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%I #29 Jun 09 2024 13:54:05
%S 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,
%T 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,
%U 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
%N Number of divisors of n of the form 3*k+2 that are at most sqrt(n).
%H Robert Israel, <a href="/A364357/b364357.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: Sum_{k>=0} x^((3*k+2)^2) / (1 - x^(3*k+2)).
%p N:= 100: # for a(1) .. a(N)
%p M:= floor((sqrt(N)-3)/2):
%p G:= series(add(x^((3*k+2)^2)/(1-x^(3*k+2)),k=0..M),x,N+1):
%p seq(coeff(G,x,i),i=1..N); # _Robert Israel_, Jun 05 2024
%t Table[Count[Divisors[n], _?(# <= Sqrt[n] && MemberQ[{2}, Mod[#, 3]] &)], {n, 100}]
%t nmax = 100; CoefficientList[Series[Sum[x^(3 k + 2)^2/(1 - x^(3 k + 2)), {k, 0, nmax}], {x, 0, nmax}], x] // Rest
%Y Cf. A001822, A038548, A364209.
%K nonn
%O 1,30
%A _Ilya Gutkovskiy_, Jul 21 2023
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