A070314 a(n) = P(n!+1)-P(2^n+1) where P(x) is the largest prime factor in x.

0, -1, -2, 4, -12, 0, 90, 28, 404, 250, 329850, 39916118, 2834088, 75021616, 3790360374, 46271010, 993974, 956666, 123610842, 1713311273189068, 117876621366, 2703875810364, 93799610095767534, 148139754734068388, 765041185860961083618, 38681321803817920155550

Offset: 0

Author

Benoit Cloitre, May 12 2002

Keywords

• sign

Comments

Is it always true that a(n) > 0 for n > 5? More generally, if m is an integer > 2, is there always an integer f(m) such that P(n!+1) > P(m^n+1) for n > f(m) (it seems that f(2) = 5, f(3) = 7, f(4) = 17, ...).

Links

• Amiram Eldar, Table of n, a(n) for n = 0..139

Crossrefs

Cf. A002583, A002587, A006530.

Programs

Mathematica

gpf[n_] := FactorInteger[n][[-1,1]]; a[n_] := gpf[n!+1] - gpf[2^n+1]; Array[a, 26, 0] (* Amiram Eldar, Apr 23 2025 *)

Formula

a(n) = A002583(n) - A002587(n). - Amiram Eldar, Apr 23 2025

Extensions

Offset changed to 0 and a(0) prepended by Amiram Eldar, Apr 23 2025