Editing Pascal's triangle (section)

== Patterns and properties ==
Pascal's triangle has many properties and contains many patterns of numbers.
[[File:Pascal's Triangle animated binary rows.gif|thumb|upright=1|Each frame represents a row in Pascal's triangle. Each column of pixels is a number in binary with the least significant bit at the bottom. Light pixels represent 1 and dark pixels 0.]]
[[File:pascal_triangle_compositions.svg|thumb|upright=1|The numbers of [[composition (combinatorics)|compositions]] of ''n''+1 into ''k''+1 ordered partitions form Pascal's triangle.]]

=== Rows ===
* The sum of the elements of a single row is twice the sum of the row preceding it. For example, row&nbsp;0 (the topmost row) has a value of 1, row&nbsp;1 has a value of 2, row&nbsp;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&nbsp;<math> n</math> equals to <math> 2^n</math>.
*Taking the product of the elements in each row, the sequence of products {{OEIS|id=A001142}} is related to the base of the natural logarithm, ''[[E (mathematical constant)|e]]''.<ref>{{citation
 | last = Brothers | first = H. J.
 | doi = 10.4169/math.mag.85.1.51
 | journal = [[Mathematics Magazine]]
 | pages = 51
 | title = Finding e in Pascal's triangle
 | volume = 85
 | year = 2012| issue = 1
 | s2cid = 218541210
 }}.</ref><ref>{{citation
 | last = Brothers | first = H. J.
 | doi =10.1017/S0025557200004204
 | journal = [[The Mathematical Gazette]]
 | pages = 145–148
 | title = Pascal's triangle: The hidden stor-''e''
 | volume = 96
 | year = 2012| issue = 535
 | s2cid = 233356674
 }}.</ref> Specifically, define the sequence <math> s_{n}</math>  for all  <math>n \ge 0</math> as follows: <math>s_{n} = \prod_{k = 0}^{n} {n \choose k} = \prod_{k = 0}^{n} \frac{n!}{k!(n-k)!}</math> {{pb}} Then, the ratio of successive row products is <math display="block">\frac{s_{n+1}}{s_{n}} = 
\frac{ \displaystyle (n+1)!^{n+2} \prod_{k = 0}^{n + 1} \frac{1}{k!^2}}{\displaystyle n!^{n+1}\prod_{k=0}^{n}{\frac{1}{k!^2}}} = \frac{(n + 1)^n}{n!}</math> and the ratio of these ratios is <math display="block">\frac{s_{n + 1} \cdot s_{n - 1}}{s_{n}^{2}} = 
\left( \frac{n + 1}{n} \right)^n, ~ n\ge 1.</math> The right-hand side of the above equation takes the form of the limit definition of [[e (mathematical constant)|<math>e</math>]] <math display="block">e =\lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^{n}.</math>
* [[pi|<math>\pi</math>]] can be found in Pascal's triangle by use of the [[Nilakantha Somayaji|Nilakantha]] infinite series.<ref>{{citation
 | last = Foster | first = T.
 | doi = 10.5951/mathteacher.108.4.0246
 | journal = [[Mathematics Teacher]]
 | pages = 247
 | title = Nilakantha's Footprints in Pascal's Triangle
 | volume = 108
 | year = 2014}}</ref> <math display="block">\pi = 3 + \sum_{n = 1}^{\infty} (-1)^{n + 1} \frac{{2n + 1 \choose 1}}{{2n + 1 \choose 2}{2n + 2 \choose 2}}</math>
* 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&nbsp;{{mvar|n}} equals the middle element of row&nbsp;{{math|2''n''}}. For example, {{math|1=1<sup>2</sup> + 4<sup>2</sup> + 6<sup>2</sup> + 4<sup>2</sup> + 1<sup>2</sup> = 70}}. In general form, <math display="block">\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}.</math>
* In any even row <math>n=2m</math>, the middle term minus the term two spots to the left equals a [[Catalan number]], specifically <math>C_{m-1} = \tbinom{2m}{m} - \tbinom{2m}{m-2}</math>. For example, in row&nbsp;4, which is 1, 4, 6, 4, 1, we get the 3rd Catalan number <math>C_3 = 6-1 = 5 </math>.
* In a row&nbsp;{{mvar|p}}, where {{mvar|p}} is a [[prime number]], all the terms in that row except the 1s are divisible by {{mvar|p}}. This can be proven easily, from the multiplicative formula <math>\tbinom pk = \tfrac{p!}{k!(p-k)!} </math>. Since the denominator <math>k!(p-k)!</math> can have no prime factors equal to {{mvar|p}}, so {{mvar|p}} remains in the numerator after integer division, making the entire entry a multiple of {{mvar|p}}.<!--
[[Image:Exp binomial grey.svg|thumb|upright=1.25|left|Binomial matrix as matrix exponential (illustration for 5×5 matrices). All the dots represent 0.]]  Comment: picture hidded since useless rescaling, also wrong paragraph for the pic -->
* ''Parity'': To count [[odd number|odd]] terms in row&nbsp;{{mvar|n}}, convert {{mvar|n}} to [[binary numeral system|binary]]. Let {{mvar|x}} be the number of 1s in the binary representation. Then the number of odd terms will be {{math|2<sup>''x''</sup>}}. These numbers are the values in [[Gould's sequence]].<ref>{{citation
 | last = Fine | first = N. J.
 | doi = 10.2307/2304500
 | journal = [[American Mathematical Monthly]]
 | mr = 0023257
 | pages = 589–592
 | title = Binomial coefficients modulo a prime
 | volume = 54
 | issue = 10
 | year = 1947| jstor = 2304500
 }}. See in particular Theorem 2, which gives a generalization of this fact for all prime moduli.</ref>
* Every entry in row 2<sup>''n''</sup>&nbsp;−&nbsp;1, ''n''&nbsp;≥&nbsp;0, is odd.<ref>{{citation
 | last = Hinz | first = Andreas M.
 | doi = 10.2307/2324061
 | issue = 6
 | journal = The American Mathematical Monthly
 | mr = 1166003
 | pages = 538–544
 | title = Pascal's triangle and the Tower of Hanoi
 | volume = 99
 | year = 1992| jstor = 2324061
 }}. Hinz attributes this observation to an 1891 book by [[Édouard Lucas]], ''Théorie des nombres'' (p.&nbsp;420).</ref>
*''Polarity'': When the elements of a row of Pascal's triangle are alternately added and subtracted together, the result is 0. For example, row 6 is 1,&nbsp;6,&nbsp;15,&nbsp;20,&nbsp;15,&nbsp;6,&nbsp;1, so the formula is 1&nbsp;−&nbsp;6&nbsp;+&nbsp;15&nbsp;−&nbsp;20&nbsp;+&nbsp;15&nbsp;−&nbsp;6&nbsp;+&nbsp;1&nbsp;=&nbsp;0.

