Let $f(n)$ be an arbitrary function such that $f(n)\in\mathbb{Z}$.
Let $a(n)$ be an arbitrary integer sequence such that $a(0)=1$.
Let $b(n)$ be an integer sequence such that $$b(2^m(2n+1))=\sum\limits_{k=0}^{m}f(m-k)b(2^kn), b(0)=1$$
Let $s(n,m)$ be an integer sequence such that $$s(n,m)=\sum\limits_{k=0}^{2^n-1}b(2^mk)$$
Now we use the results from my previous question. Here they will be considered true.
To simplify computing of $s(n,0)$ we can use $$s(n,m)=s(n-1,m+1)+\sum\limits_{k=0}^{m}f(m-k)s(n-1,k), s(0,m)=1$$ Also if we consider $s(n,0)=a(n)$, then to compute $f(n)$ we can use $$f(n)=s(1,n)-1-\sum\limits_{k=0}^{n-1}f(k)$$ where $s(1,n)$ can be computed via $$s(n,m) = s(n+1,m-1) - \sum\limits_{k=0}^{m-1}f(m-k-1)s(n,k), s(0,m)=1$$ However, the above results can be improved. Define the operator $\operatorname{SR}$, which is associated with the series reversion.
Let $c(n)$ be an integer sequence such that $$c(n)=-\sum\limits_{k=1}^{n}c(n-k)a(k-1), c(0)=1$$ Let $$g(x)=\sum\limits_{k=0}^{\infty}c(k)x^k$$ Let $$h(x)=1+\frac{x}{\operatorname{SR}(g(x)-1)}$$ I conjecture that $$f(n)=(-1)^n[x^{n+1}]h(x)$$ Also if $f(n)$ is known we can compute $a(n)$ backwards.
Let $$g_1(x)=x\sum\limits_{k=0}^{\infty}f(k)(-x)^k$$ Let $$h_1(x)=1+\operatorname{SR}(\frac{x}{g_1(x)-1})$$ Let $d(n)$ be an integer sequence such that $$d(n)=-\sum\limits_{k=1}^{n}d(n-k)[x^k]h_1(x), d(0)=1$$ I conjecture that $$a(n)=d(n+1)$$ Here is the PARI prog to verify these conjectures:
a(n)=numbpart(n+1)
f(n)=1
f1(n)=my(v, v1, v2); v=vector(n+2,i,if(i==1,vector(n+2,j,a(j-1)),vector(n-i+3,j,(j==1)))); v1=vector(n+1,i,0); v1[1]=a(1)-1; for(i=2,n+1,for(j=2,i,v[j][i-j+2]=v[j-1][i-j+3]-sum(k=1,j-1,v1[j-k]*v[k][i-j+2])); v1[i]=v[i][2]-sum(j=1,i-1,v1[j])-1); v1
s1(n)=my(v, v1, v2, v3); v=vector(n+1,i,f(i-1)); v1=vector(n+1,i,1); v2=v1; v3=vector(n+1,i,0); v3[1]=1; for(i=1,n,for(j=1,n-i+1,v2[j]=v1[j+1]+sum(q=1,j,v[j-q+1]*v1[q])); v1=v2; v3[i+1]=v1[1];); v3
f2(n)=my(A, v, v1, v2); v=vector(n+3,i,a(i-1)); v1=vector(n+3,i,0); v1[1]=1; for(i=1,n+2,v1[i+1]=-sum(k=1,i,v1[i-k+1]*v[k])); A=1+x/serreverse(sum(k=0,#v-1,v1[k+1]*x^k)-1+x^2*O(x^(n+2))); v2=vector(n+1,i,polcoeff(A,i)*(-1)^(i-1))
s2(n)=my(A, v, v1, v2); v=vector(n+3,i,f(i-1)); v1=vector(n+3,i,0); v1[1]=1; A=1+serreverse(x/(sum(k=0,#v-1,v[k+1]*x^(k+1)*(-1)^k)-1+x^2*O(x^(n+2)))); v2=vector(n+2,i,polcoeff(A,i)); for(i=1,n+2,v1[i+1]=-sum(k=1,i,v1[i-k+1]*v2[k])); v1=vector(n+1,i,v1[i+1])
test1(n)=f1(n)==f2(n)
test2(n)=s1(n)==s2(n)
Is there a way to prove it?
