# Pascal's triangle

(Redirected from Pascal triangle)
Rows zero to five of Pascal's triangle

In mathematics, Pascal's triangle is a triangular array of the binomial coefficients. In much of the Western world it is named after French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in India,[1] Iran, China, Germany, and Italy.[2]

The rows of Pascal's triangle (sequence A007318 in OEIS) are conventionally enumerated starting with row n = 0 at the top (the 0th row). The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. Having the indices of both rows and columns start at zero makes it possible to state that the binomial coefficient $\tbinom{n}{k}$ appears in the nth row and kth column of Pascal's triangle. A simple construction of the triangle proceeds in the following manner. In row 0, the topmost row, the entry is $\tbinom{0}{0} = 1$ (the entry in the zeroth row and zeroth column). Then, to construct the elements of the following rows, add the number above and to the left with the number above and to the right of a given position to find the new value to place in that position. If either the number to the right or left is not present, substitute a zero in its place. For example, the initial number in the first (or any other) row is 1 (the sum of 0 and 1), whereas the numbers 1 and 3 in the third row are added to produce the number 4 in the fourth row.

This construction is related to the binomial coefficients by Pascal's rule, which says that if

$(x+y)^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^{k}$

then

${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$

for any non-negative integer n and any integer k between 0 and n.[3]

Pascal's triangle has higher dimensional generalizations. The three-dimensional version is called Pascal's pyramid or Pascal's tetrahedron, while the general versions are called Pascal's simplices.

Each number in the triangle is the sum of the two directly above it.

## History

Yang Hui (Pascal's) triangle, as depicted by the Chinese using rod numerals.
Blaise Pascal's version of the triangle

The set of numbers that form Pascal's triangle was known well before Pascal's time. Pascal innovated many previously unattested uses of the triangle's numbers, uses he described comprehensively in what is perhaps the earliest known mathematical treatise to be specially devoted to the triangle, his Traité du triangle arithmétique (1653). Centuries before, discussion of the numbers had arisen in the context of Indian studies of combinatorics and of binomial numbers and Greeks' study of figurate numbers.[4]

From later commentary, it appears that the binomial coefficients and the additive formula for generating them, $\tbinom{n}{r}=\tbinom{n-1}{r}+\tbinom{n-1}{r-1}$, were known to Pingala in or before the 2nd century BC.[5][6] While Pingala's work only survives in fragments, the commentator Varāhamihira, around 505, gave a clear description of the additive formula,[6] and a more detailed explanation of the same rule was given by Halayudha, around 975. Halayudha also explained obscure references to Meru-prastaara, the "Staircase of Mount Meru", giving the first surviving description of the arrangement of these numbers into a triangle.[6] In approximately 850, the Jain mathematician Mahāvīra gave a different formula for the binomial coefficients, using multiplication, equivalent to the modern formula $\tbinom{n}{r}=\tfrac{n!}{r!(n-r)!}$.[6] In 1068, four columns of the first sixteen rows were given by the mathematician Bhattotpala, who was the first recorded mathematician to equate the additive and multiplicative formulas for these numbers.[6]

At around the same time, it was discussed in Persia (Iran) by the Persian mathematician, Al-Karaji (953–1029).[7] It was later repeated by the Persian poet-astronomer-mathematician Omar Khayyám (1048–1131); thus the triangle is referred to as the Khayyam-Pascal triangle or Khayyam triangle in Iran. Several theorems related to the triangle were known, including the binomial theorem. Khayyam used a method of finding nth roots based on the binomial expansion, and therefore on the binomial coefficients.[8]

Pascal's triangle was known in China in the early 11th century through the work of the Chinese mathematician Jia Xian (1010–1070). In the 13th century, Yang Hui (1238–1298) presented the triangle and hence it is still called Yang Hui's triangle in China.[9]

