A061265 Number of squares between n-th prime and (n+1)st prime.

0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Amarnath Murthy, Apr 24 2001

Keywords

• nonn
• base

Comments

If n-th prime is a member of A053001 then a(n) is at least 1. If not, then a(n) = 0.

Legendre's conjecture (still open) that there is always a prime between n^2 and (n+1)^2 is equivalent to conjecturing that a(n) <= 1 for all n. - Vladeta Jovovic, May 01 2003

a(A038107(n)) = 1 for n > 1; a(A221056(n)) = 0. - Reinhard Zumkeller, Apr 15 2013

Examples

			a(3) = 0 as there is no square between 5, the third prime and 7, the fourth prime. a(4) = 1, as there is a square (9) between the 4th prime 7 and the 5th prime 11.
		

Links

• Harry J. Smith, Table of n, a(n) for n = 1..2000
• Eric Weisstein's World of Mathematics, Legendre's Conjecture
• Wikipedia, Legendre's conjecture

Crossrefs

Cf. A053001.

Cf. A038107.

Cf. A014085.

Programs

Haskell

a061265 n = a061265_list !! (n-1)
a061265_list = map sum $
   zipWith (\u v -> map a010052 [u..v]) a000040_list $ tail a000040_list
-- Reinhard Zumkeller, Apr 15 2013

Mathematica

ns[{a_,b_}]:=Count[Range[a+1,b-1],?(IntegerQ[Sqrt[#]]&)]; ns/@ Partition[ Prime[Range[110]],2,1] (* _Harvey P. Dale, Mar 14 2015 *)

PARI

{ n=0; q=2; forprime (p=3, prime(2001), write("b061265.txt", n++, " ", floor(sqrt(p))-floor(sqrt(q))); q=p ) } \\ Harry J. Smith, Jul 20 2009

Formula

a(n) = floor(sqrt(prime(n+1))) - floor(sqrt(prime(n))). - Vladeta Jovovic, May 01 2003

Extensions

Extended by Patrick De Geest, Jun 05 2001

Offset changed from 0 to 1 by Harry J. Smith, Jul 20 2009