[go: up one dir, main page]

login
A023136
Number of cycles of function f(x) = 4x mod n.
12
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
OFFSET
1,3
FORMULA
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
EXAMPLE
a(9) = 5 because the function 4x mod 9 has the five cycles (0),(3),(6),(1,4,7),(2,8,5).
MATHEMATICA
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}]
PROG
(PARI) a(n)=sumdiv(n>>valuation(n, 2), d, eulerphi(d)/znorder(Mod(4, d))) \\ Charles R Greathouse IV, Aug 05 2016
(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
KEYWORD
nonn
STATUS
approved