The Amazing Colors of Pascal’s Triangle - Discrete · PDF filePresented by Robert Hochberg, Discrete Teaching — Feel free to copy for classroom use VCTM 2002, Pascal — TSP #1

Presented by Robert Hochberg, Discrete Teaching — Feel free to copy for classroom use VCTM 2002, Pascal — TSP #1

The Amazing Colors

of

Pascal’s Triangle

Rob HochbergDiscrete Teaching


Pascal’s Triangle

From “Visual Patterns in Pascal’s Triangle” by DaleSeymour Publications

(an excellent resource for Pascal’s Triangle patterns)


Some Patterns in Pascal’s Triangle

C Each number is the sum of the two numbers above it

C The outside numbers are all 1

C The triangle is symmetric

C The first diagonal shows the counting numbers

C The sums of the rows give the powers of 2

C Each row gives the digits of the powers of 11.

C Each entry is an appropriate “choose number.”

C And those are the “binomial coeffecients.”

C The Fibonacci numbers are in there along diagonals.


The Choose Numbers are the BinomialCoefficients

Consider

(a + b)5

=

(a + b)·(a + b)·(a + b)·(a + b)·(a + b)

In the first expression, the coefficient of a2b3, forexample, is what we call a binomial coefficient.

To compute the coefficient of a2b3 in the secondexpression, we select a term from each of the sevenfactors, making sure that exactly 3 of them are “b”. There are ways to do that.5

3


Fibonacci Numbers in Pascal’s Triangle


The Arithmetic of Pascal’s Triangle


A Hidden Pattern in Pascal’s Triangle

If we color all the entries which are divisible by 2,black, and all the entries which are not divisible by 2,blue, then we get this Sierpinski-like coloring.

This is called the “Mod-2" coloring


Mod-2 Coloring — First 128 Rows


Coloring Mod 3 — Divisibility Only


Coloring Mod 3 — With Remainder

Happy families are all alike; every unhappyfamily is unhappy in its own way.

— Leo TolstoyAnna Karenina


An Easier Way to Color Modularly

To find the remainder when you make a sum,

you can just find the remainders of the things you’readding...and add them!

+ mod 3 0 1 20 0 1 21 1 2 02 2 0 1

For example: If you add the entries “5” and “10,” youget the sum “15.” The remainder of 15, mod 3, is 0.

Alternately, you could look at the remainders of “5” and“10,” namely “2” and “1” respectively, and add them. 2 + 1 = 3, and the remainder of “3” mod 3 is 0.

So we get the same answer two different ways!


Just Add Colors

Since each remainder was assigned a color

+ mod 3 0 1 20 0 1 21 1 2 02 2 0 1

we can forget about the numbers and remainders, andjust “add” the colors themselves!

“+” mod 3


Adding Just Colors

“+” mod 3


First 243 Rows with Remainder Mod 3










Question 1

When will we get self-similar repetition in the pattern ofmultiples (mod n)?

Only if n is prime!

Now we prove that if p is prime, then the multiples of pin Pascal’s triangle form a self-similar repeating pattern.

But first...

We ask the question: How do you quickly find the value of

nk

p

mod


A Very Nifty Trick

Let’s compute 12690

7

mod

42921556603703815855529053166825

Write 126 and 90, in base 7:

Treat the columns as binomial coefficients, and multiplythem together!

Let’s go to the computer and see if this makes sense...

343 49 7 112690

Note: whenever k > n.nk

= 0


How Many Non-Multiples in the nth Row,(mod p)?

Let’s find the number of non-multiples of 7 in row 126:

Write 126 base 7:

We need the digits of k written base 7 to be less than orequal to the corresponding digits of 126, written base 7.

In general, if bn...b3b2b1 is the base-p representation of n,then the number of colored (non-multiple) entries in thenth row is

(bn + 1)· þ ·(b3 + 1)·(b2 + 1)·(b1 + 1)


Proof that the Nifty Trick Works

First we need to show that every entry of the pr-th rowis a multiple of p, except for the first and last entries.

That is, that is a multiple of p, for k …0, pr.pk

r

Here we prove it for , with a picture:22

3

Pres

ente

d by

Rob

ert H

ochb

erg,

Dis

cret

e Te

achi

ng —

Fee

l fre

e to

cop

y fo

r cla

ssro

om u

se V

CTM

200

2, P

asca

l — T

SP #

23

Pres

ente

d by

Rob

ert H

ochb

erg,

Dis

cret

e Te

achi

ng —

Fee

l fre

e to

cop

y fo

r cla

ssro

om u

se V

CTM

200

2, P

asca

l — T

SP #

24


Proof of the Nifty Trick

Consider again:12690

7

mod

This is the coefficient of x90 in the expansion of (1 + x)126.

