Primes p producing the longest possible period in the Fibonacci 4-step (mod p) sequence.
5, 31, 43, 53, 79, 83, 109, 131, 229, 233, 239
1
We select the primes p producing the longest period, which is 2*(p^3 + p^2 + p + 1).
T. D. Noe, Plot of 11 terms
Eric W. Weisstein, MathWorld: Fibonacci n-Step Number
(Mma) n = 4; t2 = Table[p = Prime[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}]; s = t2; t2 = Table[IntegerDigits[s[[i]], Prime[i]], {i, Length[s]}]; Flatten[Prime[Position[t2, {2, 2, 2, 2}]]]
Cf. A106296.
nonn,hard
T. D. Noe, Feb 10 2015