cp's OEIS Frontend

If p > 3 is prime, then p^2 | a(p).

Note the similarity to Wolstenholme's theorem.

Conjecture: for n > 3, if n^2 | a(n), then n is prime.

Are there the weak pseudoprimes m such that m | a(m)?

Primes p such that p^3 | a(p) are probably A088164.

If p is an odd prime, then a(p+1) == A330719(p+1) (mod p).

If p > 3 is a prime, then p^2 | numerator(Sum_{k=1..p+1} F(k)), where F(n) = Sum_{k=1..n} (2^(k-1)-1)/k. Cf. A027612 (a weaker divisibility).

a(3) > 107659373057 if it exists.

Primes p such that the Fermat quotient of p base 2 (A007663) is congruent to 1/2 modulo p. - Max Alekseyev, Aug 27 2023

For n>=3, round(q^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. In particular, all terms are even.

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

For n>=4, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. In particular, all terms are even.

Morley (1894/95) proved 2^(2*p-2) == (-1)^((p-1)/2)*binomial(p-1,(p-1)/2) mod p^3 for all primes p > 3.

Morley quotients are even, since 2^(2*p-2) and binomial(p-1,(p-1)/2) are even and p^3 is odd.

For n>=5, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. In particular, all terms are even.

Of course, a(n) = 0 iff n is a power of 2 and a(n) = 1 iff n is an odd number > 1. For other n, let n = 2^t*s, t > 0, s > 1 is an odd number, then a(n) is the unique solution to x == 0 (mod 2^t) and x == 1 (mod s).