View source for 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
$$
D(n;a_1,\ldots,a_m) = D(n-a_m;a_1,\ldots,a_m) + D(n;a_1,\ldots,a_{m-1}) \ ?
$$
[[User:Richard Pinch|Richard Pinch]] ([[User talk:Richard Pinch|talk]]) 17:32, 2 January 2021 (UTC)