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A152581
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Generalized Fermat numbers: a(n) = 8^(2^n) + 1, n >= 0.
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9
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OFFSET
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0,1
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COMMENTS
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These numbers are all composite. We rewrite 8^(2^n) + 1 = (2^(2^n))^3 + 1.
Then by the identity a^n + b^n = (a+b)*(a^(n-1) - a^(n-2)*b + ... + b^(n-1)) for odd n, 2^(2^n) + 1 divides 8^(2^n) + 1. All factors of generalized Fermat numbers F_n(a,b) := a^(2^n)+b^(2^n), a >= 2, n >= 0, are of the form k*2^m+1, k >= 1, m >=0 (Riesel (1994)). - Daniel Forgues, Jun 19 2011
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LINKS
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FORMULA
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a(0)=9, a(n) = (a(n-1) - 1)^2 + 1, n >= 1.
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EXAMPLE
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For n = 3, 8^(2^3) + 1 = 16777217. Similarly, (2^8)^3 + 1 = 16777217. Then 2^8 + 1 = 257 and 16777217/257 = 65281.
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MATHEMATICA
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PROG
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(PARI) g(a, n) = if(a%2, b=2, b=1); for(x=0, n, y=a^(2^x)+b; print1(y", "))
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CROSSREFS
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Cf. A000215 (Fermat numbers: 2^(2^n) + 1, n >= 0).
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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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