<s>[INST] Number of integer partitions $p(n,m)$ In "Integer partitions" by Andrews and Eriksson, the authors provide formulas to compute $p(n,m)$, i.e., the number of partitions of $n$ into parts less than or equal to $m$, for $m=1,2,3,4,5$. As discussed in this question, it seems that no formula for $m>5$ is known.
Howevere, the fact that $p(n,m)$ for $m\leq 5$ are upper-bounded by polynomials, made me naturally wonder whether this holds for $m>5$ or not.
This is also related to the result in Section 6.4, in which the authors show that $p(n)$ is sub-exponential.
 [/INST] Via Flajolet-Sedgewick:   $p_m(z)=\frac{1}{\left(1-z\right) \left(1-z^2\right)\text{...} \left(1-z^m\right)}$ is an ordinary generating function. For general k, the coefficients are "polynom + some small (O(1)) periodic function". Using Mathematica,
$$ p(n,5) = \frac{30 n \left(15 \left(3 (-1)^n+85\right)+n (n (n+30)+310)\right)+10125 (-1)^n+5400 \sin \left(\frac{\pi  n}{2}\right)+5400 \cos \left(\frac{\pi  n}{2}\right)+6400 \cos \left(\frac{2 \pi  n}{3}\right)+6912 \cos \left(\frac{2 \pi  n}{5}\right)+6912 \cos \left(\frac{4 \pi  n}{5}\right)+50651}{86400}$$
and for k=6
$$ \frac{(2 n+21) (2 n (n+21) (3 n (n+21)+434)+28511)}{1036800}+\frac{1}{32} \sin \left(\frac{\pi  n}{2}\right)+\frac{i (6 n (n+21)+581) \sin (\pi  n)}{4608}+\frac{1}{25} \sqrt{1+\frac{2}{\sqrt{5}}} \sin \left(\frac{2 \pi  n}{5}\right)+\frac{1}{25} \sqrt{1-\frac{2}{\sqrt{5}}} \sin \left(\frac{4 \pi  n}{5}\right)+\frac{\sin \left(\frac{2 \pi  n}{3}\right)}{81 \sqrt{3}}+\frac{1}{32} \cos \left(\frac{\pi  n}{2}\right)+\frac{1}{18} \cos \left(\frac{\pi  n}{3}\right)+\frac{1}{162} (2 n+21) \cos \left(\frac{2 \pi  n}{3}\right)+\frac{1}{25} \cos \left(\frac{2 \pi  n}{5}\right)+\frac{1}{25} \cos \left(\frac{4 \pi  n}{5}\right)+\frac{(6 n (n+21)+581) \cos (\pi  n)}{4608} $$
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