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%I #18 Feb 28 2023 02:05:47
%S 1,5,5,5,5,25,5,5,5,25,5,25,5,25,25,5,5,25,5,25,25,25,5,25,5,25,5,25,
%T 5,125,5,5,25,25,25,25,5,25,25,25,5,125,5,25,25,25,5,25,5,25,25,25,5,
%U 25,25,25,25,25,5,125,5,25,25,5,25,125,5,25,25,125,5,25,5,25,25,25,25,125
%N a(n) = Sum_{d|n} mu(d)^2*tau(d)^2.
%C More generally : sum(d|n, mu(d)^2*tau(d)^m) = (2^m+1)^omega(n).
%H Antti Karttunen, <a href="/A082476/b082476.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>
%F a(n) = 5^omega(n); multiplicative with a(p^e)=5.
%F a(n) = abs(sum(d|n, mu(d)*tau_3(d^2))), where tau_3 is A007425. - _Enrique Pérez Herrero_, Mar 29 2010
%F a(n) = tau_5(rad(n)) = A061200(A007947(n)). - _Enrique Pérez Herrero_, Jun 24 2010
%F a(n) = A000351(A001221(n)). - _Antti Karttunen_, Jul 26 2017
%F From _Vaclav Kotesovec_, Feb 28 2023: (Start)
%F Dirichlet g.f.: Product_{primes p} (1 + 5/(p^s - 1)).
%F Dirichlet g.f.: zeta(s)^5 * Product_{primes p} (1 - 10/p^(2*s) + 20/p^(3*s) - 15/p^(4*s) + 4/p^(5*s)), (with a product that converges for s=1). (End)
%t tau[1, n_] := 1; SetAttributes[tau, Listable];
%t tau[k_, n_] := Plus @@ (tau[k - 1, Divisors[n]]) /; k > 1;
%t A082476[n_] := Abs[DivisorSum[n, MoebiusMu[ # ]*tau[3, #^2] &]]; (* _Enrique Pérez Herrero_, Mar 29 2010 *)
%t (* or more easy *)
%t A082476[n_] := 5^PrimeNu[n] (* _Enrique Pérez Herrero_, Mar 29 2010 *)
%o (PARI) a(n)=5^omega(n)
%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (4*X+1)/(1-X))[n], ", ")) \\ _Vaclav Kotesovec_, Feb 28 2023
%Y Cf. A000005, A000351, A001221, A007425, A007947, A008683, A061200, A074816.
%K mult,nonn
%O 1,2
%A _Benoit Cloitre_, Apr 27 2003
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