In the west, the binomial coefficients were calculated by Gersonides in the early 14th century, using the multiplicative formula for them.[6] Petrus Apianus (1495–1552) published the full triangle on the frontispiece of his book on business calculations in 1527. This is the first record of the triangle in Europe.[10] Michael Stifel published a portion of the triangle (from the second to the middle column in each row) in 1544, describing it as a table of figurate numbers.[6] In Italy, Pascal's triangle is referred to as Tartaglia's triangle, named for the Italian algebraist Niccolò Fontana Tartaglia (1500–77), who published six rows of the triangle in 1556.[6] Gerolamo Cardano, also, published the triangle as well as the additive and multiplicative rules for constructing it in 1570.[6]

Pascal's Traité du triangle arithmétique (Treatise on Arithmetical Triangle) was published posthumously in 1665. In this, Pascal collected several results then known about the triangle, and employed them to solve problems in probability theory. The triangle was later named after Pascal by Pierre Raymond de Montmort (1708) who called it "Table de M. Pascal pour les combinaisons" (French: Table of Mr. Pascal for combinations) and Abraham de Moivre (1730) who called it "Triangulum Arithmeticum PASCALIANUM" (Latin: Pascal's Arithmetic Triangle), which became the modern Western name.[11]

## Binomial expansions

Visualisation of binomial expansion up to the 4th power

Pascal's triangle determines the coefficients which arise in binomial expansions. For an example, consider the expansion

(x + y)2 = x2 + 2xy + y2 = 1x2y0 + 2x1y1 + 1x0y2.

Notice the coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. In general, when a binomial like x + y is raised to a positive integer power we have:

(x + y)n = a0xn + a1xn−1y + a2xn−2y2 + ... + an−1xyn−1 + anyn,

where the coefficients ai in this expansion are precisely the numbers on row n of Pascal's triangle. In other words,

$a_i = {n \choose i}.$

This is the binomial theorem.

Notice that the entire right diagonal of Pascal's triangle corresponds to the coefficient of yn in these binomial expansions, while the next diagonal corresponds to the coefficient of xyn−1 and so on.

To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of (x + 1)n+1 in terms of the corresponding coefficients of (x + 1)n (setting y = 1 for simplicity). Suppose then that

$(x+1)^n=\sum_{i=0}^n a_i x^i.$

Now

$(x+1)^{n+1} = (x+1)(x+1)^n = x(x+1)^n + (x+1)^n = \sum_{i=0}^n a_i x^{i+1} + \sum_{i=0}^n a_i x^i.$
Six rows Pascal's triangle as Binomial Coefficients

The two summations can be reorganized as follows:

\begin{align} & \sum_{i=0}^{n } a_{i } x^{i+1} + \sum_{i=0}^n a_i x^i \\ & {} = \sum_{i=1}^{n+1} a_{i-1} x^{i } + \sum_{i=0}^n a_i x^i \\ & {} = \sum_{i=1}^{n } a_{i-1} x^{i } + \sum_{i=1}^n a_i x^i + a_0x^0 + a_{n}x^{n+1} \\ & {} = \sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + a_0x^0 + a_{n}x^{n+1} \\ & {} = \sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + x^0 + x^{n+1} \end{align}

(because of how raising a polynomial to a power works, a0an = 1).

We now have an expression for the polynomial (x + 1)n+1 in terms of the coefficients of (x + 1)n (these are the ais), which is what we need if we want to express a line in terms of the line above it. Recall that all the terms in a diagonal going from the upper-left to the lower-right correspond to the same power of x, and that the a-terms are the coefficients of the polynomial (x + 1)n, and we are determining the coefficients of (x + 1)n+1. Now, for any given i not 0 or n + 1, the coefficient of the xi term in the polynomial (x + 1)n+1 is equal to ai (the figure above and to the left of the figure to be determined, since it is on the same diagonal) + ai−1 (the figure to the immediate right of the first figure). This is indeed the simple rule for constructing Pascal's triangle row-by-row.

It is not difficult to turn this argument into a proof (by mathematical induction) of the binomial theorem. Since (a + b)n = bn(a/b +  1)n, the coefficients are identical in the expansion of the general case.

