login
If n = Product (p_j^k_j) then a(n) = Product (p_j + 2), with a(1) = 1.
2

%I #12 Nov 13 2022 08:39:59

%S 1,4,5,4,7,20,9,4,5,28,13,20,15,36,35,4,19,20,21,28,45,52,25,20,7,60,

%T 5,36,31,140,33,4,65,76,63,20,39,84,75,28,43,180,45,52,35,100,49,20,9,

%U 28,95,60,55,20,91,36,105,124,61,140,63,132,45,4,105,260,69,76,125,252

%N If n = Product (p_j^k_j) then a(n) = Product (p_j + 2), with a(1) = 1.

%H Amiram Eldar, <a href="/A343442/b343442.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>.

%F G.f.: Sum_{k>=1} mu(k)^2 * sigma(k) * x^k / (1 - x^k), where mu = A008683 and sigma = A000203.

%F a(n) = Sum_{d|n} mu(d)^2 * sigma(d).

%F Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/(12*zeta(3)) = 0.684216... (A335005). - _Amiram Eldar_, Nov 13 2022

%t a[1] = 1; a[n_] := Times @@ ((#[[1]] + 2) & /@ FactorInteger[n]); Table[a[n], {n, 70}]

%t nmax = 70; CoefficientList[Series[Sum[MoebiusMu[k]^2 DivisorSigma[1, k] x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x] // Rest

%o (PARI) a(n) = sumdiv(n, d, moebius(d)^2 * sigma(d)) \\ _Andrew Howroyd_, Apr 15 2021

%Y Cf. A000203, A007429, A007947, A008683, A048250, A062953, A063441, A074816, A092261, A107758, A107759, A319132, A335005.

%K nonn,mult

%O 1,2

%A _Ilya Gutkovskiy_, Apr 15 2021