|
|
A284281
|
|
a(n) = Sum_{d|n, d == 3 (mod 5)} d.
|
|
12
|
|
|
0, 0, 3, 0, 0, 3, 0, 8, 3, 0, 0, 3, 13, 0, 3, 8, 0, 21, 0, 0, 3, 0, 23, 11, 0, 13, 3, 28, 0, 3, 0, 8, 36, 0, 0, 21, 0, 38, 16, 8, 0, 3, 43, 0, 3, 23, 0, 59, 0, 0, 3, 13, 53, 21, 0, 36, 3, 58, 0, 3, 0, 0, 66, 8, 13, 36, 0, 68, 26, 0, 0, 29, 73, 0, 3, 38, 0, 94, 0, 8
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
LINKS
|
|
|
FORMULA
|
G.f.: Sum_{k>=0} (5*k + 3)*x^(5*k+3)/(1 - x^(5*k+3)). - Ilya Gutkovskiy, Mar 25 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
|
|
MATHEMATICA
|
Table[Sum[If[Mod[d, 5] == 3, d, 0], {d, Divisors[n]}], {n, 80}] (* Indranil Ghosh, Mar 24 2017 *)
|
|
PROG
|
(PARI) for(n=1, 82, print1(sumdiv(n, d, if(Mod(d, 5)==3, d, 0)), ", ")) \\ Indranil Ghosh, Mar 24 2017
(Python)
from sympy import divisors
def a(n): return sum([d for d in divisors(n) if d%5==3]) # Indranil Ghosh, Mar 24 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|