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Let $S_{1,k}(n) = n,\ S_{i,k}(n) = \sum_{\text{n terms},(k-i+2)\nmid j} S_{i-1,k}(j)$

For example, $S_{1,2}(n) = n, S_{2,2}(n) = 1 + 3 + \dots+(2n-1) = n^2$

With $k=3$, take partial sums avoiding indices that are multiples of $3$ first and then partial sums avoiding indices that are multiples of 2. The final numbers are simply the cubes.

The general conjecture is that $S_{k,k}(n) = n^k$. There are more examples with details in this reddit post.

I think I can prove it but it seems rather complicated and the result amazingly simple. Is there a simple proof for this fact, perhaps from the theory of finite differences?

Asvin
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See my comment here. Specifically, we have that the generating function of $n^k$ is given by $$\operatorname{Li}_{-k}(z) = \sum_{j=0}^k S(k, j) z^j j! / (1-z)^{j+1}$$. This implies that $$n^k = \sum_{j=0}^k S(k, j) \binom{n}{j} j! = \sum_{j=0}^k S(k, j) n! / (n-j)!,$$ which is a sum with $k$ terms each an integer multiple of a falling factorial function (finite, degree-$j$ polynomial in $n$). So we can write the perfect power as a sum of $k$ perfect powers each with predictable integer coefficients involving the Stirling numbers of both kinds: $$n^{k} = \sum_{m=0}^k\left(\sum_{j=0}^k S(k, j) s(j, m)\right) n^m.$$

mds
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