e n – k = 2k 1 – k k + 1, so n – k k + 1. This latter fact we are interested to see any divisibility of p for (n-k)!.

So if we assume k < p n then by Lemma 3 is p|n! but p isn‟t a divisor of k!. If p > n-k then p nor is divider of (n-k)!.

If p is a divisor of n! it involves that !

|!( )!

n np

k k n k

. This is true for k < p

n. So we can say:

4

|k p n k p n

n np p

k k

If we use the Lemma 2 now it is:

12n

k p n

p

At this point the previous relationship we can break it in a product of two

parts, one for p<k and one with k < p n:

p n p k k p n

p p p

As we have previously speculated k odd and less than n, the first part is less

than 4K, while the second product that we know is less than or equal to 2n-1, which is:

1 2 1 2 14 2 2 2 2k n k n k n

p n

p

Now we notes that:

2 1 2 2 1 2 1 2k n k n n k n n

So:

22 4n n

p n

p

which finally proves Lemma 4.

Lemma 5 – If n > 1 then 2 4

2

nn

n n

Proof

It is obtained by induction and taking account of Lemma 1.

Now we are interested in such primes p divide 2n

n

and with what esponent.

Lemma 6 – If p is a prime number, the largest esponent v that divides n! is:

1j

j

nv

p

(a)

The sum in (a) is always finite, because when j grows, it‟s pj > n and the ratio

tends to zero.

For example with p=2, the largest power that divides 55! is 250, infact from (a):

floor(55/2)+ floor(55/4) + floor(55/8) + floor(55/16) + floor(55/32) =

5

= 27 + 13 + 6 + 3 + 1 = 50.

With what we saw earlier, for example, we could indicate with vp the largest

integer exponent such that 2

|pv np

n

.

We also know that the prime factors of (2n)! are all less or equal than 2n; so:

2

2pv

p n

np

n

For example with PARI/GP for 34

17

is:

? r = binomial(34,17)

%17 = 2333606220

? factor( r )

%18 =

[2 2]

[3 3]

[5 1]

[11 1]

[19 1]

[23 1]

[29 1]

[31 1]

The factorization on the right of the brackets shows the exponents vp we are

looking for, the binomial coefficient is decomposed into factors, then, that we

are finding; so the binomial coefficient is: 2^2*3^3*5*11*19*23*29*31 with v2=2,

v3=3, v5=1, v11=1, v19=23, v23=1, v29=1, v31=1.

Now we also know that:

2 2 !

! !

n n

n n n

(b)

Then, by Lemma 6, we have:

Lemma 7 – For each n,

1

22p j j

j

n nv

p p

(c)

The (c) comes by (a), where instead of n there is 2n

n

, and by (b). In the (c)

the differences are 0 if the ratio is even, 1 otherwise.

Lemma 8 – 2pvp n

Proof

If t is the maximum exponent for which

6

2tp n

It means that for jt+1 pj > 2n, from t+1 onwards all the exponents are zero in

the (c). On the other hand, in the(c) the terms are 0 or 1 so vp t and

2pv tp p n . This proves the Lemma 8.

The Lemma 8 suggests a technique for the evaluation of exponents: not only must

be p 2n but also that pvp 2n.

For example if n=17 then we can establish that v3 cannot be greater than 3

because 2n=34 3^3 34; consequently for each 5 < p < 34 we have that vp cannot be greater than 1.

With this technical we can work on the exponents and introduce the last Lemma.

Lemma 9 – We suppose n 3

1. If 2p n then vp 1

2. If 2n/3 < p n then vp = 0

3. If n < p 2n then vp = 1

Now we have almost all the elements to demonstrate the Theorem.

Theorem (Bertrand„s Postulate)

For each n>1 is (2 ) ( ) 0n n

Corollary

For each n1 1 2n np p

The corollary is understandable if we consider the sequence of prime numbers

3,5,7,11,13 ... If we take in fact twice a prime number, we see that the next is

less than double of that. For example, 7 is less than 2 * 5.

