A284097 a(n) = Sum_{d|n, d == 1 (mod 5)} d.
1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 12, 7, 1, 1, 1, 17, 1, 7, 1, 1, 22, 12, 1, 7, 1, 27, 1, 1, 1, 7, 32, 17, 12, 1, 1, 43, 1, 1, 1, 1, 42, 28, 1, 12, 1, 47, 1, 23, 1, 1, 52, 27, 1, 7, 12, 57, 1, 1, 1, 7, 62, 32, 22, 17, 1, 84, 1, 1, 1, 1, 72, 43, 1, 1, 1, 77, 12, 33, 1
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] == 1, d, 0], {d, Divisors[n]}], {n, 80}] (* Indranil Ghosh, Mar 21 2017 *)
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PARI
for(n=1, 82, print1(sumdiv(n, d, if(Mod(d, 5)==1, 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==1]) # Indranil Ghosh, Mar 21 2017
Formula
G.f.: Sum_{k>=0} (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