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Product of the exponents in the prime factorization of n in the ring of Gaussian integers.
3

%I #25 Oct 15 2022 08:09:53

%S 1,2,1,4,1,2,1,6,2,2,1,4,1,2,1,8,1,4,1,4,1,2,1,6,4,2,3,4,1,2,1,10,1,2,

%T 1,8,1,2,1,6,1,2,1,4,2,2,1,8,2,8,1,4,1,6,1,6,1,2,1,4,1,2,2,12,1,2,1,4,

%U 1,2,1,12,1,2,4,4,1,2,1,8,4,2,1,4,1,2

%N Product of the exponents in the prime factorization of n in the ring of Gaussian integers.

%C a(n) is also the number of divisors of n in Gaussian integers that are powerful (A335851).

%H Amiram Eldar, <a href="/A335852/b335852.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/GaussianInteger.html">Gaussian Integer</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Gaussian_integer">Gaussian integer</a>.

%F Multiplicative with a(p^e) = 2*e if p = 2, e if p == 3 (mod 4) and e^2 if p == 1 (mod 4).

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = (5/2) * Product_{p prime == 3 (mod 4)} (p^2 - p + 1)/(p*(p-1)) * Product_{p prime == 3 (mod 1)} (p^4 - 3*p^3 + 6*p^2 - 5*p + 1)/(p*(p-1)^3) = 3.73805905189... . - _Amiram Eldar_, Oct 15 2022

%e a(2) = 2 since 2 = -i * (1 + i)^2 and the Gaussian prime (1 + i) has an exponent 2.

%e a(100) = 16 since 100 = (1 + i)^4 * (1 + 2*i)^2 * (2 + i)^2 and 4*2*2 = 16.

%t a[n_] := Times @@ FactorInteger[n, GaussianIntegers -> True][[All, 2]]; Array[a, 100]

%o (PARI) a(n) = my (f = factor(n*I)); f[1,1] /= I; prod(k=1, #f~, f[k,2]); \\ _Michel Marcus_, Jun 28 2020

%Y Cf. A005361, A062327, A335851.

%K nonn,mult

%O 1,2

%A _Amiram Eldar_, Jun 26 2020