A393224 a(n) = gcd(A395171(n)^(p_n*k) - 1, A395171(n)! - 1), independently of the integer k >= 1 (where p_n denotes the n-th prime number).

140929, 121, 43, 23, 53, 43691, 174763, 277, 59

Offset

2,1

Comments

By definition of A395171, gcd(A395171(n)^(p_n*k) - 1, A395171(n)! - 1) > 1 holds for any integer k >= 1.

Furthermore, gcd(A395171(n)^(p_n*k) - 1, A395171(n)! - 1) does not depend on k (i.e., gcd(A395171(n)^(p_n*1) - 1, A395171(n)! - 1) = gcd(A395171(n)^(p_n*2) - 1, A395171(n)! - 1) = ...).

Since 700000 < A395171(11), it follows that a(11) >= 700229.

Links

Table of n, a(n) for n=2..10.

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

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

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

Example

a(2) = 140929 since gcd(2283^(3*1) - 1, 2283! - 1) = gcd(2283^(3*2) - 1, 2283! - 1) = ... = 140929 > 1.

Crossrefs

Cf. A000978, A005384, A395115, A395171, A395286, A395944.

Sequence in context: A252017 A203833 A395944 * A032747 A212373 A195438

Adjacent sequences: A393221 A393222 A393223 * A393225 A393226 A393227

Keyword

nonn,hard,more

Author

Marco Ripà, May 31 2026

Status

approved