An interesting consequence of the binomial theorem is obtained by setting both variables x and y equal to one. In this case, we know that (1 + 1)n = 2n, and so

${n \choose 0} + {n \choose 1} + \cdots +{n \choose n-1} + {n \choose n} = 2^n.$

In other words, the sum of the entries in the nth row of Pascal's triangle is the nth power of 2.

## Combinations

A second useful application of Pascal's triangle is in the calculation of combinations. For example, the number of combinations of n things taken k at a time (called n choose k) can be found by the equation

$\mathbf{C}(n,k) = \mathbf{C}_k^n= {_nC_k} = {n \choose k} = \frac{n!}{k!(n-k)!}.$

But this is also the formula for a cell of Pascal's triangle. Rather than performing the calculation, one can simply look up the appropriate entry in the triangle. For example, suppose a basketball team has 10 players and wants to know how many ways there are of selecting 8. Provided we have the first row and the first entry in a row numbered 0, the answer is entry 8 in row 10: 45. That is, the solution of 10 choose 8 is 45.

## Relation to binomial distribution and convolutions

When divided by 2n, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases. This can also be seen by applying Stirling's formula to the factorials involved in the formula for combinations.

This is related to the operation of discrete convolution in two ways. First, polynomial multiplication exactly corresponds to discrete convolution, so that repeatedly convolving the sequence {..., 0, 0, 1, 1, 0, 0, ...} with itself corresponds to taking powers of 1 + x, and hence to generating the rows of the triangle. Second, repeatedly convolving the distribution function for a random variable with itself corresponds to calculating the distribution function for a sum of n independent copies of that variable; this is exactly the situation to which the central limit theorem applies, and hence leads to the normal distribution in the limit.

## Patterns and properties

Pascal's triangle has many properties and contains many patterns of numbers.

### Rows

• The sum of the elements of a single row is twice the sum of the row preceding it. For example, row 0 (the topmost row) has a value of 1, row 1 has a value of 2, row 2 has a value of 4, and so forth. This is because every item in a row produces two items in the next row: one left and one right. The sum of the elements of row n is equal to 2n.
• Taking product of the elements in each row, the sequence of products (sequence A001142 in OEIS) is related to the base of the natural logarithm, e.[12][13] Specifically, define the sequence sn as follows:
$s_n=\prod_{k=0}^{n}\binom{n}{k} = \prod_{k=0}^{n}\frac{n!}{k!(n-k)!}~, ~ n \geq 0.$
Then, the ratio of successive row products is
$\frac{s_{n+1}}{s_{n}}=\frac{(n+1)!^{(n+2)}\prod_{k=0}^{n+1}{k!^{-2}}}{n!^{(n+1)}\prod_{k=0}^{n}{k!^{-2}}} =\frac{(n+1)^n}{n!}$
and the ratio of these ratios is
$\frac{(s_{n+1})(s_{n-1})}{(s_n)^2}=\left(\frac{n+1}{n}\right)^n, ~ n\geq 1.$
The right-hand side of the above equation takes the form of the limit definition of e
$\textit{e} =\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}.$
• The value of a row, if each entry is considered a decimal place (and numbers larger than 9 carried over accordingly) is a power of 11 ( 11n, for row n). Thus, in row 2, ⟨1, 2, 1⟩ becomes 112, while ⟨1, 5, 10, 10, 5, 1⟩ in row five becomes (after carrying) 161,051, which is 115. This property is explained by setting x = 10 in the binomial expansion of (x + 1)n, and adjusting values to the decimal system. But x can be chosen to allow rows to represent values in any base.
• In base 3: 1 2 13 = 42 (16)
• ⟨1, 3, 3, 1⟩ → 2 1 0 13 = 43 (64)
• In base 9: 1 2 19 = 102 (100)
•               1 3 3 19 = 103 (1000)
• ⟨1, 5, 10, 10, 5, 1⟩ → 1 6 2 1 5 19 = 105 (100000)
In particular (see previous property), for x = 1 place value remains constant (1place=1). Thus entries can simply be added in interpreting the value of a row.
• Some of the numbers in Pascal's triangle correlate to numbers in Lozanić's triangle.
• The sum of the squares of the elements of row n equals the middle element of row 2n. For example, 12 + 42 + 62 + 42 + 12 = 70. In general form:
$\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}.$
• Another interesting pattern is that on any row n, where n is even, the middle term minus the term two spots to the left equals a Catalan number, specifically the (n/2 + 1)th Catalan number. For example: on row 4, 6 − 1 = 5, which is the 3rd Catalan number, and 4/2 + 1 = 3.
• Another interesting property of Pascal's triangle is that in a row p where p is a prime number, all the terms in that row except the 1s are multiples of p. This can be proven easily, since if $p\in \mathbb{P}$, then p has no factors save for 1 and itself. Every entry in the triangle is an integer, so therefore by definition $(p-k)!$ and $k!$ are factors of $p!\,$. However, there is no possible way p itself can show up in the denominator, so therefore p (or some multiple of it) must be left in the numerator, making the entire entry a multiple of p.
• Parity: To count odd terms in row n, convert n to binary. Let x be the number of 1s in the binary representation. Then number of odd terms will be 2x.[14]
• Every entry in row 2n-1, n≥0, is odd.[citation needed]
• Polarity: Yet another interesting pattern, when rows of Pascal's triangle are added and subtracted together sequentially, every row with a middle number, meaning rows that have an odd number of integers, they are always equal 0. Example, row 4 is, 1 4 6 4 1, so the formula would be 6 - (4+4) + (1+1) = 0, row 6 is 1 6 15 20 15 6 1, so the formula would be 20 - (15+15) + (6+6) - (1+1) = 0. So every even row of the Pascal triangle equals 0 when you take the middle number, then subtract the integers directly next to the center, then add the next integers, then subtract, so on and so forth until you reach the end of the row.

