Problems & Puzzles:
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:
Send the next entries for this Table.
Contributions came from J. K. Andersen,
Jan van Delden, Torbjörn Alm & Emmanuel Vantieghem.
Andersen & Hakan Summakoğlu wrote:
Farideh Firoozbakht says:
"a(n) is the smallest natural number m such that 2^(2^k) + m is prime
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
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.
Torbjörn Alm wrote:
First solution n>8: n=9 k= +14899339905. There is at least one solution
for n=9 above 2*10^10
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.