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A222361 Fibonacci-Legendre quotients: (Fibonacci(p) - L(p/5)) / p, where p = prime(n) and L(p/5) is the Legendre symbol.

%I A222361 #17 Nov 02 2024 03:21:56
%S A222361 1,1,1,2,8,18,94,220,1246,17732,43428,652914,4038540,10081266,
%T A222361 63217342,1005967758,16215627560,41061160360,670829406162,
%U A222361 4338894664368,11048157986978,183194101578180,1195118711985006,19999768719154092,862073644225241474,5674731128849674100,14568160545698020226,96118885585174929102,247025215671874138312,1633201998168434481118
%N A222361 Fibonacci-Legendre quotients: (Fibonacci(p) - L(p/5)) / p, where p = prime(n) and L(p/5) is the Legendre symbol.
%C A222361 Fibonacci(p) == L(p/5) mod p, where the Legendre symbol L(p/5) equals 0, +1, -1 according as p = 5, 5*k+-1, 5*k+-2 for some k.
%C A222361 Not to be confused with Fibonacci(p - L(p,5)) / p, which is A092330.
%H A222361 Vincenzo Librandi, <a href="/A222361/b222361.txt">Table of n, a(n) for n = 1..200</a>
%H A222361 Wikipedia, <a href="http://en.wikipedia.org/wiki/Fibonacci_number#Prime_divisors_of_Fibonacci_numbers">Prime divisors of Fibonacci numbers</a>
%F A222361 For n>=4, a(n) = (Fibonacci(prime(n)) +/- 1)/prime(n), where '+' is chosen if prime(n)== 2 or 3 (mod 5), '-' is chosen otherwise. For n>=2, a(n) = round(Fibonacci(prime(n))/prime(n)). - _Vladimir Shevelev_, Mar 12 2014
%e A222361 Prime(4) = 7, so a(4) = (Fibonacci(7)-L(7/5))/7 = (13-(-1))/7 = 14/7 = 2.
%t A222361 Table[p = Prime[n]; (Fibonacci[p] - JacobiSymbol[p, 5])/p, {n, 1, 30}]
%Y A222361 Cf. A000045, A092330.
%K A222361 nonn
%O A222361 1,4
%A A222361 _Jonathan Sondow_, Feb 23 2013