To demonstrate the Theorem we must use the Prime Numbers Theorem (PNT) and see

the upper and lower boundaries. On these types of reasoning Chebiscev was just a

true master.

Lemma 10 – For each n>1 is:

7(2 ) ( )

3log(2 ) 5log( )

n nn n

n n

Now we go back at the binomial coefficient, break it in two products:

2 2

2p p pv v v

p n p n n p n

np p p

n

(d)

We use Lemma 9, 3th point, the exponent values 1, so the (d) is:

2

2pv

p n n p n

np p

n

(e)

7

As 2nd point by Lemma 9, if 2n/3 < p n then vp = 0 so the (e) is:

2 /3 2

2pv

n n

p n n p n

np p Q P

n

(†)

At this point we want to find an upper boundary for Qn. We can break Qn in two

products to use the Lemma 9: the product of prime numbers for p 2n/3 where vp

= 1 and the product of prime numbers p 2n/3 where vp > 1.

For 1st point by Lemma 9 if vp > 1 then 2p n so:

2 2

2 pv

n p n p n

p n p p

Then:

2 /3 2 /3 2 /3 2 /3 2

p p pv v v

n

p n p n p n p n p n

Q p p p p p

(††)

From this we must do some considerations:

1. If we use Lemma 4 is 2 /3

2 /3

4 n

p n

p

.

2. For the second product 2pvp n . So

2

(2 )pv

p n

p n

where is the number of

prime numbers 2p n .

3. The prime numbers are odd and the number of prime numbers less than Y is

certainly less than Y/2 so we can write that: 2 / 2n and that:

2 / 2

2

(2 )pv n

p n

p n

.

If we use all the considerations, we obtain by (††):

2 /3 2 / 24 (2 )n n

nQ n

1 2 /3 2 / 24 (2 )n n

nQ n

If we go back at (†) and we remember the Lemma 5, we obtain:

1 2 /3 2 / 22 2 2

4 (2 )n n

n n n n

n n nQ P P Q n

n n n

so:

1

2 /3 2 / 2 /3 2 / 244 (2 ) 4 2 (2 )

2

nn n n n

nP n n nn

If we say:

8

1( ) /3 2 / 22 4 2 (2 )

n n nn n n

so:

1/3 2 / 2log 4 2 (2 )

( )log 2

n nn n

nn

(*)

If we remember the (†)

2

n

n p n

P p

(+)

So we talk of the prime numbers between n and 2n, referenced from Betrand‟s

Postulate and we know by PNT that they are exactly (2 ) ( )n n , so:

(2 ) ( )

2 2n n

nP n n

(2 ) ( ) ( ) (2 ) ( )

2 2 2 2n n n n n

n nP n n n P n

then

( ) (2 ) ( )n n n (++)

If we go back to (*) we say that:

( ) log 2 / 3log(4) log(2 ) 2 / 2 log 2n n n n n n

3log 2

( ) log 2 / 3 log(4) 3/ log(2 )2

nn n n n n

n

(**)

where 3log 2

( ) log(4) 3/ log(2 )2

xg x x x

x

If we consider lim ( ) log(4) 1.38629...x

g x

because the log(x) tends to infinite

slower than xt and:

log(2 )lim 0

log 2lim 0

2

x

x

x

x

x

x

So it exists an x0 which g(x)>1 and exists a n0 which part in brackets of (**)

is > 1. We also see that g(2190)0.999995 and g(2191) 1.00006. So by (**):

for n > 2190

( )3log 2

nn

n

By (++) is

(2 ) ( )3log 2

nn n

n .

If we go back to (+) and if we use log, it is: 2

log( ) logn

n p n

P p

.

