%I #29 Jul 30 2024 01:52:09
%S 0,0,1,0,1,0,1,1,1,0,1,0,2,0,2,0,1,1,1,0,1,1,2,0,2,0,1,1,2,0,1,0,2,0,
%T 2,1,2,1,1,0,1,0,3,0,2,0,1,1,2,1,1,1,2,0,2,0,2,1,1,0,2,0,2,1,2,0,2,1,
%U 2,0,2,0,3,0,2,0,1,2,1,0,1,1,2,0,4,1,1,1
%N Number of divisors of n of the form 5k+2; a(0) = 0.
%H T. D. Noe, <a href="/A001877/b001877.txt">Table of n, a(n) for n = 0..10000</a>
%H R. A. Smith and M. V. Subbarao, <a href="https://doi.org/10.4153/CMB-1981-005-3">The average number of divisors in an arithmetic progression</a>, Canadian Mathematical Bulletin, Vol. 24, No. 1 (1981), pp. 37-41.
%F G.f.: Sum_{n>=0} x^(5n+2)/(1-x^(5n+2)).
%F G.f.: Sum_{n>=1} x^(2*n)/(1-x^(5*n)). - _Joerg Arndt_, Jan 30 2011
%F Sum_{k=1..n} a(k) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(2,5) - (1 - gamma)/5 = A256780 - (1 - A001620)/5 = 0.105832... (Smith and Subbarao, 1981). - _Amiram Eldar_, Nov 25 2023
%t Join[{0}, Table[d = Divisors[n]; Length[Select[d, Mod[#, 5] == 2 &]], {n, 100}]] (* _T. D. Noe_, Aug 10 2012 *)
%t Table[Count[Divisors[n],_?(Mod[#,5]==2&)],{n,0,90}] (* _Harvey P. Dale_, May 20 2017 *)
%o (PARI) a(n) = if (n==0, 0, sumdiv(n, d, (d % 5)==2)); \\ _Michel Marcus_, Feb 28 2021
%Y Cf. A001876, A001878, A001899.
%Y Cf. A001620, A256780.
%K nonn,easy
%O 0,13
%A _N. J. A. Sloane_