Problems & Puzzles: Puzzles

Puzzle 608. 2^(2^n)+k

Last week my friend Jaime Ayala phone me and told me that 2^(2^n)+15 produces 6 primes in a row (from n=0 to 5) while the well known Fermat's model 2^(2^n)+1 produces just 5 primes in a row (from n=0 to 4).

He challenged me to look for other k values for f(k)=2^(2^n)+k, such that f(k) is prime for all 0<=n<=N, for N>5

Using a simple code in Ubasic I got the following results:

 k N Author 1 4 Fermat 15 5 Ayala 66747 6 Rivera 475425 7 Rivera 12124167 8 Rivera ? >8 Who?

Q. Send the next entries for this Table.

Contributions came from J. K. Andersen, Hakan Summakoğlu, Jan van Delden, Torbjörn Alm & Emmanuel Vantieghem.

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Andersen & Hakan Summakoğlu wrote:

http://oeis.org/A129613 by Farideh Firoozbakht says:
"a(n) is the smallest natural number m such that 2^(2^k) + m is prime
for k=0,1,...,n.
1, 1, 1, 1, 1, 15, 66747, 475425, 12124167, 14899339905, 8073774344085"

As a matter of fact, Andersen found the last term and submitted it to Puzzle 540.

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Jan wrote:

For n=9 we have k=14899339905.

I tested k equal to 15,27 mod 30 (necessary), extended by a pretest comparing the residues of k and 2^(2^n)  [precomputed] modulo the first 100 primes.

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Torbjörn Alm wrote:

First solution n>8: n=9   k= +14899339905. There is at least one solution for n=9 above 2*10^10

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Emmanuel wrote:

For  k = 14899339905, we find  k+2^(2^n)  prime  for  0 <= n <= 9.
There are not more than 10 values for n when k <= 3x10^10.

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