%I #17 Jul 28 2018 09:17:26
%S 1,1,3,5,12,18,40,60,121,186,344,524,955,1432,2484,3756,6352,9493,
%T 15750,23414,38128,56513,90406,133312,211194,309657,484214,708267,
%U 1097159,1597290,2454245,3560444,5430091,7854174,11894335,17151394,25838413,37145198,55648059
%N Twice partitioned numbers where the latter partitions are constant.
%C Also number of ways to choose a divisor of each part of an integer partition of n.
%H Alois P. Heinz, <a href="/A279784/b279784.txt">Table of n, a(n) for n = 0..5000</a>
%H Gus Wiseman, <a href="/A063834/a063834.txt">Sequences enumerating triangles of integer partitions</a>
%F G.f.: exp(Sum_{k>=1} Sum_{j>=1} d(j)^k*x^(j*k)/k), where d(j) is the number of the divisors of j (A000005). - _Ilya Gutkovskiy_, Jul 17 2018
%F From _Vaclav Kotesovec_, Jul 28 2018: (Start)
%F a(n) ~ c * 2^(n/2), where
%F c = 203.986136154799274492709451797084688042886818134781591... if n is even and
%F c = 201.491703180375661735217350021245093454724452720559762... if n is odd.
%F In closed form, a(n) ~ ((2 + sqrt(2)) * Product_{k>=3} (1/(1 - tau(k) / 2^(k/2))) + (-1)^n * (2 - sqrt(2)) * Product_{k>=3} (1/(1 - (-1)^k * tau(k) / 2^(k/2)))) * 2^(n/2 - 1), where tau() is A000005. (End)
%e The a(4)=12 twice-partitions are:
%e ((4)), ((3)(1)), ((2)(2)), ((22)),
%e ((2)(1)(1)), ((2)(11)), ((11)(2)),
%e ((1)(1)(1)(1)), ((11)(1)(1)), ((11)(11)), ((111)(1)), ((1111)).
%p b:= proc(n, i) option remember; `if`(n=0 or i=1, 1,
%p b(n, i-1)+`if`(i>n, 0, numtheory[tau](i)*b(n-i, i)))
%p end:
%p a:= n-> b(n$2):
%p seq(a(n), n=0..50); # _Alois P. Heinz_, Dec 20 2016
%t nn=20;CoefficientList[Series[Product[1/(1-DivisorSigma[0,n]x^n),{n,nn}],{x,0,nn}],x]
%Y Cf. A000005, A063834, A145515, A270995.
%K nonn
%O 0,3
%A _Gus Wiseman_, Dec 18 2016