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%I #15 Aug 23 2023 08:41:56
%S 1,2,2,2,2,2,2,3,2,2,2,2,2,4,2,2,2,2,2,4,2,2,2,2,3,4,2,2,2,2,2,4,2,2,
%T 2,4,2,4,2,2,2,2,2,4,2,2,4,2,2,4,2,3,2,2,2,4,2,4,2,2,2,4,2,2,2,2,2,6,
%U 4,2,2,2,2,4,2,2,2,2,2,6,2,2,2,4,2,4,2,2,3,2
%N Number of divisors of 5*n-4 of form 5*k+1.
%H Seiichi Manyama, <a href="/A359238/b359238.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A001876(5*n-4).
%F G.f.: Sum_{k>0} x^k/(1 - x^(5*k-4)).
%t a[n_] := DivisorSum[5*n-4, 1 &, Mod[#, 5] == 1 &]; Array[a, 100] (* _Amiram Eldar_, Aug 23 2023 *)
%o (PARI) a(n) = sumdiv(5*n-4, d, d%5==1);
%o (PARI) my(N=100, x='x+O('x^N)); Vec(sum(k=1, N, x^k/(1-x^(5*k-4))))
%Y Cf. A001876, A359233, A359236, A359237.
%K nonn,easy
%O 1,2
%A _Seiichi Manyama_, Dec 22 2022
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