%I #34 Dec 30 2022 03:55:41

%S 1,3,4,6,6,12,8,12,13,18,12,24,14,24,24,24,18,39,20,36,32,36,24,48,31,

%T 42,40,48,30,72,32,48,48,54,48,78,38,60,56,72,42,96,44,72,78,72,48,96,

%U 57,93,72,84,54,120,72,96,80,90,60,144,62,96,104,96,84,144,68

%N Sum of divisors d of n such that n/d is not congruent to 0 mod 4.

%H Seiichi Manyama, <a href="/A285895/b285895.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f.: Sum_{k>=1} k*x^k*(1 + x^k + x^(2*k))/(1 - x^(4*k)). - _Ilya Gutkovskiy_, Sep 12 2019

%F a(n) = A050460(n) + A002131(n/2) + A050464(n), where A002131(.)=0 for non-integer argument. - _R. J. Mathar_, May 25 2020

%F From _Amiram Eldar_, Oct 30 2022: (Start)

%F Multiplicative with a(2^e) = 3*2^(e-1) and a(p^e) = (p^(e+1)-1)/(p-1) if p > 2.

%F Sum_{k=1..n} a(k) ~ c * n^2, where c = 5*Pi^2/64 = 0.7710628... . (End)

%F Dirichlet g.f.: zeta(s)*zeta(s-1)*(1-1/4^s). - _Amiram Eldar_, Dec 30 2022

%e The divisors of 8 are 1, 2, 4, and 8. 8/1 == 0 (mod 4) and 8/2 == 0 (mod 4). Hence, a(8) = 4 + 8 = 12.

%t f[p_, e_] := If[p == 2, 3*2^(e-1), (p^(e+1)-1)/(p-1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Oct 30 2022 *)

%o (PARI) a(n)=sumdiv(n, d, if(n/d%4, d, 0)); \\ _Andrew Howroyd_, Jul 20 2018

%Y Cf. A002131 (k=2), A078708 (k=3), this sequence (k=4), A285896 (k=5).

%Y Cf. A002131, A050460, A050464.

%K nonn,mult

%O 1,2

%A _Seiichi Manyama_, Apr 28 2017