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%I #39 Feb 28 2020 22:09:20
%S 103,2297860813
%N Odd primes p that divide a Lucas quotient studied by H. C. Williams: A001353(p - (3/p))/p, where (3/p) is a Jacobi symbol.
%C The condition for an odd prime p to be a member of this sequence is that p^2 divides A001353(p - (3/p)).
%C Neither this quotient, nor the Lucas sequence U(4, 1) on which it is based, has a common name; but its fundamental discriminant of 3 places it between the quotient based on the Pell sequence U(2, -1) with discriminant 2 (A000129), and that based on the Fibonacci sequence U(1, -1) with discriminant 5 (A000045). Values of p dividing the Pell quotient will be found under A238736, while for the Fibonacci quotient it is known that there is no such p < 9.7*10^14.
%C The interest in this family of number-theoretic quotients derives from H. C. Williams, "Some formulas concerning the fundamental unit of a real quadratic field," p. 440, which proves a formula connecting the present quotient with the Fermat quotient base 2 (A007663), the Fermat quotient base 3 (A146211), and the harmonic number H(floor(p/12)) (see the Formula section below). As is well known, the vanishing of each of these Fermat quotients is a necessary condition for the failure of the first case of Fermat's Last Theorem (see discussions under A001220 and A014127); and a corresponding result concerning this type of harmonic number was proved by Dilcher and Skula. Thus, the vanishing mod p of the quotient based on U(4, 1) is also a necessary condition for the failure of the first case of Fermat's Last Theorem.
%C The pioneering computation for this quotient appears to be that of Elsenhans and Jahnel, "The Fibonacci sequence modulo p^2," p. 5, who report 103 as the only value of a(n) < 10^9. Extending the search to p < 2.5*10^10 has found only one further solution, 2297860813.
%C Let LucasQuotient(p) = A001353(p - (3/p))/p, q_2 = (2^(p-1) - 1)/p = A007663(p) be the corresponding Fermat quotient of base 2, q_3 = (3^(p-1) - 1)/p = A146211(p) be the corresponding Fermat quotient of base 3, H(floor(p/12)) be a harmonic number. Then Williams (1991) shows that 6*(3/p)*LucasQuotient(p) == -6*q_2 - 3*q_3 - 2*H(floor(p/12)) (mod p).
%C Also with an initial 2, primes p such that p^2 divides A001353(p - Kronecker(12,p)) (note that 12 is the discriminant of the characteristic polynomial of A001353, x^2 - 4x + 1). - _Jianing Song_, Jul 28 2018
%H John Blythe Dobson, <a href="/A238490/b238490.txt">Table of n, a(n) for n = 1..2</a>
%H Karl Dilcher and Ladislav Skula, <a href="http://dx.doi.org/10.1090/S0025-5718-1995-1248969-6">A new criterion for the first case of Fermat's Last Theorem</a>, Mathematics of Computation, 64 (1995) 363-392.
%H Andreas-Stephan Elsenhans and Jörg Jahnel, <a href="http://arxiv.org/abs/1006.0824">The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 1014</a>, arxiv 1006.0824 [math.NT], 2010.
%H H. C. Williams, <a href="https://doi.org/10.1016/0012-365X(91)90298-G">Some formulas concerning the fundamental unit of a real quadratic field,</a> Discrete Mathematics, 92 (1991), 431-440.
%e LucasQuotient(103) = 103*851367555454046677501642274766916900879231854719584128208.
%t The following criteria are equivalent:
%t PrimeQ[p] &&
%t Mod[(MatrixPower[{{1,2},{1,3}}, p-JacobiSymbol[3,p]-1].{{1},{1}})[[2,1]], p^2]==0
%t PrimeQ[p] && Mod[Last[LinearRecurrence[{4,-1},{0,1}, p-JacobiSymbol[3,p]+1]], p^2]==0
%o (PARI) isprime(p) && (Mod([2, 2; 1, 0], p^2)^(p-kronecker(3, p)))[2, 1]==0 \\ This test, which was used to find the second member of this sequence, is based on the test for A238736 devised by _Charles R Greathouse IV_
%Y Cf. A001353, A238736.
%K nonn,hard,more,bref
%O 1,1
%A _John Blythe Dobson_, Mar 28 2014
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