A253806 One half of the maximal values of the length of the period for Fibonacci numbers modulo p (A001175(p)) for primes p > 5, according to Wall's Theorems 6 and 7.

8, 5, 14, 18, 9, 24, 14, 15, 38, 20, 44, 48, 54, 29, 30, 68, 35, 74, 39, 84, 44, 98, 50, 104, 108, 54, 114, 128, 65, 138, 69, 74, 75, 158, 164, 168, 174

Offset: 1

Author

Wolfdieter Lang, Jan 16 2015

Keywords

• nonn
• easy

Examples

			a(1) = 8 = 7 + 1 because prime(4) = 7 == 7 (mod 10). The length of the period for 7 is 2*8 = 16 = A001175(7).
a(2) = 5 = (11 - 1)/2 because prime(4) = 11 = 1 (mod 10). The length of the period for 11 is 10 = A001175(11).
		

Links

• D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.

Crossrefs

Cf. A001175, A222413, A222414.

Formula

a(n) = (prime(n+3) - 1)/2 if prime(n+3) == 1 or 9 (mod 10) and a(n) = (prime(n+3) + 1) if

prime(n+3) == 3 or 7 (mod 10), n >= 1.