login
A364357
Number of divisors of n of the form 3*k+2 that are at most sqrt(n).
2
0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 1, 1, 0, 1, 0, 2, 0, 1, 0, 2, 1, 1, 0, 1, 0, 2, 0, 2, 0, 1, 1, 1, 0, 1, 0, 3, 0, 1, 0, 1, 1, 1, 0, 2, 0, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 2
OFFSET
1,30
LINKS
FORMULA
G.f.: Sum_{k>=0} x^((3*k+2)^2) / (1 - x^(3*k+2)).
MAPLE
N:= 100: # for a(1) .. a(N)
M:= floor((sqrt(N)-3)/2):
G:= series(add(x^((3*k+2)^2)/(1-x^(3*k+2)), k=0..M), x, N+1):
seq(coeff(G, x, i), i=1..N); # Robert Israel, Jun 05 2024
MATHEMATICA
Table[Count[Divisors[n], _?(# <= Sqrt[n] && MemberQ[{2}, Mod[#, 3]] &)], {n, 100}]
nmax = 100; CoefficientList[Series[Sum[x^(3 k + 2)^2/(1 - x^(3 k + 2)), {k, 0, nmax}], {x, 0, nmax}], x] // Rest
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Jul 21 2023
STATUS
approved