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 A130729 Fermat numbers of order 3 or F(n,3) = 2^(2^n)+3. 4
 5, 7, 19, 259, 65539, 4294967299, 18446744073709551619, 340282366920938463463374607431768211459, 115792089237316195423570985008687907853269984665640564039457584007913129639939 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified April 18 18:41 EDT 2024. Contains 371781 sequences. (Running on oeis4.)