A144256 a(n) = smallest prime in the range [n^K, (n+1)^K], where K = log(127)/log(16) = 1.74717117169304146332...

2, 5, 7, 13, 17, 23, 31, 41, 47, 59, 67, 79, 89, 101, 127, 127, 149, 157, 173, 191, 211, 223, 241, 263, 277, 307, 317, 347, 359, 383, 409, 431, 457, 479, 499, 541, 557, 577, 607, 631, 659, 691, 719, 751, 787, 809, 839, 877, 907, 937, 967, 997, 1031, 1069

Offset: 1

Author

Alexander R. Povolotsky, Sep 16 2008

Keywords

• nonn

Comments

T. D. Noe submitted to primepuzzles.net the following conjecture #60, which is stronger than the Legendre's conjecture: For n>0 and k>=K, there is always a prime between n^k and (n+1)^k, where K = log(127)/log(16) = 1.74717117169304146332...

One could see that calculated terms for n=15 and n=16 yield the same value: 127, which make this conjecture (as originally defined) to be questionable. If this conjecture is modified to k>K, then there will be a distinct prime between 15^k and 16^k. It appears that the relatively large prime gap between 113 and 127 is the largest gap to overcome. Another way to correct/clarify the conjecture is to mention that both boundaries of the interval are included and that the same prime value may appear in two neighboring intervals. Of course the last version of the modified definition makes this conjecture to be different from the original Legendere conjecture (rather than to be an improvement of the original Legendere conjecture). [Alexander R. Povolotsky, Sep 26 2008]

The constant log(127)/log(16) is A194361. - John W. Nicholson, Dec 13 2013

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• Carlos Rivera, Conjecture 60: Generalization of Legendre's Conjecture

Crossrefs

Cf. A000040, A144831, A143935.

Programs

Mathematica

k = Log[127]/Log[16]; Table[Select[Range[Ceiling[n^k], Floor[(n + 1)^k]], PrimeQ, 1][[1]], {n, 100}] (* T. D. Noe, Jan 30 2012 *)

PARI

i=[];for(n=1, 500, for(j=ceil(n^(log(127)/log(16))), ceil((n+1)^(log(127)/log(16))), if(isprime(j), i=concat(i,j)); if(isprime(j), break)));i

Extensions

Extended by T. D. Noe, Jan 30 2012