A190585 E.g.f. Product_{n>=1} (1 - x^n)^(-u(n)/n) where u(n) is the unitary Moebius function (A076479).

1, 1, 1, 1, -5, -29, -89, -209, -9239, -120455, -801359, -3674879, 15450931, 505760971, 4925214295, 30957618511, -3280733667119, -49063880680079, -327527326905119, -1087577476736255, 97366167074820331, 1723137650565888691, 13360549076712501511

Offset: 0

Author

Joerg Arndt, May 13 2011

Keywords

• sign

Comments

The corresponding sequence for the (usual) Moebius function is the constant sequence a(n)=1 (A000012).

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.

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.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..65

Crossrefs

Cf. A076479.

Programs

PARI

N=66;  /* that many terms */
/* First compute the unitary Moebius function */
mu=vector(N); mu[1]=1;
{ for (n=2,N,
    s = 0;
    fordiv (n,d,
        if (gcd(d,n/d)!=1, next() ); /* unitary divisors only */
        s += mu[d];
    );
    mu[n] = -s;
); };
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 +-... */
Vec(serlaplace(egf)) /* show terms */