structural:reformatting
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. In 1964, Robert L. Morton presented the more generalized argument that each row n {\displaystyle n} can be read as a radix a {\displaystyle a} numeral, where lim n → ∞ 11 a n {\displaystyle \lim _{n\to \infty }11_{a}^{n}} is the hypothetical terminal row, or limit, of the triangle, and the rows are its partial products. He proved the entries of row n {\displaystyle n}, when interpreted directly as a place-value numeral, correspond to the binomial expansion of (a + 1) n = 11 a n {\displaystyle (a+1)^{n}=11_{a}^{n}}. More rigorous proofs have since been developed. To better understand the principle behind this interpretation, here are some things to recall about binomials:

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

with compound digits (delimited by ":") in radix twelve. The digits from k = n − 1 {\displaystyle k=n-1} through k = 1 {\displaystyle k=1} are compound because these row entries compute to values greater than or equal to twelve. To normalize the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient (n n − 1) {\displaystyle {n \choose n-1}} 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 01 {\displaystyle 01} for all n {\displaystyle n}. The leftmost digit is 2 {\displaystyle 2} for n > 2 {\displaystyle n>2}, which is obtained by carrying the 1 {\displaystyle 1} of 10 n {\displaystyle 10_{n}} at entry k = 1 {\displaystyle k=1}. It follows that the length of the normalized value of 11 n n {\displaystyle 11_{n}^{n}} is equal to the row length, n + 1 {\displaystyle n+1}. The integral part of 1.1 n n {\displaystyle 1.1_{n}^{n}} contains exactly one digit because n {\displaystyle n} (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of 1.1 1234 1234 {\displaystyle 1.1_{1234}^{1234}}. Compound digits remain in the value because they are radix 1234 {\displaystyle 1234} residues represented in radix ten: