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1 vote
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Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$

Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have $$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$ Then the sum in question can be easily computed: \begin{split} & a(1,2^{n-1}+n)+\...
3 votes

Weirdness in the sequence "the number of divisors for a weird number"

I guess you mean $15$ proper divisors; it's less confusing to say $16$ divisors, and counting all the divisors has the nicest formula. The sequence of weird numbers is A006037 on the OEIS; looking at ...
6 votes
Accepted

A number sequence problem involving binomial transform

Unfortunately the argument that I originally posted contained a gap at the end. The gap is explained at the end of the proof, where I also state 3 partial results. Let $$f(z)=\sum_{n=0}^\infty b_nz_n/...

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