Pascal Triangle

Olier Raby

2005-09-05

Contents

1 Introduction 2

2 Development 3

2.1 Algebra Usage . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 Building the Triangle one Row at a Time . . . . . . . . . . . 62.4 Building the Triangle one Column at a Time . . . . . . . . . 72.5 Newton Binomial Series . . . . . . . . . . . . . . . . . . . . . 82.6 Computing a Square Root . . . . . . . . . . . . . . . . . . . . 132.7 Computing Natural Logarithm . . . . . . . . . . . . . . . . . . 152.8 Pascal vs Newton . . . . . . . . . . . . . . . . . . . . . . . . . 172.9 Graphics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.10 Prime Paving . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.11 Curiosities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.12 Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.13 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Conclusion 29

4 Recommended Readings 30

5 Copyright 32

1

Chapter 1

Introduction

I started to write this text to document the properties of Pascal pyramids.During my journey, I discovered lots of fascinating properties in Pascal tri-angle. I decided to document it in one place, since I did not find anythingvery interesting at one place on the Web.

In this text, I explore different properties of Pascal triangle, and one ofits generalizations, Newton binomial coefficients.

2

Chapter 2

Development

Take note that he was not the first to discover it, a Chinese mathematiciandid long before him. However, he was the first to list many of its properties.

2.1 Algebra Usage

You can use Pascal triangle to compute the coefficients of, say,

(x + y)5 = 1 × a5 + 5 × a4b + 10 × a3b2 + 10 × a2b3 + · · ·

The coefficients 1, 5, 10, 10,... appear within Pascal triangle:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

...

As you probably know, each term in the triangle is the sum of the twonumbers above it. For instance, 6 = 3 + 3.

Those coefficients are important in algebra, but they have many otherusages, like computing probabilities, calculating the Fibonacci terms, andcreating the Sierpinski sieve. The Newton’s generalizations of Pascal triangle

3

allows to compute the square root of a number, and to determine the naturallogarithm of a number.

Pascal found usage of the triangle within the probabilities theory. I spareyou the details.

2.2 Properties

Before discussing supplementary properties, allow me to present it in thisform:

11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 11 6 15 20 15 6 11 7 21 35 35 21 7 1... ... ... ... ... ... ... ...

Each term is obtained by adding the term right above it and the termabove it, at the left. For instance, 3 + 3 = 6.

This is what I call the “addition rule”, where the numbers 1 in the firstcolumn are not subject.

Let me again display the preceding triangle, but this time giving an orderto each row.

0 : 11 : 1 12 : 1 2 13 : 1 3 3 14 : 1 4 6 4 15 : 1 5 10 10 5 16 : 1 6 15 20 15 6 17 : 1 7 21 35 35 21 7 1... ... ... ... ... ... ... ... ...

Starting at row zero eases different computations later.

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The sum of the terms in any row is 2r, with r the row order: 0, 1, 2,...For instance, the sum of row 4 is

1 + 4 + 6 + 4 + 1 = 16 = 24

Since there is a formula for computing the sum of the rows, maybe thereis one for the columns? This is the case. Before displaying it, let us orderthe columns to help understand what we are counting.

0 1 2 3 4 5 6 70 : 11 : 1 12 : 1 2 13 : 1 3 3 14 : 1 4 6 4 15 : 1 5 10 10 5 16 : 1 6 15 20 15 6 17 : 1 7 21 35 35 21 7 1... ... ... ... ... ... ... ... ...

Again, starting at column zero eases different computations later.The sum of the terms in column 0 is trivial:

(r + 1)

If you remember some formulas from high school, you know that the sumof column 1 is

(r + 1) r

2The sum of column 2 is a little harder to get. It is

(r + 1) r (r − 1)

6

Comparing the results for columns 1 and 2, you may infer the sum ofcolumn 3. It is

(r + 1) r (r − 1) (r − 2)

24If you are familiar with factorials, then you see where this is heading.

The sum of terms in column c ending at row r is

(r + 1) × r × · · · × (r − c + 1)

(c + 1)!(2.1)

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You can find the sum with another method. Suppose we are interestedin adding terms in column 3, the fourth column, with the last row being 6.There are four terms at the numerator, starting with (r + 1) = 6 + 1 = 7,and the numerator is (c + 1)! = (3 + 1) = 4 × 3 × 2 × 1. The terms atnumerator and denominator are decreasing. In short, fourth column, fourterms at numerator, four terms at denominator, all decreasing.

We added the terms within the rows and the columns. It is possible toadd along the diagonals. Let us pick the values along the diagonals from leftto right.