But each p of the summation is p>n so log p > log n, then it is:

9

2

log( ) log (2 ) ( ) logn

n p n

P p n n n

It is also:

2n

nP

n

and by Lemma 1:

2

log log log(4 ) log(4) (2 ) ( ) log log(4)n

n

nP n n n n n

n

So:

log(4)(2 ) ( )

log

nn n

n

Now 7/5=1.4 and log(4)1.38629… so 7

(2 ) ( )5log

nn n

n

This is the demonstration of the Lemma 10, but it is also the demonstration of

Betrand‟s Postulate, because (2 ) ( ) 0n n , in fact

7

(2 ) ( )3log 2 5log

n nn n

n n .

For example n=5 2n=10, (10) (5) 2 so 5/3*(1/log(10))=0.7238… and

7*5/5*(1/log(5))=4.3493…

PART TWO

Generalized Betrand‟s Postulate and Legendre's conjecture

Now let's see if we can understand more if k = n in the Generalized Betrand‟s

Postulate and see if we can find the upper and lower boundaries such that we can

say:

2 21, ( ) ( ) 0n n n n

If it is true, then we will have:

Corollary

2 2 2 2 2 21, ( ) ( ) (( 1) ) ( ) (( 1) ) ( ) 0n n n n n n n n So it will be true the Legendre‟s Conjecture.

Lemma 11 – 2 2

2

24 2n n

p n

p

Proof

As Lemma 4.

10

Lemma 12 – ( 1)vpp n n

Proof

It‟s a consequence of the Lemma 8, generalized to n(n+1).

Lemma 13 – per n>1, ( 1)

2

n(n+1)2

n

n n

Proof

With m =(n+1)n is:

( 1)( 1)

2 20 0

( 1) ( 1)( 1)2 (1 1) 2

n nmm m n n

h h

n n n nm n n

h h n n

They are true also Lemma 3, Lemma 4 and Lemma 6, means that for

n2 < p n(n+1), p is a divisor of n(n+1)! Consequently 2

( 1)|pv n n

pn

.

So we can break the binomial coefficient in the product of prime numbers in this

mode:

2 22

( 1) ( 1)

( 1)p p pv v v

p n n p n n p n n

n np p p

n

(***)

Lemma 14 – If n > 1 then è 2

( 1)2n

n n

n

Proof

By Lemma 13, if we divide by 2n2, we have:

2( 1)

2

n(n+1)2 2 2

n

n n n n

.

Example

n=3

(n+1)n=12

n2=9

binomial(12,9)=220

2n=23=8

Lemma 15 - If n>1,

12

1

1

( )!

n

ip j j

j

n n in

vp p

Proof

By definition

2

1

( 1)

p jj

n n

nv

p

; in fact it is:

11

12

2 2 2 2 2

1

2 22

( )( 1) ( )! ! ( )( 1)...( ( 1))

! ! ! ! !

n

i

n n in n n n n n n n n n n n

n n n n nn

The Lemma 8 here also suggests a technique for the evaluation of exponents: not

only must be p n(n+1) but also that pvp n(n+1).

If n=3 yhen we can say that v3 cannot be greater than 2 because n(n+1)=12

3^2 12; consequently for each 9 < p 12 we have also that vp cannot be

greater then 2.

With this, we have the last Lemma.

Lemma 16 – If n 3

1. If ( 1)p n n then vp 1

2. If n2 < p n(n+1) then vp = 1

Example

n=5

n2=25

n(n+1)=30

radice = sqrt(n(n+1))= 5.477225575051661134569697828 p=7

n(n+1)/3=10

by 1) v7=1 otherwise 72>30

by 2) 25 < 29 < 30 v29=1 otherwise 292>30

By (***) we have:

2 22

( 1) ( 1)

( 1)p p pv v v

p n n p n n p n n

n np p p

n

By Lemma 15 3-th point, we can write:

2 2 ( 1)

pv

n n

p n n p n n

p p Q P

Where:

2

pv

n

p n

Q p

2 ( 1)

n

n p n n

P p

(^)

We see now if Qn has got an upper boundary.

