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 $m\geq 1, n\geq 1$."