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For arbitrary coefficients $a_{n,k}$, I am attempting to group by equivalent powers of $x$.

$$\sum_{n=1}^{N} \sum_{k=0}^{n} a_{n,k} x^{n+k}=\sum_{n=1}^{2N} c_n x^n$$

I wish to express $c_n$ as a sum of finitely many $a_{n,k}$.

Thank you all very much.

Talmsmen
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1 Answers1

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$$ \sum_{n=0}^{N} \sum_{k=0}^{N} a_{n,k} x^{n+k}=\sum_{i=0}^{2N} \left(\sum_{n+k=i} a_{n,k}\right) x^i $$

Leox
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  • Do you have a simplification for $\sum_{n+k=i}$? My difficulty primarily resides in the efficient characterization of all solutions for $n+k=i$. I had entertained an analogy of counting coefficients in a binomial expansion, but I failed to make substantial progress. – Talmsmen May 05 '22 at 18:00
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    $$ \sum_{n+k=i}a_{n,k}=\sum_{n=0}^i a_{n,i-n}$$ – Leox May 05 '22 at 18:30
  • That was a clever use of a inhomogeneous Diophantine equation. Thank you. – Talmsmen May 05 '22 at 18:35