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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 whether the sum of the binary digits of the exponents in the prime decomposition of n is even or odd.
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10
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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
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OFFSET
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1,1
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COMMENTS
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Also analog of Liouville's function (A008836) in Fermi-Dirac arithmetic, where the terms of A050376 play the role of primes (see examples). - Vladimir Shevelev, Oct 28 2013.
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REFERENCES
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V. S. Shevelev, Multiplicative functions in the Fermi-Dirac arithmetic, Izvestia Vuzov of the North-Caucasus region, Nature sciences 4 (1996), 28-43 (in Russian)
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LINKS
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FORMULA
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Sum_{d runs through i-divisors of n} a(d)=1 if n=1, or 0 if n>1; Sum_{d runs through i-divisors of n} a(d)/d = A091732(n)/n.
Infinitary Moebius inversion:
If Sum_{d runs through i-divisors of n} f(d)=F(n), then f(n) = Sum_{d runs through i-divisors of n} a(d)*F(n/d). (End)
Let k=k(n) be the number of terms of A050376 that divide n with odd maximal exponent. Then a(n) = (-1)^k. For example, if n=96, then the maximal exponent of 2 that divides 96 is 5, for 3 it is 1, for 4 it is 2, for 16 it is 1. Thus k(96)=3 and a(96)=-1. - Vladimir Shevelev, Oct 28 2013
a(n^2) = a(n).
(End)
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EXAMPLE
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G.f. = x - x^2 - x^3 - x^4 - x^5 + x^6 - x^7 + x^8 - x^9 + x^10 - x^11 + x^12 + ...
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.
A unique representation of 48 over distinct terms of A050376 is 3*16. Since it contains even factors, then a(48)=1; for 54 such a representation is 2*3*9, thus a(54)=-1. - Vladimir Shevelev, Oct 28 2013
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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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PROG
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(PARI) {a(n) = my(A, p, e); if( n<1, 0, A = factor(n); prod(k=1, matsize(A)[1], [p, e] = A[k, ]; (-1) ^ subst( Pol( binary(e)), x, 1)))}; /* Michael Somos, Jan 08 2008 */
(PARI) a(n) = if (n==1, 1, (-1)^omega(core(n)) * a(core(n, 1)[2])) \\ Peter Munn, Mar 16 2022
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CROSSREFS
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KEYWORD
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sign,mult,easy
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AUTHOR
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STATUS
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approved
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