The OEIS is supported by the many generous donors to the OEIS Foundation.
%I #46 Nov 26 2025 15:59:39
%S 1,5,26,10,312,130,342,20,78,1560,120,130,84,1710,312,40,4912,390,
%T 6858,1560,4446,120,12166,260,1560,420,234,1710,280,1560,61568,80,
%U 1560,24560,17784,390,1368,34290,1092,1560,240,22230,162800,120,312,60830,103822
%N Period of the Lucas 4-step sequence A073817 mod n.
%C This sequence is the same as the period of Fibonacci 4-step sequence (A000078) mod n for n<563 because the discriminant of the characteristic polynomial x^4-x^3-x^2-x-1 is -563. The two sequences differ only at n that are multiples of 563.
%H Chai Wah Wu, <a href="/A106295/b106295.txt">Table of n, a(n) for n = 1..1486</a>
%H Marcellus E. Waddill, <a href="https://www.fq.math.ca/Scanned/30-3/waddill.pdf">Some Properties of the Tetranacci Sequence Modulo m</a>, The Fibonacci Quarterly, vol. 30, no. 3, 232-238 (1992).
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>
%F Let the prime factorization of n be p1^e1...pk^ek. Then a(n) = lcm(a(p1^e1), ..., a(pk^ek)).
%F a(2^k) = 5*2^(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 [Waddill, 1992]. - _Chai Wah Wu_, Feb 25 2022
%t n=4; Table[p=i; a=Join[Table[ -1, {n-1}], {n}]; 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, 60}]
%o (Python)
%o from itertools import count
%o def A106295(n):
%o a = b = (4%n,1%n,3%n,7%n)
%o s = sum(b) % n
%o for m in count(1):
%o b, s = b[1:] + (s,), (s+s-b[0]) % n
%o if a == b:
%o return m # _Chai Wah Wu_, Feb 22-27 2022
%Y Cf. A000078, A073817, A106273 (discriminant of the polynomial x^n-x^(n-1)-...-x-1).
%K nonn
%O 1,2
%A _T. D. Noe_, May 02 2005