OFFSET
0,1
COMMENTS
As for standard Fermat numbers 2^(2^n) + 1, a number (2b)^m + 1 (with b > 1) can only be prime if m is a power of 2. On the other hand, out of the first 12 base-10 Fermat numbers, only the first two are primes.
Also, binary representation of Fermat numbers (in decimal, see A000215).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..9 (shortened by N. J. A. Sloane, Jan 13 2019)
Anders Björn and Hans Riesel, Factors of Generalized Fermat Numbers, Mathematics of Computation, Vol. 67, No. 221, Jan., 1998, pp. 441-446.
C. K. Caldwell, "Top Twenty" page, Generalized Fermat Divisors (base=10).
Wilfrid Keller, GFN10 factoring status.
Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670, 2012 - From N. J. A. Sloane, Jun 13 2012
Eric Weisstein's World of Mathematics, Generalized Fermat Number.
OEIS Wiki, Generalized Fermat numbers.
FORMULA
a(0) = 11; a(n) = (a(n - 1) - 1)^2 + 1.
a(n) = 9*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 9*(empty product, i.e., 1)+ 2 = 11 = a(0). - Daniel Forgues, Jun 20 2011
Sum_{n>=0} 2^n/a(n) = 1/9. - Amiram Eldar, Oct 03 2022
EXAMPLE
a(0) = 10^1 + 1 = 11 = 9*(1) + 2 = 9*(empty product) + 2.
a(1) = 10^2 + 1 = 101 = 9*(11) + 2.
a(2) = 10^4 + 1 = 10001 = 9*(11*101) + 2.
a(3) = 10^8 + 1 = 100000001 = 9*(11*101*10001) + 2.
a(4) = 10^16 + 1 = 10000000000000001 = 9*(11*101*10001*100000001) + 2.
a(5) = 10^32 + 1 = 100000000000000000000000000000001 = 9*(11*101*10001*100000001*10000000000000001) + 2.
MATHEMATICA
Table[10^2^n + 1, {n, 0, 6}] (* Arkadiusz Wesolowski, Nov 02 2012 *)
PROG
(Magma) [10^(2^n) + 1: n in [0..8]]; // Vincenzo Librandi, Jun 20 2011
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jens Voß, Feb 04 2003
EXTENSIONS
Edited by Daniel Forgues, Jun 19 2011
STATUS
approved