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A307380
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Number of quintic primitive Dirichlet characters modulo n.
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6
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1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0
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OFFSET
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1,11
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COMMENTS
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a(n) is the number of primitive Dirichlet characters modulo n such that all entries are 0 or a fifth-power root of unity.
Mobius transform of A319099. Every term is 0 or a power of 4.
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LINKS
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FORMULA
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Multiplicative with a(p^e) = 4 if p^e = 25 or p == 1 (mod 5) and e = 1, otherwise 0.
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EXAMPLE
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Let w = exp(2*Pi/5). For n = 11, the 4 quintic primitive Dirichlet characters modulo n are:
Chi_1 = [0, 1, w, w^3, w^2, w^4, w^4, w^2, w^3, w, 1];
Chi_2 = [0, 1, w^2, w, w^4, w^3, w^3, w^4, w, w^2, 1];
Chi_3 = [0, 1, w^3, w^4, w, w^2, w^2, w, w^4, w^3, 1];
Chi_4 = [0, 1, w^4, w^2, w^3, w, w, w^3, w^2, w^4, 1],
so a(11) = 4.
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MATHEMATICA
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f[5, 2] = 4; f[p_, e_] := If[Mod[p, 5] == 1 && e == 1, 4, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 16 2020 *)
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PROG
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(PARI) a(n)=sum(d=1, n, if(n%d==0, moebius(n/d)*sum(i=1, d, if((i^5-1)%d, 0, 1)), 0))
(PARI) A307380(n) = sumdiv(n, d, moebius(n/d)*sum(i=1, d, if((i^5-1)%d, 0, 1))); \\ (Slightly speeding the program above) - Antti Karttunen, Aug 22 2019
(PARI) A307380(n) = { my(f=factor(n)); prod(i=1, #f~, if(((5==f[i, 1])&&(2==f[i, 2]))||((1==(f[i, 1]%5))&&(1==f[i, 2])), 4, 0)); }; \\ (After the multiplicative formula, much faster) - Antti Karttunen, Aug 22 2019
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CROSSREFS
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Cf. A319099 (number of solutions to x^5 == 1 (mod n)).
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KEYWORD
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nonn,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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