1 = 11 = 11 + 1 = 21 + 2 = 31 + 3 + 1 = 51 + 4 + 3 = 81 + 5 + 6 + 1 = 131 + 6 + 10 + 4 + 1 = 21... = ...

The numbers on the right form the famous Fibonacci series.

2.3 Building the Triangle one Row at a Time

As you know, we can build each row of Pascal triangle by adding the termsfrom the preceding row, two at a time. How about building any row withoutknowing the previous one?

We know that the first term in any row is 1. Here are some computationswith row 4 that will help you figure out the solution.

4 : 1 4 6 4 1

1 × 4 − 0

1 + 0= 4

4 × 4 − 1

1 + 1= 6

6 × 4 − 2

1 + 2= 4

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4 × 4 − 3

1 + 3= 1

1 × 4 − 4

1 + 4= 0

· · ·

This behavior respects what I call the “row rule”, since it applies alongthe row elements. If you know the row order and the column order of a term,then you can compute any neighbour.

(

r

c + 1

)

=

(

r

c

)

× r − c

1 + cwith r ≥ 0, c ≥ 0 (2.2)

Equally, we can compute any term in any row with

(

r

c

)

=r!

c!(r − c)!with r ≥ 0, c ≥ 0

You remember that the columns are ordered 0, 1, 2... You can “blame”this formula, since it only works if the column order starts at zero.

2.4 Building the Triangle one Column at a

Time

As you remember, we can build columns 0, 1, 2,... of Pascal triangle by usingformula 2.1 seen before. How about building any column without knowingthat formula?

We know that the first term in any column is 1. Here are some compu-tations with column 4 that will help you figure out the solution.

5 : 1 5 15 35 · · ·

1 × 4 + 1

4 + 1 − 4= 5

5 × 5 + 1

5 + 1 − 4= 15

15 × 6 + 1

6 + 1 − 4= 35

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35 × 7 + 1

7 + 1 − 4= 70

· · ·This behavior respects what I call the “column rule”. If you know the

column and the row order of a term, then you can compute any neighbourvertically.

Mathematically, this formula is(

r + 1

c

)

=r + 1

r + 1 − c

(

r

c

)

with r ≥ 0, c ≥ 0 (2.3)

Equally, we can compute any term in any row with(

r

c

)

=r!

c!(r − c)!with r ≥ 0, c ≥ 0

You remember that the rows are ordered 0, 1, 2... You can “blame” thisformula, since it only works if the row order starts at zero.

2.5 Newton Binomial Series

Newton, the famous physicist, studied the properties of the Pascal triangle.He found two generalizations for it. Below, I study those two.

He determined that Pascal triangle extends along the negative axis.To better understand how he did this, let us rewrite the triangle, com-

pleting the rows with zeroes.

0 : 1 0 0 0 0 0 0 0 ...1 : 1 1 0 0 0 0 0 0 ...2 : 1 2 1 0 0 0 0 0 ...3 : 1 3 3 1 0 0 0 0 ...4 : 1 4 6 4 1 0 0 0 ...5 : 1 5 10 10 5 1 0 0 ...6 : 1 6 15 20 15 6 1 0 ...7 : 1 7 21 35 35 21 7 1 ...... ... ... ... ... ... ... ... ... ...

In this form, the Pascal triangle is called the figurate number triangleJakob Bernoulli is the father of this mathematical object. In the followingtext, I will call it the Bernoulli triangle.

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This array of numbers still respects the addition, the column and the rowrules.

Using the row rule, we are ready to compute the terms in row -1.

−1 : 1 −1 1 −1 1 −1 1 −1 ...0 : 1 0 0 0 0 0 0 0 ...1 : 1 1 0 0 0 0 0 0 ...2 : 1 2 1 0 0 0 0 0 ...3 : 1 3 3 1 0 0 0 0 ...4 : 1 4 6 4 1 0 0 0 ...5 : 1 5 10 10 5 1 0 0 ...6 : 1 6 15 20 15 6 1 0 ...7 : 1 7 21 35 35 21 7 1 ...... ... ... ... ... ... ... ... ... ...

At first, it is strange to see all those 1 and −1. They seem out of contextsince they do not form a triangle.

However, each term in this row respects the addition, the column and therow rules. Examples will convince you this is the case.

In row -1 , select the terms at columns 2 and 3. Add them.

1 + −1 = 0

Compute the row -1 terms with the row rule.

−1 : 1 −1 1 −1 1 −1 1 ...

