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A130729
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Fermat numbers of order 3 or F(n,3) = 2^(2^n)+3.
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1
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5, 7, 19, 259, 65539, 4294967299, 18446744073709551619, 340282366920938463463374607431768211459, 115792089237316195423570985008687907853269984665640564039457584007913129639939
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| This is equivelant to F(n)+2 or 2^(2^n)+ 1 + 2. Conjecture: If n is odd, 7 is a divisor of F(n,3). This is equivelant to 7 divides 4^(4^m) +3 for all m. Subtracting 7 we get 7 divides 4^(4^m)-4 or 7 divides 4^(4^m-1)-1 for all m.
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FORMULA
| F(n,m): The n-th Fermat number of order m = 2^(2^n)+ m. The traditional Fermat numbers are F(n,1) or Fermat numbers of order 1.
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PROG
| (PARI) fplusm(n, m)= { local(x, y); for(x=0, n, y=2^(2^x)+m; print1(y", ") ) }
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CROSSREFS
| Sequence in context: A096636 A101588 A062654 * A117321 A192422 A120035
Adjacent sequences: A130726 A130727 A130728 * A130730 A130731 A130732
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KEYWORD
| nonn
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AUTHOR
| Cino Hilliard (hillcino368(AT)hotmail.com), Jul 05 2007
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