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%I #31 Mar 14 2023 11:35:52
%S 1,3,3,7,3,9,3,15,7,9,3,21,3,9,9,31,3,21,3,21,9,9,3,45,7,9,15,21,3,27,
%T 3,63,9,9,9,49,3,9,9,45,3,27,3,21,21,9,3,93,7,21,9,21,3,45,9,45,9,9,3,
%U 63,3,9,21,127,9,27,3,21,9,27,3,105,3,9,21,21,9
%N Inverse Moebius transform of A061142.
%C a(n) is multiplicative with a(p^e) = -1 + 2^(e+1).
%C If s is squarefree then a(s) = A048691(s).
%C More generally: Let a_q(n) be multiplicative with a_q(p^e) = (q^(e+1)-1)/ (q-1) for prime p, e >= 0 and some fixed integer q. Then a_q(n) is the inverse Moebius transform of the completely multiplicative sequence b_q(n) = q^bigomega(n) with b_q(p) = q and b_q(1) = 1. For q = 1 see a_q(n) = A000005(n) and b_q(n) = A000012(n), for q = 0 see a_q(n) = A000012(n) and b_q(n) = A000007(n) with offset 1, and for q = -1 see a_q(n) = A010052(n) with offset 1 and b_q(n) = A008836(n). - _Werner Schulte_, Feb 20 2019
%H Enrique Pérez Herrero, <a href="/A182139/b182139.txt">Table of n, a(n) for n = 1..5000</a>
%F a(n) = Sum_{d|n} 2^Omega(d) = Sum_{d|n} 2^A001222(d) = Sum_{d|n} A061142(d).
%F Dirichlet g.f.: zeta(s)^3 * Product_{p prime} 1/(1 - 1/(p^s - 1)^2).
%e a(12) = a(2^2 * 3^1) = (-1 + 2^(2+1)) * (-1 + 2^(1+1)) = 7 * 3 = 21; or, using the divisors set {1,2,3,4,6,12}: 2^0 + 2^1 + 2^1 + 2^2 + 2^2 + 2^3 = 21.
%t t[n_] := DivisorSum[n, 2^PrimeOmega[#]&]; Table[t[n], {n,100}]
%o (PARI) for(n=1, 100, print1(direuler(p=2, n, 1/(1 - X)/(1 - 2*X))[n], ", ")) \\ _Vaclav Kotesovec_, Mar 14 2023
%Y Cf. A001222, A061142, A048691.
%K nonn,mult
%O 1,2
%A _Enrique Pérez Herrero_, Apr 14 2012
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