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Search: id:A064179
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| A064179 |
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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. |
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+0
5
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| 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, 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, -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
(list; graph; listen)
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OFFSET
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1,1
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EXAMPLE
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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.
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MATHEMATICA
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iMoebiusMu[n_] := Switch[MoebiusMu[n], 1, 1, -1, -1, 0, If[OddQ[Plus@@(DigitCount[Last[Transpose[FactorInteger[n]]], 2, 1])], -1, 1]];
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 ])
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PROGRAM
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(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 */
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CROSSREFS
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Cf. A064175, A064176, A000028, A000379.
Sequence in context: A114523 A000012 A008836 this_sequence A106400 A112865 A121241
Adjacent sequences: A064176 A064177 A064178 this_sequence A064180 A064181 A064182
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KEYWORD
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sign
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AUTHOR
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Wouter Meeussen (wouter.meeussen(AT)pandora.be), Sep 20 2001
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