Let $a(n,m,k)$ be an integer sequence such that $$a(n,m,k)=\sum\limits_{i=0}^{n}{n\brace i}(m-k)^{n-i}\prod\limits_{j=0}^{i-1}(kj+1)$$ Here ${n\brace k}$ is the Stirling number of the second kind.
There are many sequences in the OEIS that are special cases of a given sequence family:
- $a(n,1,1)$ - A000142
- $a(n,1,2)$ - A014307
- $a(n,2,1)$ - A000670
- $a(n,2,2)$ - A001147
- $a(n,2,3)$ - A136727
- $a(n,2,4)$ - A276371
- $a(n,3,1)$ - A122704
- $a(n,3,2)$ - A305404
- $a(n,3,3)$ - A007559
- $a(n,3,4)$ - A136728
- $a(n,4,1)$ - A255927
- $a(n,4,2)$ - A352117
- $a(n,4,3)$ - A346982
- $a(n,4,4)$ - A007696
- $a(n,4,5)$ - A136729
- $a(n,5,1)$ - A326324
- $a(n,5,4)$ - A346983
- $a(n,5,5)$ - A008548
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 $n$ 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 $$b(n,m,k)=m(\ell(n)-\operatorname{wt}(n)+2)b(f(n),m,k)+\sum\limits_{j=0}^{\ell(n)} k(1-T(n,j))b(f(n)+2^j(1-T(n,j)),m,k)$$ Here $b(n,1,1)$ is A329369.
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:
a(n, m, k) = sum(i=0, n, stirling(n, i, 2)*(m-k)^(n-i)*prod(j=0, i-1, k*j + 1))
s(n, m, k) = my(v, v1); v=vector(2^n, i, 0); v[1]=1; for(i=1, #v-1, my(L=logint(i, 2), A=i-2^L); v[i+1]=m*(L - hammingweight(i) + 2)*v[A+1] + sum(j=0, L, my(B=bittest(i, j)); k*(1-B)*v[A + 2^j*(1-B) + 1])); v1=[1]; for(i=1, n, v1=concat(v1, sum(j=0, 2^i-1, v[j+1]))); v1
[n, m, k] = [10, 1, 1]
x = s(n, m, k)
x1 = vector(n+1, i, a(i, m, k))
test = x==x1
Is there a way to prove it?
