A241014 Let p be the n-th prime, then a(n) = A/p where A is the smallest number (in absolute value) such that F_{p-(p/5)} == A (mod p^2) with F_n = A000045(n) and (p/5) the Legendre symbol.
1, 1, 1, 3, 5, 3, -1, 3, -8, -3, -6, 13, -2, -4, 16, -25, 10, -13, 7, -16, -15, -30, 21, 5, 37, -4, 22, 24, 26, -53, 13, 64, 58, -22, -29, 60, 44, -3, 44, -43, -5, -50, 94, 31, -56, 5, -99, 3, -73, 18, 29, 5, -59, -1, 2
Offset: 1
Keywords
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- R. J. McIntosh and E. L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, Math. Comp., 76 (2007), 2087-2094.
Programs
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Mathematica
Array[(#3 - #2 Boole[#3 > #2/2])/#1 & @@ {#, #^2, Mod[Fibonacci[# - KroneckerSymbol[#, 5]], #^2]} &@ Prime[#] &, 55] (* Michael De Vlieger, Sep 08 2021 *)
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PARI
forprime(p=2, 1e2, a=fibonacci(p-kronecker(p, 5))%p^2; if(a>p^2/2, a-=p^2); a=a/p; print1(a, ", "))
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PARI
a(n)=my(p=prime(n)); centerlift(((Mod([1, 1; 1, 0], p^2))^(p-kronecker(p,5))))[1, 2]/p \\ Charles R Greathouse IV, Aug 21 2014
Comments