Problems & Puzzles: Puzzles

 Puzzle 540. b^(2^n)+c Robin García sent the following puzzle: Consecutive primes of the form b^(2^n)+c for n>n0 n0>=0 Search to n=10, b=1000, c=+2000    (b,c)=(2,1) gives 5 consecutive primes for n=0,1,2,3,4 called Fermat primes   (2,15) 6 consecutive primes, n0=0 : 17,19,31,271,65551,4294967311 (2,27) 5 consecutive , n0=0 (2,81) 6 consecutive, n0=2 and n=0 (7 primes) (2,93) 6 consecutive, n0=1 (3,440) 6 consecutive n0=0 (5,1518) 6 consecutive n0=0 (6,1081) 6 consecutive n0=0 (67,240) 6 consecutive n0=0 (14,405) 6 consecutive n0=0 (20,1071) 6 consecutive n0=0 I did not find (if I do not forget any) any more 6 consecutive primes of this form in a range n=0,10;b=2,1000;c=+1,+2000 Q. Find 7 consecutive primes of this kind not those already found (see Sloane A090872):

Contributions came from J. K. Andersen, Farideh Firoozbakht & Seiji Tomita

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J. K. Andersen wrote:

Robin García also posted this to

http://www.research.att.com/~njas/sequences/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"

The listed terms are a(0) to a(9). The next is given by:
2^(2^k) + 8073774344085 is prime for k = 0,1,...,10.

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Farideh wrote:

For b=2,  n0=0  & c=14899339905 we get 10 primes 2^2^n + 14899339905,
n = 0,1, ... & 9 (see A.129613).

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Seiji Tomita wrote:

Search condition: 2<=b<=10,1<=c<10^8,0<=n<=10

8 consecutive primes: 47 numbers were found.

9 consecutive primes: (b,c)=(2,12124167),n=0..8

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