Generating binomial coefficients in a row of Pascal's triangle from extensions of powers of eleven

Abstract

Sir Isaac Newton noticed that the values of the first five rows of Pascal's triangle are each formed by a power of 11, and claimed that subsequent rows can also be generated by a power of 11. Literally, the claim is not true for the $5^{th}$ row and onward. His genius mind might have suggested a deep relation between binomial coefficients and a power of some integer that resembles the number 11 in some form. In this study, we propose and prove a general formula to generate the values in any row of Pascal's triangle from the digits of ${\left(1\underset{\text{Θ zeros}}{\underbrace{0\cdots 0}}1\right)}^n$ . It can be shown that the numbers in the cells in $n^{th}$ row of Pascal's triangle may be achieved from $\Theta +1$ partitions of the digits of the number ${\left(1\underset{\text{Θ zeros}}{\underbrace{0\cdots 0}}1\right)}^n$ , where Θ is a non-negative integer. That is, we may generate the number in the cells in a row of Pascal's triangle from a power of 11, 101, 1001, or 10001 and so on. We briefly discuss how to determine the number of zeros Θ in relation to n, and then empirically show that the partition really gives us binomial coefficients for several values of n. We provide a formula for Θ and prove that the ${(n+1)}^{th}$ row of Pascal's triangle is simply $\Theta +1$ partitions of the digits of ${\left(1\underset{\text{Θ zeros}}{\underbrace{0\cdots 0}}1\right)}^n$ from the right.

Keywords: Binomial coefficients; Logarithm; Modular arithmetic; Pascal's triangle.