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=== 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
*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.
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| 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>
\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>.
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| title = Nilakantha's Footprints in Pascal's Triangle
| volume = 108
| year = 2014}}</ref> <math display="block">\pi = 3 + \
* 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 {{mvar|n}} equals the middle element of row {{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 numbern|Catalan number]], specifically <math>C_{m-1} = \tbinom{2m}{m} - \tbinom{2m}{m-2}</math>. For example, in row 4, which is 1, 4, 6, 4, 1, we get the 3rd Catalan number <math>C_3 = 6-1 = 5 </math>.
* In a row {{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 -->
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