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%I #18 Sep 08 2022 08:46:24
%S 1,5,9,21,25,45,49,85,81,125,121,189,169,245,225,341,289,405,361,525,
%T 441,605,529,765,625,845,729,1029,841,1125,961,1365,1089,1445,1225,
%U 1701,1369,1805,1521,2125,1681,2205,1849,2541,2025,2645,2209,3069,2401,3125,2601,3549,2809,3645,3025
%N G.f. A(x) satisfies: A(x) = A(x^2) + x * (1 + x) / (1 - x)^3.
%H Amiram Eldar, <a href="/A328407/b328407.txt">Table of n, a(n) for n = 1..10000</a>
%F G.f.: Sum_{k>=0} x^(2^k) * (1 + x^(2^k)) / (1 - x^(2^k))^3.
%F G.f.: (1/3) * Sum_{k>=1} J_2(2*k) * x^k / (1 - x^k), where J_2() is the Jordan function (A007434).
%F Dirichlet g.f.: zeta(s-2) / (1 - 2^(-s)).
%F a(2*n) = a(n) + 4*n^2, a(2*n+1) = (2*n + 1)^2.
%F a(n) = Sum_{d|n} A209229(n/d) * d^2.
%F Product_{n>=1} (1 + x^n)^a(n) = g.f. for A023871.
%F Sum_{k=1..n} a(k) ~ 8*n^3/21. - _Vaclav Kotesovec_, Oct 15 2019
%F Multiplicative with a(2^e) = (4^(e+1)-1)/3, and a(p^e) = p^(2*e) for an odd prime p. - _Amiram Eldar_, Oct 25 2020
%t nmax = 55; CoefficientList[Series[Sum[x^(2^k) (1 + x^(2^k))/(1 - x^(2^k))^3, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
%t a[n_] := If[EvenQ[n], a[n/2] + n^2, n^2]; Table[a[n], {n, 1, 55}]
%t Table[DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] #^2 &], {n, 1, 55}]
%t f[p_, e_] := If[p == 2, (4^(e + 1) - 1)/3, p^(2*e)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* _Amiram Eldar_, Oct 25 2020 *)
%o (Magma) [n eq 1 select 1 else IsOdd(n) select n^2 else Self(n div 2)+n^2 :n in [1..55]]; // _Marius A. Burtea_, Oct 15 2019
%Y Cf. A000290, A001511, A007434, A016754, A023871, A129527, A209229, A328408.
%K nonn,mult,easy
%O 1,2
%A _Ilya Gutkovskiy_, Oct 14 2019