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A059917
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a(n) = (3^(2^n) + 1)/2 = A059919(n)/2, n >= 0.
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4
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OFFSET
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0,1
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COMMENTS
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Average of first 2^(n+1) powers of 3 divided by average of first 2^n powers of 3.
Numerator of b(n) where b(n) = 1/2*(b(n-1) + 1/b(n-1)), b(0)=2. - Vladeta Jovovic, Aug 15 2002
From Daniel Forgues, Jun 22 2011: (Start)
Since for the generalized Fermat numbers 3^(2^n)+1 (A059919), we have a(n) = 2*a(n-1)*a(n-2)*...*a(1)*a(0) + 2, n >= 0, where for n = 0, we get 2*(empty product, i.e. 1)+ 2 = 4 = a(0). This formula implies that the GCD of any pair of terms of A059919 is 2, which means that the terms of (3^(2^n)+1)/2 (A059917) are pairwise coprime.
2, 5, 41, 21523361, 926510094425921 are prime. 3281 = 17*193. (End)
a(0), a(1), a(2), a(4), a(5), and a(6) are prime. Conjecture: a(n) is composite for all n > 6. - Thomas Ordowski, Dec 26 2012
This may be a primarity test for Mersenne number. a(3)=41 == -1 mod 7 (=M3), a(5)=21523361 == 30 == -1 mod 31 (=M5). However, a(11) is not ==-1 mod M11. - Nobuyuki Fujita, May 16 2015
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LINKS
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Harry J. Smith, Table of n, a(n) for n=0..11
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FORMULA
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a(n) = a(n-1)*(3^(2^(n-1)) + 1) - 3^(2^(n-1)) = A059723(n+1)/A059723(n) = A059918(n) + 1 = a(n-1)*A059919(n-1) - A011764(n-1).
a(0) = 2; a(n) = ((2*a(n-1) - 1)^2 + 1)/2, n >= 1. - Daniel Forgues, Jun 22 2011
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EXAMPLE
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a(2) = Average(1,3,9,27,81,243,729,2187)/Average(1,3,9,27) = 410/10 = 41.
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MATHEMATICA
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Table[(3^(2^n) + 1)/2, {n, 0, 10}] (* Vincenzo Librandi, May 16 2015 *)
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PROG
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(PARI) { for (n=0, 11, write("b059917.txt", n, " ", (3^(2^n) + 1)/2); ) } \\ Harry J. Smith, Jun 30 2009
(MAGMA) [(3^(2^n)+1)/2: n in [0..10]]; // Vincenzo Librandi, May 16 2015
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CROSSREFS
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Cf. A059918, A059919.
Sequence in context: A126469 A054859 A076725 * A255963 A093625 A042447
Adjacent sequences: A059914 A059915 A059916 * A059918 A059919 A059920
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KEYWORD
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nonn,frac
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AUTHOR
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Henry Bottomley, Feb 08 2001
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STATUS
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approved
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