A238490 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.

103, 2297860813

Offset: 1

Author

John Blythe Dobson, Mar 28 2014

Keywords

• nonn
• hard
• more
• bref

Comments

The condition for an odd prime p to be a member of this sequence is that p^2 divides A001353(p - (3/p)).

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.

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.

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.

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).

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

Examples

			LucasQuotient(103) = 103*851367555454046677501642274766916900879231854719584128208.
		

Links

• John Blythe Dobson, Table of n, a(n) for n = 1..2
• Karl Dilcher and Ladislav Skula, A new criterion for the first case of Fermat's Last Theorem, Mathematics of Computation, 64 (1995) 363-392.
• Andreas-Stephan Elsenhans and Jörg Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 1014, arxiv 1006.0824 [math.NT], 2010.
• H. C. Williams, Some formulas concerning the fundamental unit of a real quadratic field, Discrete Mathematics, 92 (1991), 431-440.

Crossrefs

Cf. A001353, A238736.

Programs

Mathematica

The following criteria are equivalent:
PrimeQ[p] &&
  Mod[(MatrixPower[{{1,2},{1,3}}, p-JacobiSymbol[3,p]-1].{{1},{1}})[[2,1]], p^2]==0
PrimeQ[p] && Mod[Last[LinearRecurrence[{4,-1},{0,1}, p-JacobiSymbol[3,p]+1]], p^2]==0

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