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 A023139 Number of cycles of function f(x) = 7x mod n. 7
 1, 2, 3, 3, 2, 6, 1, 5, 5, 4, 2, 9, 2, 2, 6, 9, 2, 10, 7, 7, 3, 4, 2, 15, 7, 4, 7, 3, 5, 12, 3, 13, 6, 4, 2, 15, 5, 14, 6, 13, 2, 6, 8, 7, 10, 4, 3, 27, 1, 14, 6, 7, 3, 14, 5, 5, 21, 10, 3, 21, 2, 6, 5, 17, 7, 12, 2, 7, 6, 4, 2, 25, 4, 10, 21, 21, 2, 12, 2, 25, 9, 4, 3, 9, 7, 16, 15, 13, 2, 20, 2, 7, 9, 6 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Number of factors in the factorization of the polynomial x^n-1 over the integers mod 7. - T. D. Noe, Apr 16 2003 REFERENCES R. Lidl and H. Niederreiter, Finite Fields, Addison-Wesley, 1983, p. 65. LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 FORMULA a(n) = Sum_{d|m} phi(d)/ord(7, d), where m is n with all factors of 7 removed. - T. D. Noe, Apr 19 2003 a(n) = (1/ord(7,m))*Sum_{j = 0..ord(7,m)-1} gcd(7^j - 1, m), where m is n with all factors of 7 removed. - Nihar Prakash Gargava, Nov 14 2018 EXAMPLE a(8) = 5 because (1) the function 7x mod 8 has the five cycles (0),(4),(1,7),(2,6),(3,5) and (2) the factorization of x^8-1 over integers mod 7 is (1+x) (6+x) (1+x^2) (1+3x+x^2) (1+4x+x^2), which has five unique factors. Note that the length of the cycles is the same as the degree of the factors. a(10) = 2 because the function 8x mod 10 has the two cycles (0),(2,6,8,4). MATHEMATICA Table[Length[FactorList[x^n - 1, Modulus -> 7]] - 1, {n, 100}] CountFactors[p_, n_] := Module[{sum=0, m=n, d, f, i}, While[Mod[m, p]==0, m/=p]; d=Divisors[m]; Do[f=d[[i]]; sum+=EulerPhi[f]/MultiplicativeOrder[p, f], {i, Length[d]}]; sum]; Table[CountFactors[7, n], {n, 100}] CROSSREFS Cf. A000005, A000374. Cf. A023135, A023136, A023137, A023138, A023140, A023141, A023142. Sequence in context: A091813 A238114 A291300 * A328484 A319442 A299772 Adjacent sequences:  A023136 A023137 A023138 * A023140 A023141 A023142 KEYWORD nonn AUTHOR STATUS approved

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Last modified September 20 16:31 EDT 2020. Contains 337265 sequences. (Running on oeis4.)