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a(0) = 4; for n >= 0, a(n+1) = number of partitions of a(n).
2

%I #15 Apr 05 2021 18:45:39

%S 4,5,7,15,176,476715857290

%N a(0) = 4; for n >= 0, a(n+1) = number of partitions of a(n).

%C a(6) is too large to include (even in a b-file).

%C a(6) = 26680 54494 90278 28685 82404 ... {769101 digits} ... 72432 87250 16775 65100 60652. - _Robert G. Wilson v_, Sep 30 2009

%p a:= n-> `if`(n=0, 4, combinat[numbpart](a(n-1))):

%p seq(a(n), n=0..5); # _Alois P. Heinz_, Apr 05 2021

%t NestList[PartitionsP, 4, 5]

%o (Python)

%o from itertools import accumulate

%o from sympy.ntheory import npartitions

%o def f(an, _): return npartitions(an)

%o print(list(accumulate([4]*6, f))) # _Michael S. Branicky_, Apr 05 2021

%Y Cf. A000041, A072239.

%K nonn

%O 0,1

%A _Jeff Burch_, Jul 04 2002

%E Edited by _N. J. A. Sloane_, May 05 2007