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%I #16 Jan 03 2021 14:07:45
%S 1,1,1,1,-5,-29,-89,-209,-9239,-120455,-801359,-3674879,15450931,
%T 505760971,4925214295,30957618511,-3280733667119,-49063880680079,
%U -327527326905119,-1087577476736255,97366167074820331,1723137650565888691,13360549076712501511
%N E.g.f. Product_{n>=1} (1 - x^n)^(-u(n)/n) where u(n) is the unitary Moebius function (A076479).
%C The corresponding sequence for the (usual) Moebius function is the constant sequence a(n)=1 (A000012).
%C Log(e.g.f.) = x - (1/4)*x^4 - (1/4)*x^8 - (1/9)*x^9 - (3/16)*x^16 - (1/25)*x^25 - (2/27)*x^27 - (1/8)*x^32 + (1/36)*x^36 - (1/49)*x^49 - (5/64)*x^64 +- ...; the corresponding function for the usual Moebius function is log(exp(x)) = x.
%C Log(g.f.) = x + (1/2)*x^2 + (1/3)*x^3 - (23/4)*x^4 - (119/5)*x^5 - (359/6)*x^6 - (839/7)*x^7 +- ...; the corresponding function for the usual Moebius function if Sum_{n>=1} h(n)*x^n where h(n) = Sum_{k=1..n} 1/k is a harmonic number.
%H Vincenzo Librandi, <a href="/A190585/b190585.txt">Table of n, a(n) for n = 0..65</a>
%o (PARI)
%o N=66; /* that many terms */
%o /* First compute the unitary Moebius function */
%o mu=vector(N); mu[1]=1;
%o { for (n=2,N,
%o s = 0;
%o fordiv (n,d,
%o if (gcd(d,n/d)!=1, next() ); /* unitary divisors only */
%o s += mu[d];
%o );
%o mu[n] = -s;
%o ); };
%o egf=prod(n=1,N,(1-x^n)^(-mu[n]/n)); /* = 1 +x +1/2*x^2 +1/6*x^3 -5/24*x^4 +-... */
%o Vec(serlaplace(egf)) /* show terms */
%Y Cf. A076479.
%K sign
%O 0,5
%A _Joerg Arndt_, May 13 2011
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