A178428 5 followed by the generalized Fermat numbers 6^(2^n)+1 (A078303).

5, 7, 37, 1297, 1679617, 2821109907457, 7958661109946400884391937, 63340286662973277706162286946811886609896461828097

Offset: 0

Author

Roger L. Bagula, May 27 2010

Keywords

• nonn
• less

Comments

If a(0) = 3, the recursion formula gives the Fermat numbers (A000215).

With a(0) = 3 instead of 5; a(n) = 2 + product_{i=0..n-1} a(i), n >= 1.

The recurrence equation for generalized Fermat numbers F_n(a) = a^(2^n)+1,

a >= 2, n >= 0, is F_{n}(a) = (F_{n-1}(a)-1)^2 + 1. - Daniel Forgues, Jun 22 2011

Crossrefs

Cf. A000215, A178427, A178426.

Programs

Mathematica

a[0] := 5;
a[n_] := a[n] = Product[a[i], {i, 0, n - 1}] + 2;
Table[a[n], {n, 0, 10}]

Formula

a(0) = 5; a(n) = 2 + product_{i=0..n-1} a(i), n >= 1.

From Daniel Forgues, Jun 22 2011: (Start)

The motivation for this sequence comes from the recurrence for generalized Fermat numbers 6^(2^n)+1 (A078303)

a(n) = 5*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 5*(empty product, i.e., 1)+ 2 = 7 = a(0). This implies that the terms are pairwise coprime. (End)

Extensions

Definition simplified by the Assoc. Eds. of the OEIS - May 28 2010

Edited by Daniel Forgues, Jun 22 2011