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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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5
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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
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 primality test for Mersenne numbers. a(2) = 41 == -1 mod 7 (=M3), a(4) = 21523361 == 30 == -1 mod 31 (=M5). However, a(10) is not == -1 mod M11. - Nobuyuki Fujita, May 16 2015
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LINKS
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FORMULA
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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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MAPLE
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MATHEMATICA
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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
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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