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  • Let $a(n,m)$ be the number of partitions of $m^n-1$ into powers of $m$. In other words, $$a(n,m)=[z^{m^n-1}] \prod\limits_{k\geqslant 0} \frac{1}{1-z^{m^k}}$$
  • Let $$ R(n,m,q)=\sum\limits_{j=0}^{m(q+1)-1}R(n-1,m,j),\\ R(0,m,q)=1 $$

I conjecture that $$R(n,m,0)=a(n+1,m)$$

Here is the PARI/GP prog to check it numerically:

a(n, q)=local(A=Mat(1), B); if(n<0, 0, for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i, B[i, j]=1, if(j==1, B[i, j]=(A^q)[i-1, 1], B[i, j]=(A^q)[i-1, j-1])); )); A=B); return(A[n+1, 1]))
R_upto(n, m)=my(v1, v2, v3); v1=vector(m^(n+1), i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, m^(n-i), v2[q+1]=sum(j=0, m*(q+1)-1, v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
test(n, m)=R_upto(n, m)==vector(n+1, i, a(i, m))

Is there a way to prove it?

Improve this question
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    $\begingroup$ Is there any information on the compositional inverse $g:=f^{\circ(-1)}$ of $$f(z):=z\prod_{k\ge0}\frac1{1-z^{m^k}},\quad ?$$ By the Lagrange inversion formula then $$a(n,m)=[z^{m^n}]f=\frac1{m^n}\text{Res}(g^{-m^n}).$$ Maybe the numbers $R(n, m, q)$ can be related to some recurrence scheme for $g$. $\endgroup$
    – Pietro Majer
    Commented Jul 3, 2023 at 7:22
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    $\begingroup$ For $m=2$, the number of such partitions is very well understood, see the following paper and refs therein: math.ucla.edu/~pak/papers/CayleyComp7.pdf $\endgroup$
    – Igor Pak
    Commented Jul 5, 2023 at 3:44

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