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A078303
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Generalized Fermat numbers: 6^(2^n) + 1, n >= 0.
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13
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OFFSET
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0,1
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COMMENTS
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The next term is too large to include.
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 13 base-6 Fermat numbers, only the first three are primes.
Either the sequence of (standard) Fermat numbers contains infinitely many composite numbers or the sequence of base-6 Fermat numbers contains infinitely many composite numbers (cf. https://mathoverflow.net/a/404235/1593). - José Hernández, Nov 09 2021
Since all powers of 6 are congruent to 6 (mod 10), all terms of this sequence are congruent to 7 (mod 10). - Daniel Forgues, Jun 22 2011
There are only 5 known Fermat primes of the form 2^(2^n) + 1: {3, 5, 17, 257, 65537}. There are only 2 known base-10 generalized Fermat primes of the form 10^(2^n) + 1: {11, 101}. - Alexander Adamchuk, Mar 17 2007
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LINKS
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FORMULA
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a(0) = 7, a(n) = (a(n-1)-1)^2 + 1, n >= 1.
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. - Daniel Forgues, Jun 20 2011
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EXAMPLE
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a(0) = 6^1+1 = 7 = 5*(1)+2 = 5*(empty product)+2;
a(1) = 6^2+1 = 37 = 5*(7)+2;
a(2) = 6^4+1 = 1297 = 5*(7*37)+2;
a(3) = 6^8+1 = 1679617 = 5*(7*37*1297)+2;
a(4) = 6^16+1 = 2821109907457 = 5*(7*37*1297*1679617)+2;
a(5) = 6^32+1 = 7958661109946400884391937 = 5*(7*37*1297*1679617*2821109907457)+2;
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MATHEMATICA
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PROG
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CROSSREFS
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Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
Cf. A019434 (Fermat primes of the form 2^(2^n) + 1).
Cf. A123669, A123599, A056993, A126032, A178428, A059919, A199591, A078304, A152581, A080176, A199592, A152585.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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