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%I #32 Jun 26 2016 13:23:59
%S 3,2,5,3,1,2,3,1,5,1,1,2,1,2,5,2,1,1,2,1,5,1,2,1,5,1,3,2,1,3,3,1,3,1,
%T 1,1,2,2,5,2,1,1,1,5,2,1,1,3,2,1,3,1,1,1,3,5,1,1,2,1,2,2,2,1,5,2,1,5,
%U 2,1,3,1,3,2,1,5,1,2,1,1,1,1,1,5,1,2,1
%N Smallest k > 0 such that F_{p-(k/p)} == 0 (mod p), where p = prime(n), F_i = A000045(i) and (a/b) is the Kronecker symbol.
%C a(n) <= 5 for all n (cf. Sun, Sun, 1992, p. 372).
%H Z. H. Sun and Z. W. Sun, <a href="http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-aav60i4p371bwm">Fibonacci numbers and Fermat's last theorem</a>, Acta Arithmetica, Vol. 60, No. 4 (1992), 371-388.
%e Prime(9) = 23 and Kronecker symbol (5/23) = -1. 23-(-1) = 24 and A000045(24) == 0 (mod 23). Since 5 is the smallest k such that A000045(23-(k/23)) == 0 (mod 23), a(9) = 5.
%t Table[Function[p, k = 1; While[! Divisible[Fibonacci[p - KroneckerSymbol[k, p]], p], k++]; k]@ Prime@ n, {n, 120}] (* _Michael De Vlieger_, Jun 23 2016 *)
%o (PARI) a(n) = my(k=1, p=prime(n)); while(Mod(fibonacci(p-kronecker(k, p)), p)!=0, k++); k
%Y Cf. A000045.
%K nonn
%O 1,1
%A _Felix Fröhlich_, Jun 21 2016