1 × −1 − 0

1 + 0= −1

−1 × −1 − 1

1 + 1= 1

1 × −1 − 2

1 + 2= −1

−1 × −1 − 3

1 + 3= 1

1 × −1 − 4

1 + 4= −1

· · ·

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Compute the column 2 terms with the column rule.

2 : 1 0 0 1 3 6 10 . . .

1 × −1 + 1

−1 + 1 − 2= 0

0 × 0 + 1

0 + 1 − 2= 0

0 × 1 + 1

1 + 1 − 2= 1

(Since 0 × 2/0 is indeterminate, I set the result to 1 ro respect the columnrule. This is consequent with the meaning of indeterminate. In other words,until supplementary information is available, any value fits the bill.)

1 × 2 + 1

2 + 1 − 2= 3

3 × 3 + 1

3 + 1 − 2= 6

6 × 4 + 1

4 + 1 − 2= 10

· · ·Remarkable, don’t you think?

We are ready to stretch the triangle further along the negative axis.

... ... ... ... ... ... ... ... ...−4 : 1 −4 10 −20 35 −56 84 ...−3 : 1 −3 6 −10 15 −21 28 ...−2 : 1 −2 3 −4 5 −6 7 ...−1 : 1 −1 1 −1 1 −1 1 ...

0 : 1 0 0 0 0 0 0 ...1 : 1 1 0 0 0 0 0 ...2 : 1 2 1 0 0 0 0 ...3 : 1 3 3 1 0 0 0 ...4 : 1 4 6 4 1 0 0 ...5 : 1 5 10 10 5 1 0 ...... ... ... ... ... ... ... ... ...

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As you know, Newton did not stop there. He asked himself if it waspossible to compute a row with order 1

2!

And the answer is yes.Just using one of the three rules we have seen before, can you compute

the terms of such row? Take 1 minute to think about this.

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Let me again take the example where I used the row rule to compute theterms in the fifth row.

4 : 1 4 6 4 1

1 × 4 − 0

1 + 0= 4

4 × 4 − 1

1 + 1= 6

6 × 4 − 2

1 + 2= 4

· · ·What is the first term in the row 1

2? By definition, it is 1. What are the

next terms? Let the row rule computes them!

1 ×12− 0

1 + 0=

1

2

1

2×

12− 1

1 + 1= −1

8

−1

8×

12− 2

1 + 2=

1

16

1

16×

12− 3

1 + 3= − 5

128· · ·

As you can see, no term in this row is going to be zero. The valuesoscillate around zero, once positive, once negative, the absolute value beingsmaller and smaller.

Does the addition rule work? We will use row 32

to test this hypothesis.

1 ×32− 0

1 + 0=

3

2

3

2×

32− 1

1 + 1=

3

8

3

8×

32− 2

1 + 2= − 1

16

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− 1

16×

32− 3

1 + 3= − 3

128· · ·

We get this augmented Bernoulli triangle.

... ... ... ... ... ... ... ... ...−4 : 1 −4 10 −20 35 −56 84 ...−3 : 1 −3 6 −10 15 −21 28 ...−2 : 1 −2 3 −4 5 −6 7 ...−1 : 1 −1 1 −1 1 −1 1 ...

0 : 1 0 0 0 0 0 0 ...12

: 1 12

−18

116

− 5128

7256

− 211024

...1 : 1 1 0 0 0 0 0 ...32

: 1 32

38

− 116

3128

− 3256

71024

...2 : 1 2 1 0 0 0 0 ...3 : 1 3 3 1 0 0 0 ...... ... ... ... ... ... ... ... ...

A quick head calculation confirms that the addition rule works, as longas the difference between row orders is 1.

Remarkable, don’t you think?

The formula 2.3 (column rule) is equally valid for fractional rows.Let me use rows 1

2and 3

2, columns 3 and 4, to show this.

1

16×

12

+ 112

+ 1 − 3= − 1

16

− 5

128×

12

+ 112

+ 1 − 4=

3

128

Remarkable, don’t you think?

2.6 Computing a Square Root

We know this equation:

(x + y)2 = x2 + 2xy + y2

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Suppose that we replace x with 2 and y with 3, we get

(2 + 3)2 = 22 + 2 × 2 × 3 + 32

This is equal to 25, or (2 + 3)2 = 52. Nothing remarkable, yet.Suppose we replace the power, an integer, with the fractional power 1

2.

We get the square when using 2 as a power. If we use 12

as power, we getthe square root. This is one of the power laws. Thus, we can compute thesquare root of a number with the triangle coefficients.

