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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:

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?

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    $\begingroup$ Note that unless $k = 2m$, $$A(x)^k = \frac{2(2m-k)(m-k+\alpha) e^{\alpha x}}{((m-k+\alpha) e^{\alpha x} + m)^2}$$ with $\alpha = \sqrt{k^2 - 2km}$ $\endgroup$
    – Peter Taylor
    Jun 6, 2023 at 15:05
  • $\begingroup$ @PeterTaylor, thank you for comment! Nice observation! Could you tell me how to calculate the square root on PARI so that when extracting the coefficient from the function, an integer result is obtained? I tried c^(1/2) but for large $k$ it still turns out to be a decimal fraction. $\endgroup$
    – Notamathematician
    Jun 7, 2023 at 14:19
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    $\begingroup$ There does appear to be support for number fields in PARI, but I don't use PARI and I don't know whether that page contains sufficient information on how to calculate with elements of number fields. $\endgroup$
    – Peter Taylor
    Jun 7, 2023 at 14:32
  • $\begingroup$ @PeterTaylor, thank you again! I'm too lazy to figure it out myself, so I'll try to ask around from familiar experts. $\endgroup$
    – Notamathematician
    Jun 7, 2023 at 14:35
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    $\begingroup$ Just take $k$th roots. You can do it by substituting and doing a lot of algebra, but it ends up in the same place. $\endgroup$
    – Peter Taylor
    Jun 8, 2023 at 7:38

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