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A004249
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a(n) = (2^2^...^2) (with n 2's) + 1.
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8
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OFFSET
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0,1
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COMMENTS
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a(0) could equally well be taken to be 1 rather than 2, which gives A007516. - N. J. A. Sloane, Sep 14 2009
A subsequence of the Fermat numbers 2^2^n + 1 = A000215.
a(0) through a(4) are primes; a(5) = 2^65536 + 1 is divisible by 825753601.
a(5) = 20035299...19156737 has 19729 decimal digits. - Alois P. Heinz, Jun 15 2022
It is unknown if a(6) = A000215(65536) is composite. - Jeppe Stig Nielsen, Jun 15 2022
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REFERENCES
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P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 73.
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LINKS
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Table of n, a(n) for n=0..4.
Y. Bugeaud and M. Queffélec, On Rational Approximation of the Binary Thue-Morse-Mahler Number, Journal of Integer Sequences, 16 (2013), #13.2.3.
Wilfrid Keller, Prime factors k.2^n + 1 of Fermat numbers F_m
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FORMULA
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a(0) = 2, a(n) = 2^a(n-1)/2 + 1 for n >= 1.
a(n) = A014221(n) + 1. - Leroy Quet, Jun 10 2009, updated by Jeppe Stig Nielsen, Jun 15 2022
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CROSSREFS
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Cf. Fermat numbers 2^2^n + 1 = A000215. A007516 is another version.
Sequence in context: A127063 A127837 A253646 * A268210 A121510 A132346
Adjacent sequences: A004246 A004247 A004248 * A004250 A004251 A004252
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane, Robert G. Wilson v, David W. Wilson
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STATUS
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approved
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