### Diagonals

The diagonals of Pascal's triangle contain the figurate numbers of simplices:

\begin{align} P_0(n) &= P_d(0) = 1, \\ P_d(n) &= P_d(n-1) + P_{d-1}(n) \\ &= \sum_{i=0}^n P_{d-1}(i) = \sum_{i=0}^d P_i(n-1). \end{align}

The symmetry of the triangle implies that the nth d-dimensional number is equal to the dth n-dimensional number.

An alternative formula that does not involve recursion is as follows:

$P_d(n)=\frac{1}{d!}\prod_{k=0}^{d-1} (n+k) = {n^{(d)}\over d!} = \binom{n+d-1}{d}$
where n(d) is the rising factorial.

The geometric meaning of a function Pd is: Pd(1) = 1 for all d. Construct a d-dimensional triangle (a 3-dimensional triangle is a tetrahedron) by placing additional dots below an initial dot, corresponding to Pd(1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find Pd(x), have a total of x dots composing the target shape. Pd(x) then equals the total number of dots in the shape. A 0-dimensional triangle is a point and a 1-dimensional triangle is simply a line, and therefore P0(x) = 1 and P1(x) = x, which is the sequence of natural numbers. The number of dots in each layer corresponds to Pd − 1(x).

### Calculating a row or diagonal by itself

There are simple algorithms to compute all the elements in a row or diagonal without computing other elements or factorials.

To compute row $n$ with the elements $\tbinom{n}{0}$, $\tbinom{n}{1}$, ..., $\tbinom{n}{n}$, begin with $\tbinom{n}{0}=1$. For each subsequent element, the value is determined by multiplying the previous value by a fraction with slowly changing numerator and denominator:

${n\choose k}= {n\choose k-1}\times \frac{n+1-k}{k}.$

For example, to calculate row 5, the fractions are  $\tfrac{5}{1}$$\tfrac{4}{2}$$\tfrac{3}{3}$$\tfrac{2}{4}$ and $\tfrac{1}{5}$, and hence the elements are  $\tbinom{5}{0}=1$,   $\tbinom{5}{1}=1\times\tfrac{5}{1}=5$,   $\tbinom{5}{2}=5\times\tfrac{4}{2}=10$, etc. (The remaining elements are most easily obtained by symmetry.)

To compute the diagonal containing the elements $\tbinom{n}{0}$, $\tbinom{n+1}{1}$, $\tbinom{n+2}{2}$, ..., we again begin with $\tbinom{n}{0}=1$ and obtain subsequent elements by multiplication by certain fractions:

${n+k\choose k}= {n+k-1\choose k-1}\times \frac{n+k}{k}.$

For example, to calculate the diagonal beginning at $\tbinom{5}{0}$, the fractions are  $\tfrac{6}{1}$$\tfrac{7}{2}$$\tfrac{8}{3}$, ..., and the elements are $\tbinom{5}{0}=1$,   $\tbinom{6}{1}=1\times\tfrac{6}{1}=6$,   $\tbinom{7}{2}=6\times\tfrac{7}{2}=21$, etc. By symmetry, these elements are equal to $\tbinom{5}{5}$, $\tbinom{6}{5}$, $\tbinom{7}{5}$, etc.

### Overall patterns and properties

Sierpinski triangle
• The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the fractal called the Sierpinski triangle. This resemblance becomes more and more accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle, assuming a fixed perimeter.[15] More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other similar patterns.
Pascal's triangle overlaid on a grid gives the number of distinct paths to each square, assuming only rightward and downward movements are considered.
• Imagine each number in the triangle is a node in a grid which is connected to the adjacent numbers above and below it. Now for any node in the grid, count the number of paths there are in the grid (without backtracking) which connect this node to the top node (1) of the triangle. The answer is the Pascal number associated to that node. The interpretation of the number in Pascal's Triangle as the number of paths to that number from the tip means that on a Plinko game board shaped like a triangle, the probability of winning prizes nearer the center will be higher than winning prizes on the edges.
• One property of the triangle is revealed if the rows are left-justified. In the triangle below, the diagonal coloured bands sum to successive Fibonacci numbers.
 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

### Construction as matrix exponential

Binomial matrix as matrix exponential. All the dots represent 0.

Due to its simple construction by factorials, a very basic representation of Pascal's triangle in terms of the matrix exponential can be given: Pascal's triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4, … on its subdiagonal and zero everywhere else.

### Number of elements of polytopes

Pascal's triangle can be used as a lookup table for the number of elements (such as edges and corners) within a polytope (such as a triangle, a tetrahedron, a square and a cube).

Let's begin by considering the 3rd line of Pascal's triangle, with values 1, 3, 3, 1. A 2-dimensional triangle has one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements (vertices, or corners). The meaning of the final number (1) is more difficult to explain (but see below). Continuing with our example, a tetrahedron has one 3-dimensional element (itself), four 2-dimensional elements (faces), six 1-dimensional elements (edges), and four 0-dimensional elements (vertices). Adding the final 1 again, these values correspond to the 4th row of the triangle (1, 4, 6, 4, 1). Line 1 corresponds to a point, and Line 2 corresponds to a line segment (dyad). This pattern continues to arbitrarily high-dimensioned hyper-tetrahedrons (known as simplices).

To understand why this pattern exists, one must first understand that the process of building an n-simplex from an (n − 1)-simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal's triangle: 1 face, 3 edges, and 3 vertices (the meaning of the final 1 will be explained shortly). To build a tetrahedron from a triangle, we position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle.

The number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells (polyhedral elements) is 0 (the original triangle possesses none) + 1 (built upon the single face of the original triangle) = 1; the number of faces is 1 (the original triangle itself) + 3 (the new faces, each built upon an edge of the original triangle) = 4; the number of edges is 3 (from the original triangle) + 3 (the new edges, each built upon a vertex of the original triangle) = 6; the number of new vertices is 3 (from the original triangle) + 1 (the new vertex that was added to create the tetrahedron from the triangle) = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal's triangle to yield the number below. Thus, the meaning of the final number (1) in a row of Pascal's triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal's triangle. Alternately, the "extra" 1 in a row can be thought of as the empty space, or field, in which the vertices, edges, faces, etc., exist.

