- Let $a(n)$ be A343685 i.e. $$ a(n)=2na(n-1)+\sum\limits_{j=0}^{n-1}\binom{n}{j}(n-j-1)!a(j), \\ a(0)=1 $$ Here the exponential generating function $A(x)$ satisfy $$ A(x)=\frac{1}{1-2x+\log(1-x)} $$
- Let $\operatorname{wt}(n)$ be A000120 i.e. number of $1$'s in binary expansion of $n$ (or the binary weight of $n$). Here $$ \operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \\ \operatorname{wt}(2n)=\operatorname{wt}(n), \\ \operatorname{wt}(0)=0 $$
- Let $f(n)$ be A025480 i.e. $$ f(2n+1)=f(n), \\ f(2n)=n $$
- Let $$ g(n,m)=f(g(n,m-1)-1), \\ g(n,0)=n $$
- Let $$ b(n)=b\left(\left\lfloor\frac{n}{2}\right\rfloor\right)+\sum\limits_{j=1}^{\operatorname{wt}(n)}\frac{(-1)^{j-1}}{j}b(g(n,j)), \\ b(0)=1 $$
- Finally, let $$ s(n)=\sum\limits_{j=0}^{2^n-1}b(j) $$
I conjecture that $$n!s(n)=a(n).$$
Here is the PARI/GP prog to check it numerically:
a_upto(n)=my(v1); v1=vector(n+1, i, 0); v1[1]=1; for(i=1, n, v1[i+1]=2*i*v1[i] + sum(j=0, i-1, binomial(i, j)*(i-j-1)!*v1[j+1])); v1
s_upto(n)=my(v1, v2); v1=vector(2^n, i, 0); v1[1]=1; v2=vector(n+1, i, 0); v2[1]=1; for(i=1, 2^n-1, my(A=0, B=i); for(j=1, hammingweight(i), B\=2^(valuation(B,2)+1); A+=(-1)^(j-1)/j*v1[B+1]); v1[i+1]=v1[i\2+1] + A); for(i=1, n, v2[i+1]=v2[i] + sum(j=2^(i-1)+1, 2^i, v1[j])); for(i=1, n, v2[i+1]*=i!); v2
test(n)=a_upto(n)==s_upto(n)
Motivation: random generalization of the conjectured recurrence for A071585 (OEIS page already contain my conjecture).
Is there a way to prove it?
