The module distribution for Pascal's triangle row

by fungarwai, Sep 13, 2019, 12:14 PM

Let $n=\sum_{k=0}^\infty n_k p^k, N_r=\{k|n_k=r\}$

For p=2,

There are $2^{|N_1|}$ different ways to choose $m$ for $\binom{n}{m}\equiv 1\pmod{2}$

For p=3,

There are $2^{|N_1|}\cdot \frac{3^{|N_2|}-1}{2}$ different ways to choose $m$ for $\binom{n}{m}\equiv 2\pmod{3}$ and
$2^{|N_1|}\cdot \frac{3^{|N_2|}+1}{2}$ different ways to choose $m$ for $\binom{n}{m}\equiv 1\pmod{3}$

Proof

For p=5,

The number of $\binom{n}{m}\equiv 1\pmod{5}$ is $\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}+0^{|N_3|}) +\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}} \cos\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$
The number of $\binom{n}{m}\equiv 2\pmod{5}$ is $\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}-0^{|N_3|}) +\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}} \sin\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$
The number of $\binom{n}{m}\equiv 3\pmod{5}$ is $\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}-0^{|N_3|}) -\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}} \sin\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$
The number of $\binom{n}{m}\equiv 4\pmod{5}$ is $\frac{1}{4}2^{|N_1|}5^{|N_4|}(3^{|N_2|}4^{|N_3|}+0^{|N_3|}) -\frac{1}{2}2^{|N_1|}5^{\frac{|N_2|}{2}}2^{\frac{3|N_3|}{2}} \cos\left(|N_2|\tan^{-1}\frac{1}{2}-|N_3|\frac{\pi}{4}\right)$

Proof

Example

This post has been edited 1 time. Last edited by fungarwai, Sep 13, 2019, 12:15 PM

number theory

combinatorics

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The module distribution for Pascal's triangle row

by fungarwai, Sep 13, 2019, 12:14 PM

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