A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of (x + 2)Row Number, instead of (x + 1)Row Number. There are a couple ways to do this. The simpler is to begin with Row 0 = 1 and Row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule:

${n \choose k} = 2\times{n-1 \choose k-1} + {n-1 \choose k}.$

That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in:

                             1
1       2
1       4       4
1       6       12      8
1       8       24      32      16
1       10      40      80      80       32
1       12      60      160     240     192       64
1       14      84      280     560     672      448       128


The other way of manufacturing this triangle is to start with Pascal's triangle and multiply each entry by 2k, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2Position Number = 6 × 22 = 6 × 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned cube (called a hypercube) can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2-dimensional cube (a square) is one, the number of 1-dimensional elements (sides, or lines) is 4, and the number of 0-dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely.

To understand why this pattern exists, first recognize that the construction of an n-cube from an (n − 1)-cube is done by simply duplicating the original figure and displacing it some distance (for a regular n-cube, the edge length) orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an n-cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher n-cube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher n-cube.

In this triangle, the sum of the elements of row m is equal to 3m. Again, to use the elements of row 4 as an example: $1 + 8 + 24 + 32 + 16 = 81$, which is equal to $3^4 = 81$.

### Fourier transform of sin(x)n+1/x

As stated previously, the coefficients of (x + 1)n are the nth row of the triangle. Now the coefficients of (x − 1)n are the same, except that the sign alternates from +1 to −1 and back again. After suitable normalization, the same pattern of numbers occurs in the Fourier transform of sin(x)n+1/x. More precisely: if n is even, take the real part of the transform, and if n is odd, take the imaginary part. Then the result is a step function, whose values (suitably normalized) are given by the nth row of the triangle with alternating signs.[16] For example, the values of the step function that results from:

$\,\mathfrak{Re}\left(\text{Fourier} \left[ \frac{\sin(x)^5}{x} \right]\right)$

compose the 4th row of the triangle, with alternating signs. This is a generalization of the following basic result (often used in electrical engineering):

$\,\mathfrak{Re}\left(\text{Fourier} \left[ \frac{\sin(x)^1}{x}\right] \right)$

is the boxcar function.[17] The corresponding row of the triangle is row 0, which consists of just the number 1.

If n is congruent to 2 or to 3 mod 4, then the signs start with −1. In fact, the sequence of the (normalized) first terms corresponds to the powers of i, which cycle around the intersection of the axes with the unit circle in the complex plane:

$\, +i,-1,-i,+1,+i,\ldots \,$

### Elementary cellular automaton

The pattern produced by an elementary cellular automaton using rule 60 is exactly Pascal's triangle of binomial coefficients reduced modulo 2 (black cells correspond to odd binomial coefficients).[18] Rule 102 also produces this pattern when trailing zeros are omitted. Rule 90 produces the same pattern but with an empty cell separating each entry in the rows.

## Extensions

Pascal's Triangle can be extended to negative row numbers.

First write the triangle in the following form:

 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = 0 1 0 0 0 0 0 ... n = 1 1 1 0 0 0 0 ... n = 2 1 2 1 0 0 0 ... n = 3 1 3 3 1 0 0 ... n = 4 1 4 6 4 1 0 ...

Next, extend the column of 1s upwards:

 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = −4 1 ... n = −3 1 ... n = −2 1 ... n = −1 1 ... n = 0 1 0 0 0 0 0 ... n = 1 1 1 0 0 0 0 ... n = 2 1 2 1 0 0 0 ... n = 3 1 3 3 1 0 0 ... n = 4 1 4 6 4 1 0 ...

