A117209 G.f. A(x) satisfies 1/(1-x) = Product_{k>=1} A(x^k).
1, 1, 0, -1, -1, -1, 0, 0, 0, 0, 1, 0, 0, -1, 0, 1, 2, -1, -1, -2, 0, 1, 3, -1, 0, -1, 1, -1, 1, -3, 1, -1, 1, -2, 3, 0, 6, -1, -1, -6, 2, -4, 4, -3, 2, -4, 6, -5, 6, -2, 7, -5, 4, -13, 5, -3, 11, -6, 8, -14, 10, -6, 9, -14, 11, -14, 15, -13, 9, -15, 24, -13, 19, -21, 12, -20, 27, -24, 21, -26, 22, -24, 33, -33, 32, -26
Offset: 0
Keywords
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Paul D. Hanna)
- N. J. A. Sloane, Transforms
Crossrefs
Programs
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Mathematica
nmax = 85; CoefficientList[ Series[ Product[ (1 - x^k)^(-MoebiusMu[k]), {k, 1, nmax} ], {x, 0, nmax} ], x ] (* Stuart Clary, Apr 15 2006 *) -
PARI
{a(n)=polcoeff(exp(sum(k=1,n+1,sumdiv(k,d,d*moebius(d))*x^k/k)+x*O(x^n)),n)}
Formula
G.f.: A(x) = exp( Sum_{n>=1} A023900(n)*x^n/n ), where A023900 is the Dirichlet inverse of Euler totient function.
Euler transform of the Möbius function A008683. - Stuart Clary, Franklin T. Adams-Watters and Vladeta Jovovic, Apr 15 2006
G.f.: A(x) = Product_{k>=1}(1 - x^k)^(-mu(k)) where mu(k) is the Möbius function, A008683. - Stuart Clary and Franklin T. Adams-Watters, Apr 15 2006
G.f.: A(x) = Product_{k>=1} (1 + x^(2*k-1))^mu(2*k-1), where mu() is the Moebius function. - Seiichi Manyama, Jul 06 2024
Comments