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Product of the number of divisors of n and the number of distinct prime divisors of n; i.e., tau(n) * omega(n).
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%I #22 Sep 19 2023 01:43:48

%S 0,2,2,3,2,8,2,4,3,8,2,12,2,8,8,5,2,12,2,12,8,8,2,16,3,8,4,12,2,24,2,

%T 6,8,8,8,18,2,8,8,16,2,24,2,12,12,8,2,20,3,12,8,12,2,16,8,16,8,8,2,36,

%U 2,8,12,7,8,24,2,12,8,24,2,24,2,8,12,12,8,24

%N Product of the number of divisors of n and the number of distinct prime divisors of n; i.e., tau(n) * omega(n).

%H G. C. Greubel, <a href="/A248577/b248577.txt">Table of n, a(n) for n = 1..5000</a>

%H Tanay Wakhare, <a href="https://arxiv.org/abs/1604.05671">Sums involving the number of distinct prime factors function</a>, arXiv:1604.05671 [math.HO], 2016-2017.

%F a(n) = A000005(n) * A001221(n).

%F If n is squarefree, then a(n) = omega(n)*2^omega(n). - _Wesley Ivan Hurt_, Jun 09 2020

%F Dirichlet g.f.: zeta(s)^2 * (2*P(s) - P(2*s)), where P(s) is the prime zeta function (Wakhare, 2016). - _Amiram Eldar_, Sep 19 2023

%e a(6) = 8; 6 has four divisors {1,2,3,6} and two distinct prime divisors {2,3}, so a(6) = 4*2 = 8.

%e a(9) = 3; 9 has three divisors {1,3,9} and 1 distinct prime divisor {3}, so a(9) = 3*1 = 3.

%e a(12) = 12; 12 has 6 divisors {1,2,3,4,6,12} and 2 distinct prime divisors {2,3}, so a(12) = 6*2 = 12.

%p with(numtheory): A248577:=n->tau(n)*nops(factorset(n)): seq(A248577(n), n=1..100);

%t Table[DivisorSigma[0, n] PrimeNu[n], {n, 100}]

%o (PARI) vector(100, n, numdiv(n)*omega(n)) \\ _Michel Marcus_, Oct 09 2014

%Y Cf. A000005 (tau), A001221 (omega).

%K nonn,easy

%O 1,2

%A _Wesley Ivan Hurt_, Oct 08 2014