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Let $a(n)$ be A060831, i.e., $\sum\limits_{k=1}^{n}\operatorname{number of odd divisors of} k$.

Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$ Let $$b(n,k)=2b(n,k-1)-2^{k-1}, b(n,0)=n$$ Let $c(n)$ be the sequence of even numbers $k$ such that $b(k,\ell(k)+1)\equiv 0 \pmod k$.

Let $d(n)$ be A082662, i.e, numbers $k$ such that the odd part of $k$ is less than $\sqrt{2k}$.

I conjecture that $c(n)$ is a subsequence of $d(n)$.

I also conjecture that $c(a(n))=2^n$.

One may verify last conjecture using the following PARI prog:

f(n)=my(A=n); for(i=0,logint(n, 2), A*=2; A-=2^i); Mod(A, n)
my(z=1, z1=0, A=0); for(k=1, 59, while(!(f(2^(z1+1)*(2^z1+z-1))==0 && z<=3*2^z1), if(z>=3*2^z1, z1=z1++; z=1, z++)); my(B=2^(z1+1)*(2^z1+z-1), C=logint(B, 2)==valuation(B, 2)); A+=C; print([k,C,A,B]); z++);

Is there a way to prove it?

Improve this question
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    $\begingroup$ The sequence $c(m)$ is made of numbers of the form, $m=2^{nl-1-\lfloor \log_2 n \rfloor}n$ where $n$ are all odd natural numbers and $l\in\mathbb N$. This satisfies the condition that $n<\sqrt{2m}$. $\endgroup$
    – Alapan Das
    Commented Jan 12, 2023 at 9:38
  • $\begingroup$ @AlapanDas, thank you for comment! Very nice observation! Do you have a proof? $\endgroup$
    – Notamathematician
    Commented Jan 12, 2023 at 14:50
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    $\begingroup$ From the recurrence relation of $b(n,k)$ we get, $b(n,k)=2^{k-1}(2n-k)$. Now, if $c(n)$ is comprised of even numbers s.t $m=2^a_1n$, where $n$ is odd, we have $l(m)=a_1+\lfloor \log_2{n} \rfloor$,then $b(m,l(m)+1)=2^{l(m)}(2^{a_1+1}n-l(m)-1)$ Now, $m \mid b(m,l(m)+1) \Rightarrow n \mid l(m)+1$, say, $l(m)=nl-1,l \in \mathbb N$. $\endgroup$
    – Alapan Das
    Commented Jan 12, 2023 at 15:21
  • $\begingroup$ @AlapanDas, thank you very much! $\endgroup$
    – Notamathematician
    Commented Jan 12, 2023 at 15:34
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    $\begingroup$ Ok. My email is: [email protected] $\endgroup$
    – Alapan Das
    Commented Jan 13, 2023 at 13:07

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