Now the rule:

${n \choose m} = {n-1 \choose m-1} + {n-1 \choose m}$

can be rearranged to:

${n-1 \choose m} = {n \choose m} - {n-1 \choose m-1}$

which allows calculation of the other entries for negative rows:

 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = −4 1 −4 10 −20 35 −56 ... n = −3 1 −3 6 −10 15 −21 ... n = −2 1 −2 3 −4 5 −6 ... n = −1 1 −1 1 −1 1 −1 ... n = 0 1 0 0 0 0 0 ... n = 1 1 1 0 0 0 0 ... n = 2 1 2 1 0 0 0 ... n = 3 1 3 3 1 0 0 ... n = 4 1 4 6 4 1 0 ...

This extension preserves the property that the values in the mth column viewed as a function of n are fit by an order m polynomial, namely

${n \choose m} = \frac{1}{m!}\prod_{k=0}^{m-1} (n-k) = \frac{1}{m!}\prod_{k=1}^{m} (n-k+1)$.

This extension also preserves the property that the values in the nth row correspond to the coefficients of (1 + x)n:

$(1+x)^n = \sum_{k=0}^\infty {n \choose k} x^k \quad |x| < 1$

For example:

$(1+x)^{-2} = 1-2x+3x^2-4x^3+\cdots \quad |x| < 1$

When viewed as a series, the rows of negative n diverge. However, they are still Abel summable, which summation gives the standard values of 2n. (In fact, the n = -1 row results in Grandi's series which "sums" to 1/2, and the n = -2 row results in another well-known series which has an Abel sum of 1/4.)

Another option for extending Pascal's triangle to negative rows comes from extending the other line of 1s:

 m = −4 m = −3 m = −2 m = −1 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = −4 1 0 0 0 0 0 0 0 0 0 ... n = −3 1 0 0 0 0 0 0 0 0 ... n = −2 1 0 0 0 0 0 0 0 ... n = −1 1 0 0 0 0 0 0 ... n = 0 0 0 0 0 1 0 0 0 0 0 ... n = 1 0 0 0 0 1 1 0 0 0 0 ... n = 2 0 0 0 0 1 2 1 0 0 0 ... n = 3 0 0 0 0 1 3 3 1 0 0 ... n = 4 0 0 0 0 1 4 6 4 1 0 ...

Applying the same rule as before leads to

 m = −4 m = −3 m = −2 m = −1 m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = −4 1 0 0 0 0 0 0 0 0 0 ... n = −3 −3 1 0 0 0 0 0 0 0 0 ... n = −2 3 −2 1 0 0 0 0 0 0 0 ... n = −1 −1 1 −1 1 0 0 0 0 0 0 .. n = 0 0 0 0 0 1 0 0 0 0 0 ... n = 1 0 0 0 0 1 1 0 0 0 0 ... n = 2 0 0 0 0 1 2 1 0 0 0 ... n = 3 0 0 0 0 1 3 3 1 0 0 ... n = 4 0 0 0 0 1 4 6 4 1 0 ...

Note that this extension also has the properties that just as

$\exp\begin{pmatrix} . & . & . & . & . \\ 1 & . & . & . & . \\ . & 2 & . & . & . \\ . & . & 3 & . & . \\ . & . & . & 4 & . \end{pmatrix} = \begin{pmatrix} 1 & . & . & . & . \\ 1 & 1 & . & . & . \\ 1 & 2 & 1 & . & . \\ 1 & 3 & 3 & 1 & . \\ 1 & 4 & 6 & 4 & 1 \end{pmatrix},$

