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a(n) = gcd_{k >= 1} gcd(A395171(n)^(prime(n)*k) - 1, A395171(n)! - 1).
1

%I #48 Jun 28 2026 16:41:20

%S 140929,121,43,23,53,43691,174763,277,59,715827883,149,83,431,283,107,

%T 2833

%N a(n) = gcd_{k >= 1} gcd(A395171(n)^(prime(n)*k) - 1, A395171(n)! - 1).

%C By definition of A395171, gcd(A395171(n)^(p_n*k) - 1, A395171(n)! - 1) > 1 holds for any integer k >= 1 (where p_n denotes the n-th prime).

%C The present sequence records the greatest common divisor, over all integers k >= 1, of these values. A395171(11) = 357913941 gives a(11) = 715827883.

%C Since A395171(18) > 10^8, it follows that a(18) >= 100000007.

%H MathOverflow, <a href="https://mathoverflow.net/questions/510302/if-gcdn3-1-n-11-must-it-be-equal-to-140929">If gcd(n^3-1, n!-1)>1, must it be equal to 140929?</a>.

%H MathOverflow, <a href="https://mathoverflow.net/questions/510656/can-gcdnk-pm-1-hspace2mm-n-pm-11-have-arbitrarily-many-but-finitel">Can gcd(n^k +- 1, n! +- 1) > 1 have arbitrarily many but finitely many solutions?</a>.

%H MathOverflow, <a href="https://mathoverflow.net/q/511242">Answer to: Universal solutions of gcd(n^{p*k}-1, n!-1)>1 for every odd prime p</a>.

%H Aldo Roberto Pessolano, <a href="https://doi.org/10.5281/zenodo.20682049">Computational certification of OEIS A395171(11) and bounds for A395171(18)</a>, Zenodo, 2026.

%e a(2) = 140929 since gcd_{k >= 1} gcd(2283^(3*k) - 1, 2283! - 1) = 140929.

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

%K nonn,hard,more,changed

%O 2,1

%A _Marco Ripà_, May 31 2026

%E Edits and a(11)-a(17) added by _Marco Ripà_, Jun 11 2026