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First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over fixed $\ell:=\mathrm{wt}(n)$ (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 $B_m^o$.

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 $B_m^o$.

First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over fixed $\ell:=\mathrm{wt}(n)$ (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 need to understand 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 0} x^m B_m^o\right)^{-1}, \end{split} which can be recognized as INVERTi transform of ordered Bell numbers $B_m^o$.

First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over fixed $\ell:=\mathrm{wt}(n)$ (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 $B_m^o$.

Let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$. Then

\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 need to understand 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 0} x^m B_m^o\right)^{-1}, \end{split} which can be recognized as INVERTi transform of ordered Bell numbers $B_m^o$.

First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over fixed $\ell:=\mathrm{wt}(n)$ (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 need to understand 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 0} x^m B_m^o\right)^{-1}, \end{split} which can be recognized as INVERTi transform of ordered Bell numbers $B_m^o$.