Restricted integer partition functions (Q2855590)

For two sets \(A\) and \(M\) of positive integers and for a positive integer \(n\), let \(p(n,A,M)\) denote the number of partitions of \(n\) with parts in \(A\) and multiplicities in \(M\), i.e., the number of representations of \(n\) in the form \(n=\sum_{a\in A} m_a\cdot a\) with \(m_a\in M\cup\{0\}\). \textit{E. R. Canfield} and \textit{H. S. Wilf} [Developments in Mathematics 23, 39--46 (2012; Zbl 1242.05018)] raised the following question. Do there exist two infinite sets \(A\) and \(M\) so that \(p(n,A,M)> 0\) for all sufficiently large \(n\) and yet \(p\) has polynomial growth? \textit{Ž. Ljujić} and \textit{M. B. Nathanson} [Integers 12, No. 6, 1279--1286, A11 (2012; Zbl 1272.11111)] showed that this cannot be the case if \(|A\cap\{1, 2,\dots, n\}|\geq\delta\log n\) for all sufficiently large \(n\), where \(\delta>0\) is any constant, and therefore they specialized the above question for \(A= \{k!\}^\infty_{k=1}\) and \(A= \{k^k\}^\infty_{k=1}\).NEWLINENEWLINE In the paper under review the author answers the original question positively in the strongest possible way: there are infinite sets \(A\) and \(M\) so that \(p(n,A,M)= 1\) for all \(n\). He also answers the specialized questions positively in the following more general form: Let \(A\) be an infinite set of positive integers with \(1\in A\) and \(|A\cap\{1,2,\dots, n\}|= (1+ 0(1)){\log n\over\log\log n}\), where the \(0(1)\) term tends to \(0\) and \(n\) tends to infinity. Then there exists \(n_0\) and an infinite set \(M\) of positive integers so that \(0< p(n,A,M)< n^{8+0(1)}\) for all \(n>n_0\).