Let $a(n,m,k)$ be an integer sequence with e.g.f. $$A(x)=\operatorname{exp}\left(x + m\int\int (A(x))^k \, dx \, dx\right)$$ I don't know much about integrals, so here's a concrete example:
- $a(n,1,3)$ - A234239
Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$ $$f(n)=n-2^{\ell(n)}$$ $$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \operatorname{wt}(2n)=\operatorname{wt}(n), \operatorname{wt}(0)=0$$ $$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$ Here $f(n)$ is the same as $n$ without the most significant bit, $\operatorname{wt}(n)$ is the binary weight of $n$ or the number of $1$'s in the binary expansion of $n$ and $T(n,k)$ is the $(k+1)$-th bit from the right side in the binary representation of $n$.
Let $b(n,m,k)$ be an integer sequence such that $$b(n,m,k)=m(\ell(n)-2\operatorname{wt}(n)+3)b(f(n),m,k)+\sum\limits_{j=0}^{\ell(n)} k(1-T(n,j))b(f(n)+2^j(1-T(n,j)),m,k)$$
Let $s(n,m,k)$ be an integer sequence such that $$s(n,m,k)=\sum\limits_{j=0}^{2^n-1}b(j,m,k)$$ I conjecture that $$s(n,m,k)=a(n+1,m,k)$$ Here is the PARI prog to verify this conjecture:
s(n, m, k) = my(v, v1); v=vector(2^n, i, 0); v[1]=1; v1=vector(n+1, i, 0); v1[1]=1; for(i=1, #v-1, my(L=logint(i, 2), A=i-2^L); v[i+1]=m*(L - 2*hammingweight(i) + 3)*v[A+1] + k*sum(j=0, L, my(B=bittest(i, j)); (B==0)*v[A + 2^j*(B==0) + 1])); for(i=1, n, v1[i+1]=v1[i]+sum(j=2^(i-1)+1, 2^i, v[j])); v1
a(n, m, k) = local(A=1+x); for(i=1, n, A=exp(x + m*intformal(intformal(A^k + x*O(x^n))))); n!*polcoeff(A, n)
test(n, m, k) = s(n, m, k)==vector(n+1, i, a(i, m, k))
Is there a way to prove it?
