Search: id:A064179 Results 1-1 of 1 results found. %I A064179 %S A064179 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, %T A064179 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, %U A064179 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1 %V A064179 1,-1,-1,-1,-1,1,-1,1,-1,1,-1,1,-1,1,1,-1,-1,1,-1,1,1,1,-1,-1,-1,1,1,1, -1,-1,-1,1,1,1, %W A064179 1,1,-1,1,1,-1,-1,-1,-1,1,1,1,-1,1,-1,1,1,1,-1,-1,1,-1,1,1,-1,-1,-1,1, 1,1,1,-1,-1,1,1, %X A064179 -1,-1,-1,-1,1,1,1,1,-1,-1,1,-1,1,-1,-1,1,1,1,-1,-1,-1,1,1,1,1,1,-1,-1, 1,1,1,-1,-1,-1 %N A064179 Infinitary version of Moebius function: Infinitary_MoebiusMu of n, equal to mu(n) iff mu(n) differs from zero, else 1 or -1 depending on the sum of the binary digits of the exponents in the prime decomposition of n being even or odd. %e A064179 mu[45]=0 but iMoebiusMu[45]=1 because 45 = 3^2 * 5^1 and the binary digits of 2 and 1 add up to 2, an even number. %t A064179 iMoebiusMu[n_] := Switch[MoebiusMu[n], 1, 1, -1, -1, 0, If[OddQ[Plus@@(DigitCount[Last[Transpose[FactorIntege\ r[n]]], 2, 1])], -1, 1]]; %t A064179 The Moebius inversion formula seems to hold for iMoebiusMu and the infinitary_divisors of n: if g[ n_ ] := Plus@@(f/@iDivisors[ n ]) for all n, then f[ n_ ]===Plus@@(iMoebiusMu[ # ]g[ n/# ])/@iDivisors[ n ]) %o A064179 (PARI) {a(n) = local(A, p, e); if( n<1, 0, A = factor(n); prod(k=1, matsize(A)[1], if(p = A[k,1], e = A[k,2]; (-1) ^ subst(Pol( binary(e)), x, 1))))} /* Michael Somos Jan 08 2008 */ %Y A064179 Cf. A064175, A064176, A000028, A000379. %Y A064179 Sequence in context: A114523 A000012 A008836 this_sequence A106400 A112865 A121241 %Y A064179 Adjacent sequences: A064176 A064177 A064178 this_sequence A064180 A064181 A064182 %K A064179 sign %O A064179 1,1 %A A064179 Wouter Meeussen (wouter.meeussen(AT)pandora.be), Sep 20 2001 Search completed in 0.001 seconds