A291139 Primes of the form floor(k^(3/2)) for some integer k.

2, 5, 11, 31, 41, 89, 103, 181, 281, 311, 353, 419, 769, 797, 811, 839, 853, 911, 1091, 1153, 1201, 1217, 1249, 1499, 1621, 1873, 2081, 2557, 2999, 3307, 3533, 3671, 3881, 3929, 4289, 5431, 6131, 6269, 6491, 6547, 7001, 7349, 7583

Offset: 1

Author

Charles R Greathouse IV, Aug 18 2017

Keywords

• nonn

Comments

While Piatetski-Shapiro proved that there are infinitely many primes of the form floor(n^e) with 1 < e < 12/11, it is not currently known if this sequence is infinite.

Examples

			a(1) = floor(2^(3/2)) = floor(2.8...) = 2.
a(2) = floor(3^(3/2)) = floor(5.1...) = 5.
floor(4^(3/2)) = floor(8) = 8 is composite.
a(3) = floor(5^(3/2)) = floor(11.1) = 11.
floor(6^(3/2)) = floor(14.6...) = 14 is composite.
floor(7^(3/2)) = floor(18.5...) = 18 is composite.
floor(8^(3/2)) = floor(22.6...) = 22 is composite.
floor(9^(3/2)) = floor(27) = 27 is composite.
a(4) = floor(10^(3/2)) = floor(31.6) = 31.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A200141.

Programs

Mathematica

Select[Floor[Range[1000]^(3/2)],PrimeQ] (* Harvey P. Dale, Jul 01 2019 *)

PARI

list(lim)=my(v=List(),t); lim\=1; for(n=2,sqrtnint((lim+1)^2,3)-ispower(lim+1,3), if(isprime(t=sqrtint(n^3)), listput(v,t))); Vec(v)

Formula

Conjecturally, a(n) ~ (1.5 n log n)^1.5 and there are ~ x^(2/3)/log x members of this sequence up to x. - Charles R Greathouse IV, Oct 14 2017