A395171 Smallest integer m > 1 such that gcd(m^(prime(n)*k) - 1, m! - 1) > 1 holds for each positive integer k.

2283, 9, 41, 4, 15, 43689, 174761, 236, 29, 357913941, 25, 41, 64, 141, 53, 2831

Offset

2,1

Comments

For several indices n such that prime(n) is a Sophie Germain prime, the equality a(n) = prime(n) holds. This is not a general rule (e.g., a(24) = 177, while prime(24) = 89).

If q is a prime divisor of (2^(prime(n)) + 1)/3, then m = q - 2 satisfies gcd(m^(prime(n)*k) - 1, m! - 1) > 1 for every positive integer k. Hence, every term exists (see Love's answer to "Universal solutions of gcd(n^{p*k}-1, n!-1)>1 for every odd prime p" in Links).

For every integer n >= 2, with m = a(n), gcd(m^(prime(n)*k) - 1, m! - 1) is divisible by a fixed integer > 1, independently of the positive integer k.

Furthermore, the author conjectures that, for every integer n >= 2, the value of gcd(a(n)^(prime(n)*k) - 1, a(n)! - 1) itself does not depend on the positive integer k.

10^8 < a(18) <= 768614336404564649 (see Pessolano (2026)). - Marco Ripà, Jun 13 2026

Links

Table of n, a(n) for n=2..17.

MathOverflow, Can gcd(n^k +- 1, n! +- 1) > 1 have arbitrarily many but finitely many solutions?.

MathOverflow, If gcd(n^3-1, n!-1)>1, must it be equal to 140929?.

MathOverflow, Answer to: Universal solutions of gcd(n^{p*k}-1, n!-1)>1 for every odd prime p.

Aldo Roberto Pessolano, Computational certification of OEIS A395171(11) and bounds for A395171(18), Zenodo, 2026.

Example

a(2) = 2283 since gcd(2283^(3*k) - 1, 2283! - 1) = 140929 > 1 for every positive integer k.

Crossrefs

Cf. A000978, A005384, A393224, A395115, A395286, A395944.

Sequence in context: A247401 A264256 A395115 * A396221 A236700 A268892

Adjacent sequences: A395168 A395169 A395170 * A395172 A395173 A395174

Keyword

nonn,hard,more,changed

Author

Marco Ripà, May 19 2026

Status

approved