OFFSET
0,1
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
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)
MATHEMATICA
a[0] := 5;
a[n_] := a[n] = Product[a[i], {i, 0, n - 1}] + 2;
Table[a[n], {n, 0, 10}]
KEYWORD
nonn,less
AUTHOR
Roger L. Bagula, May 27 2010
EXTENSIONS
Definition simplified by the Assoc. Eds. of the OEIS - May 28 2010
Edited by Daniel Forgues, Jun 22 2011
STATUS
approved