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Number of divisors of 5*n-4 of form 5*k+1.
7

%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