%I #22 Oct 08 2023 09:05:42
%S 1,1,1,3,1,1,1,3,4,1,1,3,1,1,1,3,1,4,1,3,1,1,1,3,6,1,4,3,1,1,1,3,1,1,
%T 1,12,1,1,1,3,1,1,1,3,4,1,1,3,8,6,1,3,1,4,1,3,1,1,1,3,1,1,4,11,1,1,1,
%U 3,1,1,1,12,1,1,6,3,1,1,1,3,4,1,1,3,1,1
%N The sum of exponentially odd divisors of the square root of the largest square dividing n.
%C First differs from A295295 at n = 64.
%C The sum of divisors of the square root of the largest square dividing n is A069290(n).
%C The number of these divisors is A365335(n).
%H Amiram Eldar, <a href="/A365336/b365336.txt">Table of n, a(n) for n = 1..10000</a>
%H Vaclav Kotesovec, <a href="/A365336/a365336_1.jpg">Graph - the asymptotic ratio (1000000 terms)</a>
%F a(n) = A033634(A000188(n)).
%F a(n) = 1 if and only if n is squarefree (A005117).
%F Multiplicative with a(p^e) = (p^(2*floor((e+2)/4) + 1) - p)/(p^2 - 1) + 1. [corrected by _Georg Fischer_, Oct 07 2023]
%F Dirichlet g.f.: zeta(s) * zeta(4*s-2) * Product_{p prime} (1 + 1/p^(2*s-1) - 1/p^(4*s-2)).
%F From _Vaclav Kotesovec_, Sep 02 2023: (Start)
%F Dirichlet g.f.: zeta(s)^2 * zeta(4*s-2) * Product_{p prime} (1 - 1/p^s + 1/p^(2*s-1) - 1/p^(3*s-1) - 1/p^(4*s-2) + 1/p^(5*s-2)).
%F Dirichlet g.f.: zeta(s) * zeta(2*s-1) * zeta(4*s-2) * Product_{p prime} (1 - 2/p^(4*s-2) + 1/p^(6*s-3)).
%F Let f(s) = Product_{p prime} (1 - 2/p^(4*s-2) + 1/p^(6*s-3)), then Sum_{k=1..n} a(k) ~ Pi^2/12 * n * (f(1) * (log(n) + 3*gamma - 1 + 24*zeta'(2)/Pi^2) + f'(1)), where f(1) = Product_{p prime} (1 - 2/p^2 + 1/p^3) = A065464 = 0.42824950567709444021876..., f'(1) = f(1) * Sum_{primes p} 2*(4*p-3)*log(p) / (p^3 - 2*p + 1) = 1.617322217899181826790... and gamma is the Euler-Mascheroni constant A001620. (End)
%t f[p_, e_] := (p^(2*Floor[(e+2)/4] + 1) - p)/(p^2 - 1) + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
%o (PARI) a(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i,1]^(2*((f[i,2]+2)\4) + 1) - f[i,1])/(f[i,1]^2 - 1) + 1);}
%Y Cf. A000188, A005117, A033634, A069290, A295295, A365335.
%K nonn,easy,mult
%O 1,4
%A _Amiram Eldar_, Sep 01 2023