we have

$\exp\begin{pmatrix} . & . & . & . & . & . & . & . & . & . \\ -4 & . & . & . & . & . & . & . & . & . \\ . & -3 & . & . & . & . & . & . & . & . \\ . & . & -2 & . & . & . & . & . & . & . \\ . & . & . & -1 & . & . & . & . & . & . \\ . & . & . & . & 0 & . & . & . & . & . \\ . & . & . & . & . & 1 & . & . & . & . \\ . & . & . & . & . & . & 2 & . & . & . \\ . & . & . & . & . & . & . & 3 & . & . \\ . & . & . & . & . & . & . & . & 4 & . \end{pmatrix} = \begin{pmatrix} 1 & . & . & . & . & . & . & . & . & . \\ -4 & 1 & . & . & . & . & . & . & . & . \\ 6 & -3 & 1 & . & . & . & . & . & . & . \\ -4 & 3 & -2 & 1 & . & . & . & . & . & . \\ 1 & -1 & 1 & -1 & 1 & . & . & . & . & . \\ . & . & . & . & . & 1 & . & . & . & . \\ . & . & . & . & . & 1 & 1 & . & . & . \\ . & . & . & . & . & 1 & 2 & 1 & . & . \\ . & . & . & . & . & 1 & 3 & 3 & 1 & . \\ . & . & . & . & . & 1 & 4 & 6 & 4 & 1 \end{pmatrix}$

Also, just as summing along the lower-left to upper-right diagonals of the Pascal matrix yields the Fibonacci numbers, this second type of extension still sums to the Fibonacci numbers for negative index.

Either of these extensions can be reached if we define

${n \choose k} = \frac{n!}{(n-k)! k!} \equiv \frac{\Gamma(n+1)}{\Gamma(n-k+1)\Gamma(k+1)}$

and take certain limits of the Gamma function, $\Gamma(z)$.

## References

1. ^ Maurice Winternitz, History of Indian Literature, Vol. III
2. ^ Peter Fox (1998). Cambridge University Library: the great collections. Cambridge University Press. p. 13. ISBN 978-0-521-62647-7.
3. ^ The binomial coefficient $\scriptstyle {n \choose k}$ is conventionally set to zero if k is either less than zero or greater than n (sequence A007318 in OEIS).
4. ^ Pascal's Triangle | World of Mathematics Summary
5. ^ A. W. F. Edwards. Pascal's arithmetical triangle: the story of a mathematical idea. JHU Press, 2002. Pages 30–31.
6. Edwards, A. W. F. (2013), "The arithmetical triangle", in Wilson, Robin; Watkins, John J., Combinatorics: Ancient and Modern, Oxford University Press, pp. 166–180.
7. ^
8. ^ Coolidge, J. L. (1949), "The story of the binomial theorem", The American Mathematical Monthly 56: 147–157, JSTOR 10.2307/2305028, MR 0028222.
9. ^ Weisstein, Eric W. (2003). CRC concise encyclopedia of mathematics, p.2169. ISBN 978-1-58488-347-0.
10. ^ Smith, Karl J. (2010), Nature of Mathematics, Cengage Learning, p. 10, ISBN 9780538737586.
11. ^ Fowler, David (January 1996). "The Binomial Coefficient Function". The American Mathematical Monthly 103 (1): 1–17. doi:10.2307/2975209. JSTOR 2975209. See in particular p. 11.
12. ^ Brothers, H. J. (2012), "Finding e in Pascal’s triangle", Mathematics Magazine 85: 51, doi:10.4169/math.mag.85.1.51.
13. ^ Brothers, H. J. (2012), "Pascal's triangle: The hidden stor-e", The Mathematical Gazette 96: 145–148.
14. ^ Fine, N. J. (1947), "Binomial coefficients modulo a prime", American Mathematical Monthly 54: 589–592, doi:10.2307/2304500, MR 0023257. See in particular Theorem 2, which gives a generalization of this fact for all prime moduli.
15. ^ Wolfram, S. (1984). "Computation Theory of Cellular Automata". Comm. Math. Phys. 96: 15–57. Bibcode:1984CMaPh..96...15W. doi:10.1007/BF01217347.
16. ^ For a similar example, see e.g. Hore, P. J. (1983), "Solvent suppression in Fourier transform nuclear magnetic resonance", Journal of Magnetic Resonance 55 (2): 283–300, doi:10.1016/0022-2364(83)90240-8.
17. ^ Karl, John H. (2012), An Introduction to Digital Signal Processing, Elsevier, p. 110, ISBN 9780323139595.
18. ^ Wolfram, S. (2002). A New Kind of Science. Champaign IL: Wolfram Media. pp. 870, 931–2.