This observation is not the end of the story. If the power is a fraction,then x SHOULD be equal to 1 and y MUST be smaller than or equal to 1.More precisely, |y| < 1. This has to do with series convergence, which I donot detail here.

In practice, the formula rarely applies directly. We have to do somealgebraic operations before using it.

(2 + 3)1

2 = (4 + 1)1

2 = (4 × (1 +1

4))

1

2 = 41

2 × (1 +1

4)

1

2 = 2 (1 + 0.25)1

2

Within the parenthesis, x is equal to 1, and |y| is lower than 1. Oh! Iforgot to tell you what I did!

Compute the lowest and closest integer square to (2+3). It is 4. Changethe sum in parenthesis in order to display the square. Extract the square fromthe parenthesis, keeping the square root operation for both terms. Extractthe square root of both terms, one being easy, the other using the binomialcoefficients.

Compare this recipe to the example before.There is only one operation to figure: How to extract the square root of

the parenthesis.Here, we use the binomial coefficients of order 1

2.

(1+0.25)1

2 = 1+1

2×0.25−1

8×0.25+

1

16×0.25− 5

128×0.25+

7

256×0.25− 21

1024×0.25 · · ·

(1 + 0.25)1

2 = 1 + 0.25 × (1

2− 1

8+

1

16− 5

128+

7

256− 21

1024· · · )

(1 + 0.25)1

2 ≈ 1.1013183593

The estimated square root of (2 + 3) is 2 ∗ 1.1013183593 = 2.202636718.The square of this value is 4.8516085. We can raise the precision by addingmore terms.

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In day to day applications, this formula is never used, since it takestoo many terms to get enough precision. The prefered way is the Newton-Raphson algorithm.

You can see why x should be 1: it eases the computation a lot. If you arenot convinced, try to compute the square root of (2+3) by applying directlythe binomial expansion (a = 2, b = 3) with row 1

2coefficients.

The method, with proper modifications, works for any fractional power:23, −4, 7,. . .

For instance, if you have to extract the cubic root of (3 + 9), do thefollowing

(3 + 9)1

3 = (8 + 4)1

3 = (8 × (1 +1

2))

1

3 = 81

3 × (1 +1

2)

1

3 = 2 (1 + 0.5)1

3

Compute the binomial coefficients of row 13. You know the rest.

2.7 Computing Natural Logarithm

We have seen this equation before

(x + y)2 = x2 + 2xy + y2

Let me replace some terms within it.

(1 + z)2 = 12 + 2 × 1 × z + z2

(1 + z)2 = 1 + 2 × z + z2

To compute the natural logarithm, we use the row -1 and perform somemodifications.

Using what we have just seen and the coefficients within row -1, we get

(1 + z)−1 = 1 − z + z2 − z3 + z4 − z5 + z6 + · · ·

If you did some calculus, you may remember those integrals:

· · ·∫

z1dz =z2

2

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∫

z0dz =z1

1∫

z−1dz =

∫

dz

z= ln z

∫

z−2dz =z−1

−1∫

z−3dz =z−2

−2· · ·

If not, then trust me that these results are valid.Now, let us get back to computing the natural logarithm. Integrating

both sides of

(1 + z)−1 = 1 − z + z2 − z3 + z4 − z5 + z6 + · · ·

we get

ln (1 + z) = z − z2

2+

z3

3− z4

4+

z5

5− z6

6+

z7

7+ · · · with z ∈ IR

This is a formula to compute natural logarithms. And Newton’s secondgeneralization gives that formula.

Remarkable, don’t you think?

In order to complete the calculation of the natural logarithm, we need atleast these logarithm laws:

ln (a) + ln (b) = ln (a × b)

ln (ab) = b × ln (a)

log10 (z) =ln (z)

ln (10)

I will not go any further on the logarithm subject.

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2.8 Pascal vs Newton

Take note that the sum of rows with negative orders is NOT 2r. The sumsdepend on rules applying to infinite series.

The sum of rows with order mn

is 2m

n . You should be able to easily derivethis formula from what we have seen before.

The sum of columns is invalid if we include rows with negative orders.However, it is easy to change the formula to account for them. We cannotinclude terms from fractional rows in the sum (why?).

We cannot compute negative Fibonacci numbers Fn using the formula wehave seen before. If we use the recursive formula Fn+2 = Fn+1 + Fn, we cancompute the terms with negative indexes, but the Newton’s generalizationof Pascal triangle does not seem able to give us the expected numbers (..., 5,-3, 2, -1, 1, 0, 1, 1, 2,...). This has to do with infinite series.

