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I conjecture that $$a(1,2^{n-1}+n)+\sum\limits_{i=1}^{2^{n-1}}a(i+2,2^{n-1}+n)=(2^{n+1}+1)2^{2^{n-1}+n-1}-1$$ I also conjecture that numbers of the form $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ have only $2$ partitions …

Let $a(n)$ be A227559, i.e., number of partitions of $n$ into distinct parts with boundary size $2$. Be careful here: offset is $3$. …

Let $a(n,m)$ be the number of partitions of $m^n-1$ into powers of $m$. …

Let $a(n)$ be A290383 i.e. number of set partitions of $[n]$ such that the smallest element of each block is odd. …

See also A091891 (number of partitions of $n$ into parts which are a sum of exactly as many distinct powers of $2$ as $n$ has $1$'s in its binary representation) and A091892 (numbers $k$ having only one …

Let $p(n)$ be A000041 i.e. the number of partitions of $n$ (the partition numbers). …

Let $a(n)$ be A309099 i.e. the number of partitions of $n$ avoiding the partition $(4,3,1)$. … For example, the only partitions avoiding $(2,1)$ are those whose Ferrers boards are rectangles. …

Let $p(n)$ be A000041 i.e. number of partitions of $n$ (the partition numbers). …

Let $a(n)$ be A225114 (i.e., number of skew partitions of $n$ whose diagrams have no empty rows and columns). …

Let $a(n)$ be A261041 (i.e., number of partitions of subsets of $\{1,2,\dotsc,n\}$, where consecutive integers are required to be in different parts). …