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A056993
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a(n) is the smallest k >= 2 such that k^(2^n)+1 is prime, or -1 if no such k exists.
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36
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2, 2, 2, 2, 2, 30, 102, 120, 278, 46, 824, 150, 1534, 30406, 67234, 70906, 48594, 62722, 24518, 75898, 919444
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OFFSET
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0,1
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COMMENTS
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Smallest base value yielding generalized Fermat primes. - Hugo Pfoertner, Jul 01 2003
The first 5 terms correspond with the known (ordinary) Fermat primes. A probable candidate for the next entry is 62722^131072+1, discovered by Michael Angel in 2003. It has 628808 decimal digits. - Hugo Pfoertner, Jul 01 2003
For any n, a(n+1) >= sqrt(a(n)), because k^(2^(n+1))+1 = (k^2)^(2^n)+1. - Jeppe Stig Nielsen, Sep 16 2015
Does the sequence contain any perfect squares? If a(n) is a perfect square, then a(n+1) = sqrt(a(n)). - Jeppe Stig Nielsen, Sep 16 2015
If for a particular n, a(n) exists, then a(i) exist for all i=0,1,2,...,n. No proof is known that this sequence is infinite. Such a result would clearly imply the infinitude of A002496. - Jeppe Stig Nielsen, Sep 18 2015
919444 is a candidate for a(20). See Zimmermann link. - Serge Batalov, Sep 02 2017
Now PrimeGrid has tested and double checked all b^(2^20) + 1 with b < 919444, so we have proof that a(20) = 919444. - Jeppe Stig Nielsen, Dec 30 2017
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LINKS
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FORMULA
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EXAMPLE
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The primes are 2^(2^0) + 1 = 3, 2^(2^1) + 1 = 5, 2^(2^2) + 1 = 17, 2^(2^3) + 1 = 257, 2^(2^4) + 1 = 65537, 30^(2^5) + 1, 102^(2^6) + 1, ....
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MATHEMATICA
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f[n_] := (p = 2^n; k = 2; While[cp = k^p + 1; !PrimeQ@cp, k++ ]; k); Do[ Print[{n, f@n}], {n, 0, 17}] (* Lei Zhou, Feb 21 2005 *)
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PROG
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(PARI) a(n)=my(k=2); while(!isprime(k^(2^n)+1), k++); k \\ Anders Hellström, Sep 16 2015
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CROSSREFS
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Cf. A006093, A005574, A000068, A006314, A006313, A006315, A006316, A056994, A056995, A057465, A057002, A088361, A088362, A226528, A226529, A226530, A251597, A253854, A244150, A243959, A321323.
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KEYWORD
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hard,more,nonn,changed
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AUTHOR
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EXTENSIONS
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24518 and 75898 from Lei Zhou, Feb 01 2012
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STATUS
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approved
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