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A023136
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Number of cycles of function f(x) = 4x mod n.
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12
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1, 1, 3, 1, 3, 3, 3, 1, 5, 3, 3, 3, 3, 3, 9, 1, 5, 5, 3, 3, 9, 3, 3, 3, 5, 3, 7, 3, 3, 9, 7, 1, 9, 5, 9, 5, 3, 3, 9, 3, 5, 9, 7, 3, 15, 3, 3, 3, 5, 5, 15, 3, 3, 7, 9, 3, 9, 3, 3, 9, 3, 7, 23, 1, 13, 9, 3, 5, 9, 9, 3, 5, 9, 3, 15, 3, 9, 9, 3, 3, 9, 5, 3, 9, 23, 7, 9, 3, 9, 15, 17, 3, 21, 3, 9, 3, 5, 5, 15, 5, 3, 15
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OFFSET
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1,3
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LINKS
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FORMULA
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a(n) = Sum_{d|m} phi(d)/ord(4, d), where m is n with all factors of 2 removed. The formula was developed by extending the ideas in A000374 to composite multipliers. - T. D. Noe, Apr 21 2003
Mobius transform of A133702: (1, 2, 4, 3, 4, 8, 4, 4, 9, 8, ...). = Row sums of triangle A133703. - Gary W. Adamson, Sep 21 2007
a(n) = (1/ord(4, m))*Sum_{j = 0..ord(4, m) - 1} gcd(4^j - 1, m), where m is the odd part of n (A000265). - Nihar Prakash Gargava, Nov 14 2018
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EXAMPLE
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a(9) = 5 because the function 4x mod 9 has the five cycles (0),(3),(6),(1,4,7),(2,8,5).
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MATHEMATICA
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CountFactors[p_, n_] := Module[{sum=0, m=n, d, f, i, ps, j}, ps=Transpose[FactorInteger[p]][[1]]; Do[While[Mod[m, ps[[j]]]==0, m/=ps[[j]]], {j, Length[ps]}]; d=Divisors[m]; Do[f=d[[i]]; sum+=EulerPhi[f]/MultiplicativeOrder[p, f], {i, Length[d]}]; sum]; Table[CountFactors[4, n], {n, 100}]
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PROG
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(Python)
from sympy import totient, n_order, divisors
def A023136(n): return sum(totient(d)//n_order(4, d) for d in divisors(n>>(~n & n-1).bit_length(), generator=True) if d>1)+1 # Chai Wah Wu, Apr 09 2024
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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