A086758 a(n) is the smallest m such that the integer part of the first n powers of m^(1/n) are primes.

2, 5, 13, 31, 631, 173, 409, 967, 3450844193, 39661481813, 2076849234433, 52134281654579, 14838980942616539, 260230524377962793, 4563650703502319197, 80032531899785490253, 172111744128569095516889

Offset: 1

Author

Farideh Firoozbakht, Aug 01 2003

Keywords

• nonn

Comments

All terms of this sequence must be primes because floor((a(n)^(1/n))^n) = a(n).

Floor[(a(8)^(1/8))^k] = floor[(1287/545)^k] for k=1..10 (see puzzle 227). If a(9) exists it must be greater than 22000000.

Examples

			a(5)=631 because floor(631^(1/5)) = 3, floor(631^(2/5)) = 13, floor(631^(3/5)) = 47, floor(631^(4/5)) = 173 and floor(631^(5/5)) = 631 are primes and 631 is the smallest m with this property.
a(8)=967 because the sequence {2, 5, 13, 31, 73, 173, 409, 967} consists entirely of primes, the i-th term in the sequence being floor(967^(i/8)) and 967 is the smallest integer with this property.
		

References

• R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2001; see Exercise 1.75, p. 69.

Links

• Martin Raab, Table of n, a(n) for n = 1..53
• Carlos Rivera, Puzzle 227. Research Problem 1.75, Prime Puzzles and Problems Connection.

Crossrefs

Cf. A063636, A076255, A076357.

Programs

Mathematica

Do[Print[For[m=1, Union[Table[PrimeQ[Floor[Prime[m]^(k/n)]], {k, n}]]!={True}, m++ ]; Prime[m]], {n, 8}]

Formula

For[m=1, Union[Table[PrimeQ[Floor[Prime[m]^(k/n)]], {k, n}]]!={True}, m++ ]; Prime[m]

Extensions

Terms a(9) and following from Jon E. Schoenfield, May 15 2010