A023142 Number of cycles of function f(x) = 10x mod n.

1, 1, 3, 1, 1, 3, 2, 1, 9, 1, 6, 3, 3, 2, 3, 1, 2, 9, 2, 1, 6, 6, 2, 3, 1, 3, 15, 2, 2, 3, 3, 1, 18, 2, 2, 9, 13, 2, 9, 1, 9, 6, 3, 6, 9, 2, 2, 3, 3, 1, 6, 3, 5, 15, 6, 2, 6, 2, 2, 3, 2, 3, 18, 1, 3, 18, 3, 2, 6, 2, 3, 9, 10, 13, 3, 2, 17, 9, 7, 1, 21, 9, 3, 6, 2, 3, 6, 6, 3, 9, 16, 2, 9, 2, 2, 3, 2, 3, 54, 1, 26

Offset

1,3

Links

T. D. Noe, Table of n, a(n) for n = 1..10000

Formula

a(n) = Sum_{d|m} phi(d)/ord(10, d), where m is n with all factors of 2 and 5 removed. - T. D. Noe, Apr 21 2003

Example

a(12) = 3 because the function 10x mod 12 has the three cycles (0),(1,10,4),(2,8).

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[10, n], {n, 100}]

Prog

(PARI) a(n)=n/=2^valuation(n, 2)*5^valuation(n, 5); sumdiv(n, d, eulerphi(d)/znorder(Mod(10, d))) \\ Charles R Greathouse IV, Apr 24 2013
(Python)
from sympy import totient, n_order, divisors
def A023142(n):
    m = n>>(~n & n-1).bit_length()
    a, b = divmod(m, 5)
    while not b:
        m = a
        a, b = divmod(m, 5)
    return sum(totient(d)//n_order(10, d) for d in divisors(m, generator=True) if d>1)+1 # Chai Wah Wu, Apr 09 2024

Crossrefs

Cf. A000374, A023135-A023142.

Sequence in context: A176514 A238559 A077196 * A225335 A229166 A143159

Adjacent sequences: A023139 A023140 A023141 * A023143 A023144 A023145

Keyword

nonn

Author

David W. Wilson

Status

approved