Editing Pascal's triangle (section)

=== To arbitrary bases ===
[[Isaac Newton]] once observed that the first five rows of Pascal's triangle, when read as the digits of an integer, are the corresponding powers of eleven. He claimed without proof that subsequent rows also generate powers of eleven.<ref>{{citation
 | last = Newton | first = Isaac
 | journal = The Mathematical Works of Isaac Newton
 | title = A Treatise of the Method of Fluxions and Infinite Series
 | pages = 1:31–33
 | year = 1736
 | quote = But these in the alternate areas, which are given, I observed were the same with the figures of which the several ascending powers of the number 11 consist, viz. <math>11^{0}</math>, <math>11^{1}</math>, <math>11^{2}</math>, <math>11^{3}</math>, <math>11^{4}</math>, etc. that is, first 1; the second 1, 1; the third 1, 2, 1; the fourth 1, 3, 3, 1; the fifth 1, 4, 6, 4, 1, and so on
 | url = https://www.loc.gov/item/42048007/
 }}.</ref> In 1964, Robert L. Morton presented the more generalized argument that each [[#Rows|row]] <math>n</math> can be read as a radix <math>a</math> numeral, where [[E (mathematical constant)|<math>\lim_{n \to \infty} 11^{n}_{a}</math>]] is the hypothetical terminal row, or [[Infinite product|limit]], of the triangle, and the rows are its partial products.<ref>{{citation
 | last = Morton | first = Robert L.
 | issue = 6
 | journal = The Mathematics Teacher
 | pages = 392–394
 | title = Pascal's Triangle and powers of 11
 | volume = 57
 | year = 1964
 | doi = 10.5951/MT.57.6.0392
 | jstor = 27957091
 }}.</ref> He proved the entries of row <math>n</math>, when interpreted directly as a place-value numeral, correspond to the binomial expansion of <math>(a + 1)^n = 11^{n}_{a}</math>. More rigorous proofs have since been developed.<ref>{{citation
 | display-authors = etal
 | last = Arnold | first = Robert
 | journal = Proceedings of Undergraduate Mathematics Day
 | title = Newton's Unfinished Business: Uncovering the Hidden Powers of Eleven in Pascal's Triangle
 | year = 2004
 | url = https://ecommons.udayton.edu/mth_epumd/6/
 }}.</ref><ref>{{citation
 | display-authors = etal
 | last = Islam | first = Robiul
 | title = Finding any row of Pascal's triangle extending the concept of power of 11
 | year = 2020
 | url = https://www.researchgate.net/publication/341785706
 }}.</ref> To better understand the principle behind this interpretation, here are some things to recall about binomials:
* A radix <math>a</math> numeral in [[positional notation]] (e.g. <math>14641_{a}</math>) is a univariate polynomial in the variable <math>a</math>, where the [[Monomial#Degree|degree]] of the variable of the <math>i</math>th [[Addition#summand|term]] (starting with <math>i = 0</math>) is <math>i</math>. For example, <math>14641_{a} = 1 \cdot a^{4} + 4 \cdot a^{3} + 6 \cdot a^{2} + 4 \cdot a^{1} + 1 \cdot a^{0}</math>.
* A row corresponds to the binomial expansion of <math>(a + b)^{n}</math>. The variable <math>b</math> can be eliminated from the expansion by setting <math>b = 1</math>. The expansion now typifies the expanded form of a radix <math>a</math> numeral,<ref>{{citation
 | last = Winteridge | first = David J.
 | issue = 1
 | journal = Mathematics in School
 | pages = 12–13
 | title = Pascal's Triangle and Powers of 11
 | volume = 13
 | year = 1984
 | jstor = 30213884
 }}.</ref><ref>{{citation
 | last = Kallós | first = Gábor
 | issue = 1
 | journal = Annales Mathématiques
 | pages = 1–15
 | title = A generalization of Pascal's triangle using powers of base numbers
 | volume = 13
 | year = 2006
 | doi = 10.5802/ambp.211
 | url = https://ambp.centre-mersenne.org/item/10.5802/ambp.211.pdf
 }}.</ref> as demonstrated [[#Binomial expansions|above]]. Thus, when the entries of the row are concatenated and read in radix <math>a</math> they form the numerical equivalent of <math>(a + 1)^{n} = 11^{n}_{a}</math>. If <math>c = a + 1</math> for <math>c < 0</math>, then the theorem [[Negative base|holds]] for <math>a = \{c - 1, -(c + 1)\} \;\mathrm{mod}\; 2c</math> with odd values of <math>n</math> [[Negative number#Multiplication|yielding]] negative row products.<ref>{{cite book
 | display-authors = etal
 | last = Hilton | first = P.
 | series = In International Series in Modern Applied Mathematics and Computer Science
 | pages = 89–91
 | title = Symmetry 2
 | chapter = Extending the binomial coefficients to preserve symmetry and pattern
 | publisher = Pergamon
 | year = 1989
 | doi = 10.1016/B978-0-08-037237-2.50013-1
 | isbn = 9780080372372 | chapter-url = https://www.sciencedirect.com/science/article/pii/B9780080372372500131
 }}.</ref><ref>{{citation
 | last = Mueller | first = Francis J.
 | issue = 5
 | journal = The Mathematics Teacher
 | pages = 425–428
 | title = More on Pascal's Triangle and powers of 11
 | volume = 58
 | year = 1965
 | doi = 10.5951/MT.58.5.0425
 | jstor = 27957164
 }}.</ref><ref>{{citation
 | last = Low | first = Leone
 | issue = 5
 | journal = The Mathematics Teacher
 | pages = 461–463
 | title = Even more on Pascal's Triangle and Powers of 11
 | volume = 59
 | year = 1966
 | doi = 10.5951/MT.59.5.0461
 | jstor = 27957385
 }}.</ref>

