Vtyjkyuk

If we define $S(n)$ as

$4=3+1=2+2=2+1+1=1+1+1+1$

$S(4)=3cdot1+2cdot2+2cdot1cdot1+1cdot1cdot1cdot1=3+4+2+1=10$

$5=4+1=3+2=2+2+1=2+1+1+1=1+1+1+1+1$

$S(5)=4cdot1+3cdot2+2cdot2cdot1+2cdot1cdot1cdot1+1cdot1cdot1cdot1cdot1=4+6+4+2+1=20$

$6=5+1=4+2=4+1+1=3+3=3+2+1=3+1+1+1=2+2+2=2+2+1+1=2+1+1+1+1=1+1+1+1+1+1$

$S(6)=5cdot1+4cdot2+4cdot1cdot1+3cdot3+3cdot2cdot1+ 3cdot1cdot1cdot1+2cdot2cdot2+2cdot2cdot1cdot1+2cdot1cdot1cdot1cdot1+1cdot1cdot1cdot1cdot1cdot1=5+8+4+9+6+3+8+4+2+1=50$

so what quantity of $n$ up to $41$ which $S(n)equiv0pmod5$?

Rephrasing the question: let $S(n) = sum_lambda, vdash , n,, lambda_1 < n prod_i lambda_i$. Then what is $S(n) bmod 5$?

As a calculation strategy, define $S(n, k) = sum_lambda,vdash,n,, lambda_1 le k prod_i lambda_i$ to be the corresponding sum over partitions of $n$ into parts no greater than $k$, so that $S(n) = S(n, n-1)$. Then we can consider the largest part in each partition and find the recursion $S(n, k) = sum_j=1^min(k,n) j S(n - j, j)$ with base case $S(0, k) = 1$ as there is one partition of $0$ and it has an empty product.

Using this recursion it is easy to write a computer program to output a table of $S(n, k)$, but neither the table nor the values of $n$ for which $S(n, n-1) equiv 0 pmod 5$ are in OEIS, so this may not have been previously studied.