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

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    by thedodecagon, Jan 22, 2022, 1:33 AM

  • thanks for this

    by jasperE3, Dec 3, 2021, 10:01 PM

  • I am working as accountant and studying as ACCA student now.
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    by fungarwai, Aug 28, 2021, 4:54 AM

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    by Feridimo, Jan 23, 2020, 5:05 PM

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  • rip 0 shouts and 0 comments until now

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