By setting the row's radix (the variable <math>a</math>) equal to one and ten, row <math>n</math> becomes the product <math>11^{n}_{1} = 2^{n}</math> and <math>11^{n}_{10} = 11^{n}</math>, respectively. To illustrate, consider <math>a = n</math>, which yields the row product <math>\textstyle n^n \left( 1 + \frac{1}{n} \right)^{n} = 11^{n}_{n}</math>. The numeric representation of [[Superparticular ratio|<math>11^{n}_{n}</math>]] is formed by concatenating the entries of row <math>n</math>. The twelfth row denotes the product:

: <math>11^{12}_{12} = 1:10:56:164:353:560:650:560:353:164:56:10:1_{12} = 27433a9699701_{12}</math>

with compound digits (delimited by ":") in radix twelve. The digits from <math>k = n - 1</math> through <math>k = 1</math> are compound because these row entries compute to values greater than or equal to twelve. To [[Trapdoor function|normalize]]<ref>{{citation
 | last = Fjelstad | first = P.
 | issue = 9
 | journal = Computers & Mathematics with Applications
 | pages = 3
 | title = Extending Pascal's Triangle 
 | volume = 21
 | doi = 10.1016/0898-1221(91)90119-O
 | year = 1991
 | doi-access = free
 }}.</ref> the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient <math>\textstyle {n \choose n - 1}</math> from its leftmost digit up to, but excluding, its rightmost digit, and use radix-twelve arithmetic to sum the removed prefix with the entry on its immediate left, then repeat this process, proceeding leftward, until the leftmost entry is reached. In this particular example, the normalized string ends with <math>01</math> for all <math>n</math>. The leftmost digit is <math>2</math> for <math>n > 2</math>, which is obtained by carrying the <math>1</math> of <math>10_{n}</math> at entry <math>k = 1</math>. It follows that the length of the normalized value of <math>11^{n}_{n}</math> is [[Bijection|equal]] to the row length, <math>n + 1</math>. The integral part of <math>1.1^{n}_{n}</math> contains exactly one digit because <math>n</math> (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of <math>1.1^{1234}_{1234}</math>. Compound digits remain in the value because they are radix <math>1234</math> [[Modular arithmetic#Integers modulo m|residues]] represented in radix ten:

: <math>1.1^{1234}_{1234} = 2.885:2:35:977:696:\overbrace{\ldots}^\text{1227 digits}:0:1_{1234} = 2.717181235\ldots_{10}</math>