I asked George Andrews about this problem and this was part of his reply:
"Define $g_1(n)=2n-1$ and $g_m(n)=g_{m-1}(n)(g_{m-1}(n)+1)$
Thus
$g_1(n):1,3,5,7,9,11,13,15,...$
$g_2(n):2,12,30,56,90,132,182,...$
$g_3(n):6,156,930,3192,8190,17556,...$
$g_4(n):42,24492,865830,...$
I claim that the sequence for $a$ consists of all the values of $g_m(n)$ for $m\geq 1, n\geq 1$."
