Talk:Denumerant

Euler's recurrence

The paragraph

The simplest method of computing a denumerant is by Euler's recurrence relation: $$ D(n;1,\ldots,k) - D(n-k;1,\ldots,k) = D(n;1,\ldots,k-1) \ . $$

suddenly specialises for no apparent reason to the case when the $a_i$ are $(1,\ldots,k)$. Is there any reason not to keep the context of the previous paragraphs and write $$ D(n;a_1,\ldots,a_m) = D(n-a_m;a_1,\ldots,a_m) + D(n;a_1,\ldots,a_{m-1}) \ ? $$ Richard Pinch (talk) 17:32, 2 January 2021 (UTC)