Let p(n) be the number of unrestricted partitions of n. p(0) is taken to be 1. Let set 1 and set 2 be two empty sets.
Here's an algorithm. Put p(n) into set 1. On each successive step, k=1,2,3,..., n, put p(n-k) into that set which has the smaller sum of elements, or into set 1 if the two sets have equal sums.
After the algorithm has ended will the sum of elements in set 1 always differ from the sum of elements in set 2 by at most 1?
I've checked for n<= 200 and this has always been the case.
Let a nondecreasing sequence (we'll say of positive integers) $a_n$ starting from $a_1=1$ be subdoubling if for all $n>2$ it satisfies the relation $a_n <= 2*a_{n-1}$. Let $d$ be an integer with $-a_n \leq d \leq a_n$ and put $d$ in set 2, then run your algorithm. I claim at the finish of the algorithm the difference between the sums is at most $a_1=1$.
Proof Sketch: at the start the difference between the two sums is $\left| d \right| \leq a_n$.
Because of the subdoubling property, the difference when $a_{n-1}$ is added at most
$a_{n-1}$. End of Proof Sketch.
Corollary: Given $-2*a_n\leq d \leq 2*a_n$, you can choose coefficients $\epsilon_i$ from $\{-1,+1\}$ so that $\sum_i \epsilon_i * a_i$ is within one of $d$.
If I recall correctly, your sequence $a_n$ is subdoubling. So you will end up with 1 even if you pad set 2 with a small enough value.
I used something like this for the determinant spectrum problem for $0-1$ matrices.
Gerhard "Ah, The Good Old Days" Paseman, 2016.08.27