If:

2 ( 1)

( 1) pv

p n np n

p n n p p

Some considerations are:

12

1. by Lemma 11 2 ( 1)

( 1)

2 n n

p n n

p

2. by Lemma 12 ( 1)vpp n n so it is ( 1)

( ( 1))pv

p n n

p n n

where is the

number of prime numbers ( 1)p n n .

3. The prime numbers are odd and the number of prime numbers less than Y

is certainly less than Y/2 so: ( 1) / 2n n

and( 1) / 2

( 1)

( ( 1))pv n n

p n n

p n n

.

Then:

2 2 2 2 ( 1)

p p pv v v

n

p n np n p n p n p n

Q p p p p p

2 ( 1) / 222 ( 1)

n nn

nQ n n

2 ( 1) / 21 22 ( 1)

n nnQn n n

By (***) we have:

2 22

( 1)

( 1)pv

n n

p n n p n n

n np p Q P

n

2 ( 1) / 21 2

2 2

( 1) ( 1)2 ( 1)

n nn

n n

n n n nP Q n n

n n

2

( 1)(1 )(1 )( 1) / 2

( 1)( 1)

2 ( 1)22 2 ( 1)

( 1)( 1)

n nn nn nn nn n

n n nn n

n nP n n

n nn n

If we write:

( 1)(1 )

( )

( 1)

2 ( 1)( ( 1))

( 1)

n nn n

n

n n

n nn n

n n

(=)

So we have

Lemma 17 – For each n>1 is:

2 7 ( 1)0 ( ( 1)) ( )

5log( ( 1))

n nn n n

n n

Proof

13

By (=) and if we use the log, we obtain:

( 1)(1 )

( 1)

2 ( 1)( ) log ( 1) log

( 1)

n nn n

n n

n nn n n

n n

For n the square bracket goes to zero because the numerator goes to the

infinite lower.

Now it is:

( ) 0n

otherwise the (=) is 1 for the rules of powers: a finite number x with exponent

zero is x^0 = 1. It is a absurd.

By (^) is:

2 ( 1)

log( ) logn

n p n n

P p

.

Each p in the summation is p>n(n+1) so log p > log n(n+1), then:

2

2

( 1)

log( ) log ( ( 1)) ( ) log ( 1)n

n p n n

P p n n n n n

We also see that:

2

( 1)n

n nP

n

so:

( 1) 2

2

( 1)log( ) log log(4 ) ( 1) log 4 ( ( 1)) ( ) log ( 1) ( 1) log 4n n

n

n nP n n n n n n n n n

n

Now 7/5=1.4 and log(4)1.38629… so:

2 ( 1) log 4 7 ( 1)0 ( ( 1)) ( )

log ( 1) 5log ( 1)

n n n nn n n

n n n n

This proves Lemma 17. At this point it is automatically shown that generalized

Bertrand's postulate is true and his Corollary:

Corollary of the Legendre's conjecture

2 2 2 2 2 21, ( ) ( ) (( 1) ) ( ) (( 1) ) ( ) 0n n n n n n n n

So it‟s true the Legendre's conjecture!

14

References:

[1] http://it.wikipedia.org/wiki/Dimostrazione_del_postulato_di_Bertrand

[2] M. El Bachraoui. Primes in the Interval [2n,3n]. International Journal of

Contemporary Mathematical Sciences, 1(3):617–621, 2006.

[3] La congettura di Legendre – R. Turco, M. Colonnese et al.

[4] Il postulato di Bertrand e i primi di Ramanujan - Umberto Cerruti

[5] Towards Proving Legendre‟s Conjecture – Shiva Kintali

CNR SOLAR

http://150.146.3.132/

Prof. Matthew R. Watkins

http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/tutorial.htm

ing. Rosario Turco

www.scribd.com/rosario_turco

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