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A285989 a(0) = 0, a(n) = Sum_{0 for n > 0.

%I A285989 #37 Jan 09 2023 01:46:23
%S A285989 0,1,16,82,256,626,1312,2402,4096,6643,10016,14642,20992,28562,38432,
%T A285989 51332,65536,83522,106288,130322,160256,196964,234272,279842,335872,
%U A285989 391251,456992,538084,614912,707282,821312,923522,1048576,1200644,1336352,1503652,1700608
%N A285989 a(0) = 0, a(n) = Sum_{0<d|n, n/d odd} d^4 for n > 0.
%C A285989 Multiplicative because this sequence is the Dirichlet convolution of A000035 and A000583 which are both multiplicative. - _Andrew Howroyd_, Aug 05 2018
%H A285989 Robert Israel, <a href="/A285989/b285989.txt">Table of n, a(n) for n = 0..10000</a>
%H A285989 J. W. L. Glaisher, <a href="https://books.google.com/books?id=bLs9AQAAMAAJ&amp;pg=RA1-PA1">On the representations of a number as the sum of two, four, six, eight, ten, and twelve squares</a>, Quart. J. Math. 38 (1907), 1-62 (see p. 4 and p. 8).
%H A285989 <a href="/index/Ge#Glaisher">Index entries for sequences mentioned by Glaisher</a>.
%F A285989 a(n) = A051001(n)*16^A007814(n) for n >= 1. - _Robert Israel_, Apr 30 2017
%F A285989 From _Amiram Eldar_, Nov 01 2022: (Start)
%F A285989 Multiplicative with a(2^e) = 2^(4*e) and a(p^e) = (p^(4*e+4)-1)/(p^4-1) for p > 2.
%F A285989 Sum_{k=1..n} a(k) ~ c * n^5, where c = 31*zeta(5)/160 = 0.200904... . (End)
%F A285989 Dirichlet g.f.: zeta(s)*zeta(s-4)*(1-1/2^s). - _Amiram Eldar_, Jan 08 2023
%p A285989 f:= n -> add((n/d)^4, d = numtheory:-divisors(n/2^padic:-ordp(n,2))); # _Robert Israel_, Apr 30 2017
%t A285989 {0}~Join~Table[DivisorSum[n, Mod[#, 2] (n/#)^4 &], {n, 36}] (* _Michael De Vlieger_, Aug 05 2018 *)
%o A285989 (PARI) a(n)={sumdiv(n, d, (d%2)*(n/d)^4)} \\ _Andrew Howroyd_, Aug 05 2018
%Y A285989 Sum_{0<d|n, n/d odd} d^k: A002131 (k=1), A076577 (k=2), A007331 (k=3), this sequence (k=4), A096960 (k=5), A096961 (k=7), A096962 (k=9), A096963 (k=11).
%Y A285989 Cf. A000035, A000583, A007814, A013663, A051001, A206623, A285990.
%K A285989 nonn,mult
%O A285989 0,3
%A A285989 _Seiichi Manyama_, Apr 30 2017