126 = 2×49 + 4×7 + 0×1 (the base-7 representation)

so:

(1 + x)126 = (1 + x)49@(1 + x)49@(1 + x)7@(1 + x)7@(1 + x)7@(1 + x)7

= (1 + stuff + x49)@(1 + stuff + x49) @(1 + stuff + x7)@(1 + stuff + x7)@(1 + stuff + x7) @(1 + stuff + x7)

0 0 0 0 0 049 49 7 7 7 7


Self-Similarity of Multiples of p

Now we can prove that the pattern of multiples of aprime p in Pascal’s Triangle repeats itself on ever-increasing scales.

Let’s consider the first 27 rows (mod 3):

Row 14: 14 base 3 = 1 1 2

Row 23: 23 base 3 = 2 1 2

(Maybe check out 27 rows mod 5)


An Unexpected Discovery

Suppose we change the rule for generating Pascal’sTriangle so that, instead of each term being the sum of

(left term) + (right term)

it is the sum of

2×(left term) + 3×(right term)

This gives a very different triangle full of numbers:

But what about divisibility properties?


Explanation of Unexpected Discovery

Once again, the binomial interpretation comes to ouraid:

The numbers in the triangle above result fromexpanding the binomial:

(3x + 2y)k

13x + 2y

9x2 + 12xy + 4y2

27x3 + 54x2y + 36xy2 + 8y3

( )3 2 3 2x ynk

x yn k n k k n k+ =

− −∑

Since the powers of 3 and 2 do not affect divisibility byany prime other than 2 or 3, we will always get exactlythe same divisibility conditions.


A Template for Coloring Larger Sectionsof Pascal’s Triangle

Presented by Robert Hochberg, Discrete Teaching — Feel free to copy for classroom use VCTM 2002 Pascal page 1

The Amazing Colors

of

Pascal’s Triangle

VCTM 2002Manassas, Virginia

Rob HochbergDiscrete Teaching

All slides and handouts, as well as thisresource pack, can be obtained online at:

www.discreteteaching.com Username: VCTM2002 Password: manassas


Generate Pascal’s TriangleEach entry is the sum of the two numbers directly above, and all the leftmost and rightmost entries are "1."


Some Patterns in Pascal’s TriangleC Each number is the sum of the two numbers above itC The outside numbers are all 1C The triangle is symmetricC The first diagonal shows the counting numbersC The sums of the rows give the powers of 2C Each row gives the digits of the powers of 11.C Each entry is an appropriate “choose number.”C And those are the “binomial coeffecients.”C The Fibonacci numbers are in there along diagonals.

The Arithmetic of Pascal’s TriangleC The “hockey stick” gives a quick way to add the terms in a diagonal, starting from an edge of the

triangle.C The “funnel” gives a way of adding a rectangle of numbers, where the rectangle extends to both

edges of the triangle, and contains the “1” at the top. The sum is given by one less than the term inthe circle.

C The star of David indicates that the product of the three terms in one triangle equals the product ofthe terms in the other triangle.



Black = EvenGray = Odd


An Easier Way to Color Modularly

To find the remainder when you make a sum,

you can just find the remainders of the things you’readding...and add them!

+ mod 3 0 1 20 0 1 21 1 2 02 2 0 1

For example: If you add the entries “5” and “10,” youget the sum “15.” The remainder of 15, when dividedby 3 (mod 3, as we say), is 0.

Alternately, you could look at the remainders of “5” and“10,” namely “2” and “1” respectively, and add them. 2 + 1 = 3, and the remainder of “3” mod 3 is 0.

So we get the same answer two different ways!


Just Add Colors

Since each remainder was assigned a color

+ mod 3 0 1 20 0 1 21 1 2 02 2 0 1

we can forget about the numbers and remainders,and just “add” the colors themselves!

“+” mod 3



Black = EvenGray = Odd


A Mod 4 Coloring of 256 Rows

Black = Multiple of 4White = Even, but not a multiple of 4Gray = Odd


The Sierpinski Triangle


Investigations

What are some questions you can ask about thevarious stages of the Sierpinski Triangle?