We know that(

4

2

)

=4!

2!(4 − 2)!= 6

What about(

45

)

? If you remember the Bernoulli triangle filled with zeroes,then the answer is simple: 0.

Now, what is(

−25

)

? Look at Newton’s generalization of Pascal triangle,and the answer is -6.

How about(

(1/2)5

)

? Look at Newton’s generalization of Pascal triangle,and the answer is 7

256.

Finally, what is(

2−5

)

? This is a term located in a negative column index.In this case, the answer is 0. This is because all the terms to the left ofcolumn 0 are set to zero by convention.

2.9 Graphics

If you draw an histogram of the 30th row coefficients, you get something likethe following picture.

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Suppose that you are able to compute the terms of the 1000th row inPascal triangle, would it be possible to display each of them on a graph?This is very unlikely, since it would be a very long job, unless you use aspecialized application. Using any such application, you could divide eachterm of that row by 21000, the sum of the terms within this row, then considereach computed value as a point on a continuous curve. Doing so, you wouldget the following curve.

Carl Friedrich Gauss, the famous German mathematician, studied thiscurve, named the normal curve in statistics. It is the commonest statisticallaw encountered in natural phenomenons. The properties of many statisticalcurves are compared to the normal curve properties.

In its simplest form, its equation is

P (z) = e−z2/2

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2.10 Prime Paving

Using Pascal triangle, we can create the following picture.**** ****** *** *** * * ********** *** *** * * ***** ***** * * *** ** ** *** * * * * * * ****************** *** *** * * ***** ***** * * *** ** ** *** * * * * * * ********* ********* * * *** ** ** *** * * * * * * ***** **** **** ***** * * * * * * *** ** ** ** ** ** ** *** * * * * * * * * * * * * * * ********************************** *** *** * * ***** ***** * * *** ** ** *** * * * * * * ********* ********* * * *** ** ** *** * * * * * * ***** **** **** ***** * * * * * * *** ** ** ** ** ** ** *** * * * * * * * * * * * * * * ***************** ***************** * * *** ** ** *** * * * * * * ***** **** **** ***** * * * * * * *** ** ** ** ** ** ** *** * * * * * * * * * * * * * * ********* ******** ******** ********* * * * * * * *** ** ** ** ** ** ** *** * * * * * * * * * * * * * * ***** **** **** **** **** **** **** ***** * * * * * * * * * * * * * * *** ** ** ** ** ** ** ** ** ** ** ** ** ** ** *** * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *****************************************************************

Replace each odd number with a star ’*’ and each even number witha space ’ ’. After 8 rows, you will see a pattern emerge. The picture, aSierpinski sieve, contains 64 rows.

If we extend Pascal’s triangle to infinitely many rows, and reduce thescale of our picture in half each time we double the number of rows, then theresulting design repeats itself and is called self-similar. The picture can bereproduced by taking any subtriangle and magnifying it.

Computing the odd-ness or even-ness of a number is an operation similarto applying modulo 2 (mod 2) to these numbers, 0 mapping to even and 1 toodd. This is how I manage to create Sierpinski sieve. We can apply mod 3,mod 4, mod 5,... to the same terms, replacing non-zero numbers with a star’*’ and zero with a space ’ ’. Let me show you what this gives when using11, 12 and 13.

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******************************************************************* *** ***** ******* ********* *********** ************* *************** ***************** ******************* ********************************* * *** ** ***** *** ******* **** ********* ***** *********** ****** ************* ******* *************** ******** ***************** ********* ******************* ********** *******************************************

*********************************************** ** ***** * * **************** *** *** ******* ************* ************************* ** ***** ** *********************** *** ******* **** ******* * ***** * ********* ** ** ************* *** *** ******************************** ** * ** * ** * ** *** ******** ** ******** ***************************************** *************** * * * * * * * * ********* ****** ****** ************* *** *** *** *** ********************************************* * ******** * * ** * * ******* ** * * ** ************** **** ********** * * * * * * * * ******* ** ** ** ** ********* *** *** * * * * *** *** *********** * * ******** * * ********** ** *** ** * * ** *** ** ******** ******* ** ** ******* *********** * *** * ** * * ** * *** * ******************* ***** ***** ******************************** * * ************************ ****** *** ** ** *** ****** ********* *** *** *** * *** * * *** * *** *** *** ***************************************************

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******************************************************************************************** *** ***** ******* ********* *********** ************* *************** ***************** ******************* ********************* *********************** *************************************** * *** ** ***** *** ******* **** ********* ***** *********** ****** ************* ******* *************** ******** ***************** ********* ******************* ********** ********************* *********** *********************** ************ ***************************************************

Regular tiling with the same subtriangle only happens when using oddprime numbers greater than 3: 5, 7, 11, 13... With other numbers, thepattern is less simple.

