The equality of the generalized sums S~k,j(n)\tilde{S}_{k,j}(n) and Sk,j(n)S_{k,j}(n)

From papers

Let xr:=r2k+1x_r:=r^{2k+1}, let Cj,q(n)C_{j,q}(n) be the coefficients defined by the paper, and define

sk,j(n):=q=0j(n1)Cj,q(n)xj+q.s_{k,j}(n):=\sum_{q=0}^{j(n-1)}C_{j,q}(n)x_{j+q}.

For integers j1j\geq 1, let

S~k,j(n):={λ}Bj,jn{λ12k+1+(λ2n)2k+1++(λjjn+n)2k+1},\tilde{S}_{k,j}(n):=\sum_{\{\lambda\}\in B_{j,jn}}\left\{\lambda_1^{2k+1}+(\lambda_2-n)^{2k+1}+\cdots+(\lambda_j-jn+n)^{2k+1}\right\},

where Bj,jn:={λk:1λ1λjjn}B_{j,jn}:=\{\lambda_k:1\leq\lambda_1\leq\cdots\leq\lambda_j\leq jn\}, and set

Sk,j(n):=q=0j1(j(n+1)q)sk,jq(n).S_{k,j}(n):=\sum_{q=0}^{j-1}{j(n+1)\choose q}s_{k,j-q}(n).

The generalized-sums equality. For any k0k\geq 0 and j1j\geq 1, one has

S~k,j(n)=Sk,j(n).\tilde{S}_{k,j}(n)=S_{k,j}(n).

The equality is proved in the paper for the particular case j=2j=2, while the general case is presented as a conjecture.

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Sources & referencesView supporting material

Primary source

Andrei K. Svinin, “Conjectures involving a generalization of the sums of powers of integers”, arXiv:1610.05387 (2017).

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