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

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

Offset

0,1

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.

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.

Mathematica

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

Prog

(PARI) fplusm(n, m)= { local(x, y); for(x=0, n, y=2^(2^x)+m; print1(y", ") ) }
(Magma) [2^(2^n) + 3: n in [0..11]]; // Vincenzo Librandi, Jan 09 2013

Crossrefs

Cf. A063486, A130730.

Sequence in context: A280150 A062654 A231865 * A222411 A274022 A117321

Adjacent sequences: A130726 A130727 A130728 * A130730 A130731 A130732

Keyword

nonn

Author

Cino Hilliard, Jul 05 2007

Status

approved