Here n$n$ is a positive integer and p(n)$p(n)$ is the number of unrestricted partitions. Can one always find a subset, s, of {1,2,...n} such that the number of partitions of n with parts from s is p(n)/2 if p(n) is even and is (p(n)+1)/2 or (p(n)-1)/2 if n is odd?
Can one always find a subset $s$, of $\{1,2,\ldots,n\}$ such that the number of partitions of $n$ with parts from $s$ is $p(n)/2$ if $p(n)$ is even and is $[p(n)+1]/2$ or $[p(n)-1]/2$ if $n$ is odd?
For example: if n=7$n=7$ we may choose s$s$ to be {1,3,4,5,7}$\{1,3,4,5,7\}$. The partitions of 7 which are to be counted are
7
5+1+1
4+3
3+3+1
4+1+1+1
3+1+1+1+1
1+1+1+1+1+1+1
and (p(n)-1)/2= $$ \begin{split} &7\\ &5+1+1\\ &4+3\\ &3+3+1\\ &4+1+1+1\\ &3+1+1+1+1\\ &1+1+1+1+1+1+1\\ \end{split} $$ and (15-1)/2-7$[p(n)-1]/2= (15-1)/2=7$ so we can do what was asked of us.
