A111252 Primes p such that the difference between the closest squares surrounding p is prime.

2, 3, 5, 7, 11, 13, 29, 31, 37, 41, 43, 47, 67, 71, 73, 79, 83, 89, 97, 127, 131, 137, 139, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 331, 337, 347, 349, 353, 359, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 541, 547, 557, 563

Offset: 1

Author

Cino Hilliard, Nov 12 2005

Keywords

• easy
• nonn

Comments

Conjecture: The number of terms in this sequence is infinite.

That there are infinitely many terms in this sequence would follow from the Legendre conjecture (one of the Landau problems - see the Weisstein link) that there is always a prime between n^2 and (n+1)^2. This is still an open problem. - Max Alekseyev, Apr 20 2006

Examples

			29 is a prime number. 5^2 and 6^2 are the closest squares surrounding 29. Now the difference 36-25 = 11 is prime so 29 is in the table.
		

Links

• Eric Weisstein's World of Mathematics, Landau's Problems

Programs

Mathematica

Clear[f,lst,p,n]; f[n_]:=IntegerPart[Sqrt[n]]; lst={};Do[p=Prime[n];If[PrimeQ[a=(f[p]+1)^2-f[p]^2],AppendTo[lst,p]],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Aug 05 2009 *)
Select[Prime[Range@103],PrimeQ[2*Floor[Sqrt[#]]+1]&] (* Ivan N. Ianakiev, Jul 30 2015 *)

PARI

surrsqpr(n) = { local(x,y,j,r,d); forprime(x=2,n, r=floor(sqrt(x)); d=r+r+1; if(isprime(d), print1(x, ",") ) ) }

Formula

Let p be a prime number and r = floor(sqrt(p)). Then the closest surrounding squares of p are r^2 and (r+1)^2. So d = (r+1)^2 - r^2 = 2r+1. If d is prime then list p.