A364357 Number of divisors of n of the form 3*k+2 that are at most sqrt(n).
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
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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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
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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
Formula
G.f.: Sum_{k>=0} x^((3*k+2)^2) / (1 - x^(3*k+2)).