=== Diagonals ===
[[File:Pascal_triangle_simplex_numbers.svg|thumb|upright=1.25|Derivation of [[simplex]] numbers from a left-justified Pascal's triangle]]
The diagonals of Pascal's triangle contain the [[Figurate numbers#Triangular numbers and their analogs in higher dimensions|figurate numbers]] of simplices:
* The diagonals going along the left and right edges contain only 1's.
* The diagonals next to the edge diagonals contain the [[natural number]]s in order. The 1-dimensional simplex numbers increment by 1 as the line segments extend to the next whole number along the number line.
* Moving inwards, the next pair of diagonals contain the [[triangular number]]s in order.
* The next pair of diagonals contain the [[tetrahedral number]]s in order, and the next pair give [[pentatope number]]s.

::<math>\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}</math>

The symmetry of the triangle implies that the ''n''<sup>th</sup> d-dimensional number is equal to the ''d''<sup>th</sup> ''n''-dimensional number.

An alternative formula that does not involve recursion is
<math display="block">P_d(n)=\frac{1}{d!}\prod_{k=0}^{d-1} (n+k) = {n^{(d)}\over d!} = \binom{n+d-1}{d},</math>
where ''n''<sup>(''d'')</sup> is the [[rising factorial]].

The geometric meaning of a function ''P''<sub>''d''</sub> is: ''P''<sub>''d''</sub>(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 ''P''<sub>''d''</sub>(1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find P<sub>''d''</sub>(''x''), have a total of ''x'' dots composing the target shape. P<sub>''d''</sub>(''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 ''P''<sub>0</sub>(''x'') = 1 and ''P''<sub>1</sub>(''x'') = ''x'', which is the sequence of natural numbers. The number of dots in each layer corresponds to ''P''<sub>''d''&nbsp;−&nbsp;1</sub>(''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 <math>n</math> with the elements <math>\tbinom{n}{0}, \tbinom{n}{1}, \ldots, \tbinom{n}{n}</math>, begin with <math>\tbinom{n}{0}=1</math>. For each subsequent element, the value is determined by multiplying the previous value by a fraction with slowly changing numerator and denominator:

:<math> {n\choose k}= {n\choose k-1}\times \frac{n+1-k}{k}.</math>

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

To compute the diagonal containing the elements <math>\tbinom{n}{0}, \tbinom{n+1}{1}, \tbinom{n+2}{2},\ldots,</math> begin again with <math>\tbinom{n}{0} = 1</math> and obtain subsequent elements by multiplication by certain fractions:

:<math> {n+k\choose k}= {n+k-1\choose k-1}\times \frac{n+k}{k}.</math>

For example, to calculate the diagonal beginning at <math>\tbinom{5}{0}</math>, the fractions are&nbsp; <math>\tfrac{6}{1}, \tfrac{7}{2}, \tfrac{8}{3}, \ldots</math>, and the elements are <math>\tbinom{5}{0}=1, \tbinom{6}{1}=1 \times \tfrac{6}{1}=6, \tbinom{7}{2}=6\times\tfrac{7}{2}=21</math>, etc. By symmetry, these elements are equal to <math>\tbinom{5}{5}, \tbinom{6}{5}, \tbinom{7}{5}</math>, etc.

[[File:Fibonacci in Pascal triangle.png|thumb|[[Fibonacci sequence]] in Pascal's triangle]]

=== Overall patterns and properties ===
[[File:Sierpinski Pascal triangle.svg|thumb|A level-4 approximation to a [[Sierpiński triangle]] obtained by shading the first 32 rows of a Pascal triangle white if the binomial coefficient is even and black if it is odd.]]
* The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the [[fractal]] known as the [[Sierpiński triangle]]. This resemblance becomes increasingly accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern ''is'' the Sierpiński triangle, assuming a fixed perimeter. 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.
:As the proportion of black numbers tends to zero with increasing ''n'', a corollary is that the proportion of odd binomial coefficients tends to zero as ''n'' tends to infinity.<ref>Ian Stewart, "How to Cut a Cake", Oxford University Press, page 180</ref>
<div class="thumb tright" style="clear: right; text-align: center;">
<div class="thumbinner" style="width: 230px;">
{| cellpadding="0" cellspacing="0" style="line-height: 0; background: white; font-size: 88%; border: 1px #b0b0b0 solid; padding: 0; margin: auto"
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess rll45.svg|26x26px|alt=a4 white rook|a4 white rook]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=b4 one|b4 one]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x1l45.svg|26x26px|alt=c4 one|c4 one]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=d4 one|d4 one]]
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=a3 one|a3 one]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x2l45.svg|26x26px|alt=b3 two|b3 two]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x3d45.svg|26x26px|alt=c3 three|c3 three]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x4l45.svg|26x26px|alt=d3 four|d3 four]]
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x1l45.svg|26x26px|alt=a2 one|a2 one]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x3d45.svg|26x26px|alt=b2 three|b2 three]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x6l45.svg|26x26px|alt=c2 six|c2 six]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | {{resize|150%|10}}
|- style="vertical-align: middle"
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | [[File:Chess x1d45.svg|26x26px|alt=a1 one|a1 one]]
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | [[File:Chess x4l45.svg|26x26px|alt=b1 four|b1 four]]
| style="padding: 0; vertical-align: inherit; background-color: #d18b47;" | {{resize|150%|10}}
| style="padding: 0; vertical-align: inherit; background-color: #ffce9e;" | {{resize|150%|20}}
|}
<div class="thumbcaption">
Pascal's triangle overlaid on a grid gives the number of distinct paths to each square, assuming only rightward and downward steps to an adjacent square are considered.
</div></div></div>

* In a triangular portion of a grid (as in the images below), the number of shortest grid paths from a given node to the top node of the triangle is the corresponding entry in Pascal's triangle. On a [[Plinko]] game board shaped like a triangle, this distribution should give the probabilities of winning the various prizes.

[[Image:Pascal's Triangle 4 paths.svg|center|400px]]

* If the rows of Pascal's triangle are left-justified, the diagonal bands (colour-coded below) sum to the [[Fibonacci number]]s.
::{| style="align:center;"
|- align=center
|bgcolor=red|1
|- align=center
| style="background:orange;"|1
| style="background:yellow;"|1
|- align=center
| style="background:yellow;"|1
|bgcolor=lime|2
|bgcolor=aqua|1
|- align=center
|bgcolor=lime|1
|bgcolor=aqua|3
| style="background:violet;"|3
|bgcolor=red|1
|- align=center
|bgcolor=aqua|1
| style="background:violet;"|4
|bgcolor=red|6
| style="background:orange;"|4
| style="background:yellow;"|1
|- align=center
| style="background:violet;"|1
|bgcolor=red|5
| style="background:orange;"|10
| style="background:yellow;"|10
|bgcolor=lime|5
|bgcolor=aqua|1
|- align=center
|bgcolor=red|1
| style="background:orange;"|6
| style="background:yellow;"|15
|bgcolor=lime|20
|bgcolor=aqua|15
| style="background:violet;"|6
|bgcolor=red|1
|- align=center
| style="background:orange; width:40px;"|1
| style="background:yellow; width:40px;"|7
| style="background:lime; width:40px;"|21
| style="background:aqua; width:40px;"|35
| style="background:violet; width:40px;"|35
| style="background:red; width:40px;"|21
| style="background:orange; width:40px;"|7
| style="background:yellow; width:40px;"|1
|}

=== Construction as matrix exponential ===
{{Image frame|caption=Binomial matrix as matrix exponential. All the dots represent 0.
|content=<math>\begin{align}
\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}\\
e^{\text{counting}} &= \text{binomial}
\end{align}
</math>
}}
{{See also|Pascal matrix}}
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 sub-diagonal and zero everywhere else.

===Construction of Clifford algebra using simplices===
Labelling the elements of each n-simplex matches the basis elements of [[Clifford algebra]] used as forms in [[Geometric algebra|Geometric Algebra]] rather than matrices. Recognising the geometric operations, such as rotations, allows the algebra operations to be discovered. Just as each row, {{mvar|n}}, starting at 0, of Pascal's triangle corresponds to an {{mvar|(n-1)}}-simplex, as described below, it also defines the number of named basis forms in {{mvar|n}} dimensional [[Geometric algebra]]. The [[binomial theorem]] can be used to prove the geometric relationship provided by Pascal's triangle.<ref>{{citation |url=https://github.com/GPWilmot/geoalg|title=The Algebra Of Geometry|year=2023|last=Wilmot|first=G.P.}}</ref> This same proof could be applied to simplices except that the first column of all 1's must be ignored whereas in the algebra these correspond to the real numbers, <math>\R</math>, with basis 1.

=== Relation to geometry of polytopes ===
{{more citations needed section|date=February 2025}}
Each row of Pascal's triangle gives the number of elements (such as edges and corners) of each dimension in a corresponding [[simplex]] (such as a triangle or tetrahedron).  In particular, for {{math| ''k'' > 0}}, the {{mvar|k}}th entry in the {{mvar|n}}th row is the number of {{math|(''k'' − 1)}}-dimensional elements in a {{math|(''n'' − 1)}}-dimensional simplex.  For example, a triangle (the 2-dimensional simplex) one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements ([[Vertex (graph theory)|vertices]], or corners); this corresponds to the third row 1, 3, 3, 1 of Pascal's triangle.  This fact can be explained by combining Pascal's rule for generating the triangle with the geometric construction of simplices: each simplex is formed from a simplex of one lower dimension by the addition of a new vertex, outside the space in which the lower-dimensional simplex lies.  Then each {{mvar|d}}-dimensional element in the smaller simplex remains a {{mvar|d}}-dimensional element of the higher simplex, and each {{math|(''d'' − 1)}}-dimensional element when joined to the new vertex forms a new {{mvar|d}}-dimensional element of the higher simplex.<ref>{{Cite book |last=Coxeter |first=Harold Scott Macdonald |url=https://books.google.com/books?id=iWvXsVInpgMC |title=Regular Polytopes |date=1973-01-01 |publisher=Courier Corporation |isbn=978-0-486-61480-9 |edition=3rd |pages=118–144 |language=en |chapter=Chapter VII: ordinary polytopes in higher space, 7.2: Pyramids, dipyramids and prisms}}</ref>

