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%I #8 Jan 16 2015 15:46:24
%S 8,5,14,18,9,24,14,15,38,20,44,48,54,29,30,68,35,74,39,84,44,98,50,
%T 104,108,54,114,128,65,138,69,74,75,158,164,168,174
%N 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.
%H D. D. Wall, <a href="http://www.jstor.org/stable/2309169">Fibonacci series modulo m</a>, Amer. Math. Monthly, 67 (1960), 525-532.
%F a(n) = (prime(n+3) - 1)/2 if prime(n+3) == 1 or 9 (mod 10) and a(n) = (prime(n+3) + 1) if
%F prime(n+3) == 3 or 7 (mod 10), n >= 1.
%e 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).
%e a(2) = 5 = (11 - 1)/2 because prime(4) = 11 = 1 (mod 10). The length of the period for 11 is 10 = A001175(11).
%Y Cf. A001175, A222413, A222414.
%K nonn,easy
%O 1,1
%A _Wolfdieter Lang_, Jan 16 2015
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