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$\DeclareMathOperator\wt{wt}$Let $\wt(n)$ be A000120, number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.

Also $$n=2^{t_1}(1+2^{t_2+1}(1+\dotsb(1+2^{t_{wt(n)}+1}))\dotsb).$$ Then we have an integer sequence given by $$a(n)=\sum\limits_{j=0}^{2^{\wt(n)}-1}(-1)^{\wt(n)-\wt(j)}\prod\limits_{k=0}^{\wt(n)-1}\left(1+f\left(\left\lfloor\frac{j}{2^k}\right\rfloor+1\right)\right)^{t_{k+1}+1},\quad a(0)=1.$$ Let $$s(n)=\sum\limits_{k=0}^{2^n-1}a(k).$$ Then I conjecture that $s(n)$ is A095989, INVERTi transform applied to the ordered Bell numbers.

I also conjecture that \begin{align} a(0)=a(1)&=1\\ a(2n+1) &= a(2n)\\ a(2n)& = a(n)+a(2n-2^{f(n)})+b(n-1)\\ b(2n+1) &= b(n)\\ b(2n) &= a(2n). \end{align} In other words \begin{align} a(2n) &= c(n)\\ c(0)&=1\\ c(n)& = c\left(\left\lfloor\frac{n}{2}\right\rfloor\right)+c\left(\left\lfloor\frac{2n-2^{f(n)}}{2}\right\rfloor\right)+c(g(n-1)) \end{align} where $g(n)$ is A025480, $g(2n) = n$, $g(2n+1) = g(n)$.

Is there a way to prove it? Is it possible to at least get a closed form for $s(n)$?

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1 Answer 1

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First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over $n$ of fixed weight $\ell:=\mathrm{wt}(n)$ (like in this answer): \begin{split} s(n) &= \sum_{\ell=0}^n \sum_{t_1 + \dots + t_\ell \leq n-\ell} \sum_{j=0}^{2^\ell-1} (-1)^{\ell-\mathrm{wt}(j)} \prod_{k=0}^{\ell-1} P(j,k)^{t_k+1} \\ &= [x^{n+1}]\ \sum_{\ell=0}^n \sum_{j=0}^{2^\ell-1} (-1)^{\mathrm{wt}(j)+1} \prod_{k=0}^{\ell} \frac{-P(j,k)x}{1-xP(j,k)}. \end{split}

Now, we notice that $P(j,k)$ depends on runs of unit bits in $j$. Namely, each run of length $u-1$ contributes the term $$\prod_{v=1}^{u} \frac{-vx}{1-vx} = x^{u} (-1)^{u} u! \prod_{v=1}^{u} \frac1{1-vx}.$$ Hence, introducing the number $z$ of zero bits in $j$ padded with an extra zero at the beginning (and so $\mathrm{wt}(j)=\ell+1-z$), we have \begin{split} s(n) &= [x^{n+1}]\ \sum_{\ell=0}^n [y^{\ell+1}]\ \sum_{z\geq 0} (-1)^{\ell+2-z} \left(\sum_{u\geq 1} y^u (-1)^u x^u u! \prod_{v=1}^u \frac1{1-vx}\right)^z\\ &= -[x^{n+1}]\ \sum_{\ell=0}^n (-1)^{\ell+1} [y^{\ell+1}]\ \left(1+\sum_{u\geq 1} y^{u} (-1)^u x^u u! \prod_{v=1}^u \frac1{1-vx}\right)^{-1}\\ &=- [x^{n+1}]\ \left(1+\sum_{u\geq 1} x^u u! \prod_{v=1}^u \frac1{1-vx}\right)^{-1}\\ &=- [x^{n+1}]\ \left(1+\sum_{u\geq 1} x^u u! \sum_{m\geq 0} S(m,u) x^{m-u}\right)^{-1}\\ &=- [x^{n+1}]\ \left(1+\sum_{m\geq 1} x^m B_m^o\right)^{-1}, \end{split} which can be recognized as INVERTi transform of ordered Bell numbers $B_m^o$.

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