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If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)*(k_j + 1)).
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%I #18 Sep 17 2023 10:00:33

%S 1,6,8,9,12,48,16,12,12,72,24,72,28,96,96,15,36,72,40,108,128,144,48,

%T 96,18,168,16,144,60,576,64,18,192,216,192,108,76,240,224,144,84,768,

%U 88,216,144,288,96,120,24,108,288,252,108,96,288,192,320,360,120,864,124,384,192,21,336,1152,136,324

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

%H Andrew Howroyd, <a href="/A304412/b304412.txt">Table of n, a(n) for n = 1..1000</a>

%H Ilya Gutkovskiy, <a href="/A304412/a304412.jpg">Scatter plot of a(n) up to n=50000</a>.

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.

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

%F a(n) = A000005(n)*A048250(n) = A000005(n)*A000203(A007947(n)).

%F a(p^k) = (p + 1)*(k + 1) where p is a prime and k > 0.

%F a(n) = 2^omega(n)*Product_{p|n} (p + 1) if n is a squarefree (A005117), where omega() = A001221.

%F Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 + 2/p^(s-1) - 1/p^(2*s-1)). - _Amiram Eldar_, Sep 17 2023

%e a(36) = a(2^2*3^2) = (2 + 1)*(2 + 1) * (3 + 1)*(2 + 1) = 108.

%t a[n_] := Times @@ ((#[[1]] + 1) (#[[2]] + 1) & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 68}]

%t Table[DivisorSigma[0, n] Total[Select[Divisors[n], SquareFreeQ]], {n, 68}]

%o (PARI) a(n)={numdiv(n)*sumdiv(n, d, moebius(d)^2*d)} \\ _Andrew Howroyd_, Jul 24 2018

%Y Cf. A000005, A000026, A000040, A000203, A000302 (numbers n such that a(n) is odd), A001221, A001615, A003959, A005117, A007947, A008864, A045967, A048250, A064549, A064840, A304407, A304408, A304409, A304411.

%K nonn,easy,mult

%O 1,2

%A _Ilya Gutkovskiy_, May 12 2018