A similar pattern is observed relating to [[square (geometry)|squares]], as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of {{math|(''x'' + 2)<sup>row number</sup>}}, instead of {{math|(''x'' + 1)<sup>row number</sup>}}. 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:

:<math> {n \choose k} = 2\times{n-1 \choose k-1} + {n-1 \choose k}.</math>

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:
:<math>\begin{matrix}
                           \text{  1} \\
                       \text{  1}   \quad \text{  2} \\
                   \text{  1}   \quad \text{  4}   \quad \text{  4} \\
               \text{  1}   \quad\text{  6}   \quad   \text{ 12}   \quad\text{  8} \\
           \text{  1}   \quad\text{  8}   \quad   \text{ 24}   \quad  \text{ 32}   \quad  \text{ 16} \\
       \text{  1}   \quad   \text{ 10}   \quad  \text{ 40}   \quad  \text{ 80}   \quad  \text{ 80}    \quad  \text{ 32} \\
   \text{  1}   \quad  \text{ 12}   \quad  \text{ 60}   \quad  160   \quad 240  \quad  192   \quad   \text{ 64} \\
\text{  1}   \quad  \text{ 14}   \quad  \text{ 84}   \quad  280  \quad  560  \quad  672   \quad  448   \quad   128
\end{matrix}</math>
The other way of producing this triangle is to start with Pascal's triangle and multiply each entry by 2<sup>k</sup>, 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 {{math|1=2<sup>position number</sup> = 6 × 2<sup>2</sup> = 6 × 4 = 24}}. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned [[cube (geometry)|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 {{math|1=(''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 3<sup>''m''</sup>. Again, to use the elements of row 4 as an example: {{math|1=1 + 8 + 24 + 32 + 16 = 81}}, which is equal to <math>3^4 = 81</math>.

==== Counting vertices in a cube by distance ====
Each row of Pascal's triangle gives the number of vertices at each distance from a fixed vertex in an ''n''-dimensional cube. For example, in three dimensions, the third row (1 3 3 1) corresponds to the usual three-dimensional [[cube]]: fixing a vertex ''V'', there is one vertex at distance 0 from ''V'' (that is, ''V'' itself), three vertices at distance 1, three vertices at distance {{sqrt|2}} and one vertex at distance {{sqrt|3}} (the vertex opposite ''V''). The second row corresponds to a square, while larger-numbered rows correspond to [[hypercube]]s in each dimension.

=== Fourier transform of sin(''x'')<sup>''n''+1</sup>/''x'' ===

As stated previously, the coefficients of (''x''&nbsp;+&nbsp;1)<sup>''n''</sup> are the nth row of the triangle. Now the coefficients of (''x''&nbsp;−&nbsp;1)<sup>''n''</sup> 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'')<sup>''n''+1</sup>/''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 ''n''th row of the triangle with alternating signs.<ref>For a similar example, see e.g. {{citation|title=Solvent suppression in Fourier transform nuclear magnetic resonance|first=P. J.|last=Hore|journal=Journal of Magnetic Resonance|year=1983|volume=55|issue=2|pages=283–300|doi=10.1016/0022-2364(83)90240-8|bibcode=1983JMagR..55..283H}}.</ref> For example, the values of the step function that results from:

:<math>\mathfrak{Re}\left(\text{Fourier} \left[  \frac{\sin(x)^5}{x}   \right]\right)</math>

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

:<math>\mathfrak{Re}\left(\text{Fourier} \left[ \frac{\sin(x)^1}{x}\right] \right)</math>

is the [[boxcar function]].<ref>{{citation|title=An Introduction to Digital Signal Processing|first=John H.|last=Karl|publisher=Elsevier|year=2012|isbn=9780323139595|page=110|url=https://books.google.com/books?id=9Dv1PClLZWIC&pg=PA110}}.</ref>  The corresponding row of the triangle is row 0, which consists of just the number&nbsp;1.

If n is [[congruence relation|congruent]] to 2 or to 3 mod 4, then the signs start with&nbsp;−1. In fact, the sequence of the (normalized) first terms corresponds to the powers of [[imaginary unit|i]], which cycle around the intersection of the axes with the unit circle in the complex plane: <math display="block">  +i,-1,-i,+1,+i,\ldots  </math>