In the Bernoulli triangle, if the number in column 1 (i.e., 1, 2, 3, 4, 5,...)is a prime number, then it divides all the terms within the row. For instance,row 7 contains 1, 7, 21, 35, 35,..., where number 7 divides all these terms,while row 6 contains 1, 6, 15, 20, 15,..., where 6 cannot divide some numbers.

This property creates numerical subtriangles within the Pascal triangle.Suppose that a row is the top side of such triangle. Each term in the rowbelow is the sum of the terms from this row. Adding two multiple of, say, 7gives a multiple of 7. The row below contains a shorter sub-row of multipleof 7. Computing the terms in the next row, we again get a shorter sub-rowof multiple of 7, and so on.

2.11 Curiosities

In the Graphics section, we have seen the normal curve, a way to representthe binomial coefficients as a continuous curve. In its simplest form, it hasthe equation P (z) = e−z2/2, and the area under that curve is

√2π.

The equation of this curve involves e, while its area involves π. Howcan this be possible, since the binomial coefficients are obtained by addingnatural numbers, and we divided those coefficients by a power of 2?

Applying modulo 2 to every number in Pascal triangle, we get

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1

1 1

1 0 1

1 1 1 1

1 0 0 0 1

1 1 0 0 1 1

1 0 1 0 1 0 1

...

If we view these ones and zeroes as binary numbers, we get 1, 3, 5,15, 17, 51,... These values are the number of sides found in constructiblepolygons with odd number of sides. This is good for up to row 32. A curiouscoincidence, or an unknown link?

We can compute π from the Pascal triangle:

(

2nn

)

22n=

1√nπ

For instance,(

5025

)

250=

1√25π

outputs π = 3.17316. This formula is not used in practice since n must bevery large to compute π to any decent precision.

If you stack marbles in the simplest 1D figure (along a line), then youarrange 1, 2, 3, 4,... marbles. If you stack marbles in the simplest 2D figure(triangle), then you arrange 1, 3, 6, 10,... marbles. If you stack marbles inthe simplest 3D figure (a tetrahedron or 4-sided pyramid), then you arrange1, 4, 10, 20,... marbles. If you stack marbles in the simplest 4D figure (a4-simplex), then you arrange 1, 5, 15, 35,... marbles.

These numbers appears within the columns of the Bernoulli triangle, thecolumn 1 contains the classic and basic counting numbers: 1, 2, 3, 4,... Thecolumn 2 contains the triangular numbers: 1, 3, 6, 10, 15, 21,... The column3 contains the tetrahedral numbers: 1, 4, 10, 20, 35, 56,... The column 4contains the pentatope numbers: 1, 5, 15, 35, 70, 126,...

The Pascal triange hides the Catalan number series.

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1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

...

If we divide the underlined numbers with 1, 2, 3, 4,..., then we get 1, 1, 2,5, 14,... These are the Catalan numbers, which appear in solutions of manygeometrical problems.

There is a surprising way to get the underlined numbers. Square thenumbers in each row and add them:

12 = 1

12 + 12 = 2

12 + 22 + 12 = 6

12 + 32 + 32 + 12 = 20

12 + 42 + 62 + 42 + 12 = 70

...

There is a way to directly compute the Catalan series. Take an underlinednumber. Substract its immediate neighbour. The result is a Catalan number.

Somebody found a strange property relating geometry to the numberswithin the Pascal triangle. A dot has one vertex (1). A segment has twovertices and one line (2,1). A triangle has three vertices, three edges andone surface (3, 3, 1). A triangular pyramid has four vertices, six edges, foursurfaces and one volume (4, 6, 4, 1). And so on. Thus, the element count ofa minimal regular geometric object (build from regular shapes) in 1D, 2D,3D, 4D, etc., is specified by a row in the Pascal triangle.

Bernoulli numbers, named in honor of Jacob Bernoulli who studied themintensively, arise in a wide variety of analytical and combinatorial contexts.

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They are

B0 = 1, B1 =1

2, B2 =

1

6, B3 = B5 = B7 = ... = 0

B4 = − 1

30, B6 =

1

42, B8 = − 1

30, B10 =

5

66, ...

