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A166484
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Prime sums of three Fermat numbers: primes of form 2^2^x + 2^2^y + 5.
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3
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OFFSET
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1,1
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COMMENTS
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One can have a prime sum of two Fermat Primes, starting with 2 + 3 = 5.
Hence this current sequence is a proper subset of prime sums of a Fermat prime number of Fermat numbers, which in turn is a proper subset of prime sums of a Fermat number of Fermat numbers.
According to the Maple 9 primality test, the next term is larger than 10^300 if it exists. - R. J. Mathar, Oct 16 2009
At least one of the three Fermat numbers must be 3 because all Fermat numbers greater than 3 are equal to 2 (mod 3). Hence, the sum of three Fermat numbers greater than 3 is always a multiple of 3.
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LINKS
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FORMULA
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{p = (2^(2^a) + 1) + (2^(2^b) + 1) + (2^(2^c) + 1) for nonnegative integers a, b, c, such that p is prime}.
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EXAMPLE
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PROG
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(PARI) for(x=1, 9, for(y=1, x, if(isprime(t=2^2^x+2^2^y+5), print1(t", ")))) \\ Charles R Greathouse IV, Apr 29 2016
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CROSSREFS
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KEYWORD
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hard,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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