C How many shaded triangles are in each figure?

C What is the shaded area of each figure?

C What is the total perimeter of each figure?

C What will the "end stage" look like?


The Choose Numbers are the Binomial CoefficientsConsider (a + b)5 = (a + b)·(a + b)·(a + b)·(a + b)·(a + b). In the first expression, the coefficient of a2b3, forexample, is what we call a binomial coefficient. To compute the coefficient of a2b3 in the secondexpression, we select a term from each of the seven factors, making sure that exactly 3 of them are “b”.

There are ways to do that.53

Question 1When will we get self-similar repetition in the pattern of multiples (mod n)? Only if n is prime!Now we prove that if p is prime, then the multiples of p in Pascal’s triangle form a self-similar repeating

pattern. But first... We ask the question: How do you quickly find the value of ?nk

p

mod

A Very Nifty Trick

Let’s compute . (This is 42921556603703815855529053166825), by the way:12690

7

mod

Write 126 and 90, in base 7:

Treat the columns as binomial coefficients, and multiply them together!

How Many Non-Multiples in the nth Row, (mod p)?Let’s find the number of non-multiples of 7 in row 126:Write 126 base 7:

We need the digits of k written base 7 to be less than or equal to the corresponding digits of 126, writtenbase 7. In general, if bn...b3b2b1 is the base-p representation of n, then the number of colored (non-multiple)entries in the nth row is

(bn + 1)· þ ·(b3 + 1)·(b2 + 1)·(b1 + 1)

Proof that the Nifty Trick WorksFirst we need to show that every entry of the pr-th row is a multiple of p, except for the first and last entries.

That is, that is a multiple of p, for k …0, pr.pk

r

343 49 7 1

126

90

Note: whenever k > n.nk

= 0


Here we prove it for , with a picture:22

3

The picture above shows all 28 ways to select 2 of the 8 vertices of the octagon. Since they come in groupsof either 8 or 4, when grouped by rotational equivalence, the resulting number is definitely a multiple of 2.

Here is a template for your own perusal:


Proof of the Nifty Trick

Consider again:12690

7

mod

This is the coefficient of x90 in the expansion of (1 + x)126.

126 = 2×49 + 4×7 + 0×1 (the base-7 representation)

so: (1 + x)126 = (1 + x)49@(1 + x)49@(1 + x)7@(1 + x)7@(1 + x)7@(1 + x)7

= (1 + stuff + x49)@(1 + stuff + x49)@(1 + stuff + x7)@(1 + stuff + x7)@(1 + stuff + x7)@(1 + stuff + x7)

0 0 0 0 0 0

49 49 7 7 7 7Most of the ways to obtain a product of x90 involve taking a term from some “stuff,” which will be amultiple of 7. The only way not to obtain a multiple of 7 would be to use only the first and last terms fromeach factor. But this is like expressing 90 as the sum of powers of 7.

Self-Similarity of Multiples of pNow we can prove that the pattern of multiples of a prime p inPascal’s Triangle repeats itself on ever-increasing scales.Let’s consider the first 27 rows (mod 3):Row 14: 14 base 3 = 1 1 2Row 23: 23 base 3 = 2 1 2(Maybe check out 27 rows mod 5)

An Unexpected Discovery

Suppose we change the rule for generating Pascal’s Triangle sothat, instead of each term being the sum of the left and rightterms above it is the sum of 2×(left term) + 3×(right term)This gives a very different triangle full of numbers:

But what about divisibility properties?

Explanation of Unexpected DiscoveryOnce again, the binomial interpretation comes to our aid:The numbers in the triangle above result from expanding the binomial (3x + 2y)k :1, 3x + 2y, 9x2 + 12xy + 4y2, 27x3 + 54x2y + 36xy2 + 8y3

( )3 2 3 2x ynk

x yn k n k k n k+ =

− −∑

Since the powers of 3 and 2 do not affect divisibility by any prime other than 2 or 3, we will always getexactly the same divisibility conditions. So the pattern o f number divisible by 5, for example, will beexactly the same as for the original triangle.


A Template for Coloring Larger Sectionsof Pascal’s Triangle

Documents

The Amazing Colors of Pascal’s Triangle - Discrete · PDF filePresented by Robert Hochberg, Discrete Teaching — Feel free to copy for classroom use VCTM 2002, Pascal — TSP #1