B0 and B1 are the two first terms of this series. We can compute all theterms:

B0 = 1, then B0 = 1

B1 − B0 = −1/2, then B1 = 1/2

B2 − 2B1 + B0 = B2, then B1 = 1/2

B3 − 3B2 + 3B1 − B0 = B3, then B2 = 1/6

B4 − 4B3 + 6B2 − 4B1 + B0 = B4, then B3 = 0

B5 − 5B4 + 10B3 − 10B2 + 5B1 − B0 = B5, then B4 = −1/30

...

B1 is computed twice, but this arrangement clearly shows the binomialcoefficients.

2.12 Formulas

I already presented some formulas useful when building the Pascal trian-gle. There are many more available relating to combinatorial mathematics.Below, I list some of them.

For x, y ∈ IR and n ∈ IN,

(x + y)n =n

∑

k=0

(

n

k

)

xkyn−k (2.4)

with(

n

k

)

=n!

k!(n − k)!

For n ∈ IN,n

∑

k=0

(

n

k

)

= 2n (2.5)

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For n ∈ IN,n

∑

k=0

(−1)k

(

n

k

)

= 0 (2.6)

For n ∈ IN,n

∑

k=0

2k

(

n

k

)

= 3n (2.7)

(Pascal’s identity) For n ∈ IN,

(

n + 1

k

)

=

(

n

k − 1

)

+

(

n

k

)

(2.8)

(Vandermonde’s identity) For m, n, r ∈ IN+, r ≤ m, r ≤ n,

(

m + n

r

)

=r

∑

k=0

(

m

r − k

)(

n

k

)

(2.9)

For n ∈ IN,(

2n

n

)

=n

∑

k=0

(

n

k

)2

(2.10)

For n, r ∈ IN+, r ≤ n,

(

n + 1

r + 1

)

=

n∑

j=r

(

j

r

)

(2.11)

In the preceding formulas, x, y, and z are members of IR, but they workif the variables are complex (e.g., (3 − 7i)).

Here is one formula applying to numbers within Newton’s generalizationof the Pascal triangle.

(Newton’s identity ) For x, y ∈ IR and r ∈ IR,

(x + y)r =

∞∑

k=0

(

n

k

)

xkyr−k (2.12)

with(

n

k

)

=1

k!

n−1∏

i=0

(n − k) =n(n − 1) · · · (n − k + 1)

k!

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In equation 2.12, if r is an integer, then the number of terms is stillinfinite, but almost all of them are equal to zero (those in the tail). Thisformula is a generalization of equation 2.4.

Some formulas relating to Pascal triangle, if not all, work when m, n, andr are fractional.

2.13 Generalizations

The Pascal triangle is so full of different properties that it lends itself togeneralizations. We saw two of them: negative rows and fractional rows,both being associated with Newton’s binomial coefficients.

Newton was maybe the first to extend it, but many others followed. Be-low, I challenge you to explore different generalizations.

Create another kind of Pascal triangle by replacing the addition rule witha more complex algebraic rule, like a term is equal to the sum of twice theterm above it and once the term above it, at left. In other words, (c + 1, r +1) = 2 × (c + 1, r) + (c, r).

How did I build the following triangle? Where could it be used?

1

1 1 1

1 2 3 2 1

1 3 6 7 6 3 1

...

(Hint: Play with 1 + x + x2.)Once you have found where it is used, try to insert negative and frac-

tional rows. Note: These numbers form what is sometimes called a trinomialtriangle.

You could consider Pascal triangle in 3D. How could you build such pyra-mid? (Hint: Play with a + b + c)

Steven Roman generalized the binomial coefficients into the logarithmicbinomial coefficients. Try to find out what they are. Any good Web search en-gine will return some pages about them. If not, then go to http://mathworld.wolfram.com/,and search for it.

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Niels Henrik Abel found an identity in three variables: x, y, and z. Forx, y, z ∈ IR, n ∈ IN+, and x 6= 0,

(x + y)n =

n∑

k=0

(

n

k

)

x(x − kz)k−1(y + kz)n−k

Play with it, use z = 1 or z = −1, for instance. Is it possible to map allpolynoms with integer coefficients to this generalization?

You observed that column indexes are integers. How about having frac-tional column indexes, just like fractional row indexes? Spontaneously, Ibelieve this is not possible. If these exist, the coefficients should respect thethree rules.

Moreover, the row indexes are integers or fractions. How about havingcomplex indexes (r ∈ C)? Those coefficients would allow us to compute32+5i, for instance.

