A284103 a(n) = Sum_{d|n, d == 4 (mod 5)} d.
0, 0, 0, 4, 0, 0, 0, 4, 9, 0, 0, 4, 0, 14, 0, 4, 0, 9, 19, 4, 0, 0, 0, 28, 0, 0, 9, 18, 29, 0, 0, 4, 0, 34, 0, 13, 0, 19, 39, 4, 0, 14, 0, 48, 9, 0, 0, 28, 49, 0, 0, 4, 0, 63, 0, 18, 19, 29, 59, 4, 0, 0, 9, 68, 0, 0, 0, 38, 69, 14, 0, 37, 0, 74, 0, 23, 0, 39, 79
Offset: 1
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
Table[Sum[If[Mod[d, 5] == 4, d, 0], {d, Divisors[n]}], {n, 79}] (* Indranil Ghosh, Mar 21 2017 *) Table[Total[Select[Divisors[n],Mod[#,5]==4&]],{n,80}] (* Harvey P. Dale, Sep 24 2024 *)
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PARI
for(n=1, 79, print1(sumdiv(n, d, if(Mod(d, 5)==4, d, 0)), ", ")) \\ Indranil Ghosh, Mar 21 2017
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Python
from sympy import divisors def a(n): return sum([d for d in divisors(n) if d%5==4]) # Indranil Ghosh, Mar 21 2017
Formula
G.f.: Sum_{k>=1} (5*k - 1)*x^(5*k-1)/(1 - x^(5*k-1)). - Ilya Gutkovskiy, Mar 21 2017
Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/60 = 0.164493... (A013661 / 10). - Amiram Eldar, Nov 26 2023