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%I #23 May 30 2021 22:05:12
%S 5,30,135,490,1575,4565,12310,31185,75175,173525,386155,831820,
%T 1741630,3554345,7089500,13848650,26543660,49995925,92669720,
%U 169229875,304801365,541946240,952050665,1653668665,2841940500,4835271870,8149040695,13610949895,22540654445,37027355200,60356673280,97662728555,156919475295
%N p(5n+4) where p(k) = number of partitions of k = A000041(k).
%C It is known that a(n) is divisible by 5 (see A071734).
%H Seiichi Manyama, <a href="/A213260/b213260.txt">Table of n, a(n) for n = 0..1000</a>
%H James Grime and Brady Haran, <a href="https://www.youtube.com/watch?v=NjCIq58rZ8I">Partitions</a>, Numberphile video (2016).
%H Lasse Winquist, <a href="http://dx.doi.org/10.1016/S0021-9800(69)80105-5">An elementary proof of p(11m+6) == 0 (mod 11)</a>, J. Combinatorial Theory 6 1969 56--59. MR0236136 (38 #4434). - From _N. J. A. Sloane_, Jun 07 2012
%F a(n) = A000041(A016897(n)). - _Omar E. Pol_, Jan 18 2013
%t Table[PartitionsP[5n+4],{n,0,40}] (* _Harvey P. Dale_, May 30 2013 *)
%o (PARI) a(n) = numbpart(5*n+4); \\ _Michel Marcus_, Jan 07 2015
%o (Python)
%o from sympy.ntheory import npartitions
%o def a(n): return npartitions(5*n+4)
%o print([a(n) for n in range(33)]) # _Michael S. Branicky_, May 30 2021
%Y Cf. A000041, A071734, A213256, A076394.
%K nonn
%O 0,1
%A _N. J. A. Sloane_, Jun 07 2012
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