In probability theory, we get equations like

n∑

k=0

k0

(

n

k

)

n∑

k=0

k1

(

n

k

)

n∑

k=0

k2

(

n

k

)

· · ·The summation of the first equation is easy to compute (2n), while the

second summation is a little harder (n2n−1). However, the summations fork2, k3, k4, k5, . . . are not easy to derive. You could compute the first 10 termsof each equation, then find the formula at

http://www.research.att.com/ ˜ njas/sequences/index.htmlGiven a positive integer exponent, is there a formula to compute any

summation? How about negative integer exponent or rational exponent (z ∈IR)?

When I observe the numbers in the Newton’s generalization of Pascaltriangle, I feel that the negative rows have a relationship with the fractionalrows. However, I do not have an handy formula for this. Is there any suchrelation?

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Sierpinski sieve, once we know the recipe, is easy to obtain. Computingthe odd-ness or even-ness of a number is an operation similar to applyingmodulo 2 (mod 2) to these numbers, 0 mapping to even and 1 to odd. Howabout applying mod 3, mod 4, mod 5,..., associate a color to computed values,then draw the figure with such colors? For instance, applying mod 4 to thePascal triangle will output a number from {0, 1, 2, 3}. Mapping each numberto the colors {red, black, green, violet} will output the figure...

We have seen the Bernoulli triangle. Visually, I see an array of numbers,Pij. What is Pij ×Pij? Is there any formula linking the Bernoulli triangle tothose products? What is the determinant of the matrices?

Using the Bernoulli triangle, compute the differences between every adja-cent two terms in Pascal triangle. Create a new number triangle with these.Is there any new or interesting property in this new triangle?

The normal curve is a symetric curve, unlike many other curves in sta-tistics. Using any of those asymetric curves, χ2 and Student-t for instance,find the matching asymetric “Pascal triangles”.

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Chapter 3

Conclusion

In this text, I explored different properties of Pascal triangle (e.g., Fibonaccinumbers). Along the way, I added some formulas discovered by Newton (e.g.,square root, natural logarithm).

There is much more to Pascal triangle than you and I can imagine. Wehave seen many generalizations, some evident, others not.

Lots of its properties are due to its symetrical nature along its center.There is another reason why it gives so many results: Each row starts withone, a number easy to compute with.

29

Chapter 4

Recommended Readings

The books below present supplementary information on Pascal triangle. Knuthpresents Pascal triangle with software developer in mind, Maor presents thePascal triangle in historic perspective, Weisstein lists different results re-lating to Pascal triangle, and Conway and Guy walk you through differentmathematical paths, where Pascal triangle coefficients arise unexpectedly.

Each book is well written and can give you (too) many days of intellectualpleasure if you love mathematics.

1. Donald E. Knuth, The Art of Computer Programming, vol. 1, Addison-Wesley, 1997, ISBN: 0-201-89683-4. Section 1.2.6 Binomial Coefficients.

2. Eli Maor, e: The Story of a Number, Princeton University Press, 1994,ISBN: 0-691-05854-7. Chapter 8.

3. Eric W. Weisstein, CRC Concise Encyclopedia of Mathematics, CRCPress, 1999, ISBN: 0-8493-9640-9. Article on Binomial coefficient.

4. John H. Conway, Richard K. Guy, The Book of Numbers, Springer-Verlag, 1996, ISBN: 0-387-97993-X.

Stumbling on identities involving binomial coefficients is common in math-ematics. For different reasons, mathematicians like to express them as re-currence relation (e.g., Sn+2 − 16Sn+1 + 3Sn = 5n2). Until mid-90’s, therewas no computer-generated method to find a recurrence relation if you knewthe identity. The mathematicians used identities catalogs to find it. Twomathematicians finally solved the puzzle and wrote a book on the subject.The book A=B exposes the needed tools to do so.

30

The reading is easy, but is not for the novice.

1. Marko Petkos̃ek, Herbert S. Wilf, Doron Zeilberger, A=B, A. K. Peters,Ltd., 1997, ISBN: 1-56881-063-6. It is available for free in the Web, askGoogle for “A=B”.

Web resources are available, too.

1. http://mathforum.org/workshops/usi/pascal/pascal.links.html

2. http://mathworld.wolfram.com/PascalsTriangle.html

3. http://www-math.cudenver.edu/ ˜ wcherowi/jcorn5.html

4. http://binomial.csuhayward.edu/

c© 2005 GFDL 31

Chapter 5

Copyright

This document is published under the terms of the GNU Free DocumentationLicense, v 1.2. The complete text is available from

Free Software Foundation, Inc.51 Franklin St, Fifth Floor,Boston, MA02110-1301USA

In the Web, it is available at http://www.gnu.org/copyleft/.

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