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Expanding in Fibonacci powers

The results with generating functions also suggest that a "closed formula" for $N(P_n)$ can be written as: $$N(P_n)=br^n+cs^n+dt^n$$ where $b,c,d$ and $r,s,t$ are the roots of \begin{align} ...
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93 votes

A number theory problem where pi appears surprisingly

Note that $a_k$ is always a multiple of $k$ (in particular, this structure disrupts the random model proposed in comments). We can exploit this structure to simplify the recurrence and clarify the ...
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1 vote

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)+\...
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How this expression leads to the given sequence

After the observation (and display in my earlier answer) that the coefficients occur in the Jordan-decomposition of the Stirlingmatrix 2nd kind, there is a much easier decomposition. First to make the ...
7 votes

Which $n$ have $\lvert\{2^n-2^k -1\}\cap {\mathrm{PRIMES}}\rvert=m$?

The conjecture is false, eg with $m=5, n=29$ or $m=3, n=49$. Here is some R code, which I ran online at rdrr: ...
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8 votes

(Explicit) Tauberian theorems: removing $(\log x/n)$

$\newcommand\ep\epsilon$You cannot improve the upper bound $c\sqrt\ep\,x$ on $|S(x)|$ (where $c>0$ is a universal real constant factor) by more than a universal positive real constant factor. ...

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