A130729 Fermat numbers of order 3 or F(n,3) = 2^(2^n)+3.

5, 7, 19, 259, 65539, 4294967299, 18446744073709551619, 340282366920938463463374607431768211459, 115792089237316195423570985008687907853269984665640564039457584007913129639939

Offset: 0

Author

Cino Hilliard, Jul 05 2007

Keywords

• nonn

Comments

This is equivalent to F(n)+2 or 2^(2^n)+ 1 + 2. Conjecture: If n is odd, 7 is a divisor of F(n,3).

The conjecture is true: the order of 2 mod 7 is 3, and if n is odd then 2^n == 2 mod 3 so 2^(2^n) + 3 == 2^2 + 3 == 0 mod 7. - Robert Israel, Nov 20 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..11
• Tigran Hakobyan, On the unboundedness of common divisors of distinct terms of the sequence a(n)=2^2^n+d for d>1, arXiv:1601.04946 [math.NT], 2016.

Crossrefs

Cf. A063486, A130730.

Programs

Magma

[2^(2^n) + 3: n in [0..11]]; // Vincenzo Librandi, Jan 09 2013

Mathematica

Table[(2^(2^n) + 3), {n, 0, 15}] (* Vincenzo Librandi, Jan 09 2013 *)

PARI

fplusm(n,m)= { local(x,y); for(x=0,n, y=2^(2^x)+m; print1(y",") ) }

Formula

F(n,m): The n-th Fermat number of order m = 2^(2^n)+ m. The traditional Fermat numbers are F(n,1) or Fermat numbers of order 1.