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 A106303 Period of the Fibonacci 5-step sequence A001591 mod n. 9
 1, 6, 104, 12, 781, 312, 2801, 24, 312, 4686, 16105, 312, 30941, 16806, 81224, 48, 88741, 312, 13032, 9372, 291304, 96630, 12166, 312, 3905, 185646, 936, 33612, 70728, 243672, 190861, 96, 1674920, 532446, 2187581, 312, 1926221, 13032 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS This sequence differs from the corresponding Lucas sequence (A106297) at all n that are multiples of 2 or 599 because 9584 is the discriminant of the characteristic polynomial x^5-x^4-x^3-x^2-x-1 and the prime factors of 9584 are 2 and 599. LINKS Chai Wah Wu, Table of n, a(n) for n = 1..388 Eric Weisstein's World of Mathematics, Fibonacci n-Step FORMULA Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)). Conjectures: a(5^k) = 781*5^(k-1) for k > 0. If a(p) != a(p^2) for p prime, then a(p^k) = p^(k-1)*a(p) for k > 0. - Chai Wah Wu, Feb 25 2022 MATHEMATICA n=5; Table[p=i; a=Join[{1}, Table[0, {n-1}]] a=Mod[a, p]; a0=a; k=0; While[k++; s=Mod[Plus@@a, p]; a=RotateLeft[a]; a[[n]]=s; a!=a0]; k, {i, 50}] PROG (Python) from itertools import count def A106303(n):     a = b = (0, )*4+(1 % n, )     s = 1 % n     for m in count(1):         b, s = b[1:] + (s, ), (s+s-b) % n         if a == b:             return m # Chai Wah Wu, Feb 21-27 2022 CROSSREFS Cf. A001591, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1), A106297 (period of Lucas 5-step sequence mod n). Sequence in context: A302911 A174481 A306205 * A157518 A001526 A295940 Adjacent sequences:  A106300 A106301 A106302 * A106304 A106305 A106306 KEYWORD nonn AUTHOR T. D. Noe, May 02 2005 STATUS approved

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Last modified May 16 22:37 EDT 2022. Contains 353724 sequences. (Running on oeis4.)