A050216 Number of primes between (prime(n))^2 and (prime(n+1))^2, with a(0) = 2 by convention.

2, 2, 5, 6, 15, 9, 22, 11, 27, 47, 16, 57, 44, 20, 46, 80, 78, 32, 90, 66, 30, 106, 75, 114, 163, 89, 42, 87, 42, 100, 354, 99, 165, 49, 299, 58, 182, 186, 128, 198, 195, 76, 356, 77, 144, 75, 463, 479, 168, 82, 166, 270, 90, 438, 275, 274, 292, 91, 292, 199, 99

Offset: 0

Author

Eric W. Weisstein

Keywords

• nonn
• look

Comments

The function in Brocard's Conjecture, which states that for n >= 2, a(n) >= 4.

The lines in the graph correspond to prime gaps of 2, 4, 6, ... . - T. D. Noe, Feb 04 2008

Lengths of blocks of consecutive primes in A000430 (union of primes and squares of primes). - Reinhard Zumkeller, Sep 23 2011

In the n-th step of the sieve of Eratosthenes, all multiples of prime(n) are removed. Then a(n) gives the number of new primes obtained after the n-th step. - Jean-Christophe Hervé, Oct 27 2013

More precisely, after the n-th step, one is sure to have eliminated all composites less than prime(n+1)^2, since any composite N has a prime factor <= sqrt(N). It is in exactly this (restricted) sense that a(n) yields the number of "new primes" (additional numbers known to be prime) after the n-th step. But one knows after the n-th step also that all remaining numbers between prime(n+1)^2 and prime(n+1)*(prime(n+1)+2) are prime: By construction they don't have a factor less than prime(n+1) and they don't have a factor prime(n+1) so the least prime factor could be prime(n+2) >= prime(n+1)+2. For example, after eliminating multiples of 3 in the 2nd step, one has (2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 31, 35, ...) and one knows that all remaining numbers strictly in between 5^2=25 and 5*(5+2)=35 are prime, too. - M. F. Hasler, Dec 31 2014

Numerically, the slope of the lowest "ray" m(n) = min {a(k); k>n}, seems to converge to a value somewhere in the range 1.75 < m(n)/n < 1.8; with m(n)/n > 1.7 for n > 900, m(n)/n > 1.75 for n > 2700. - M. F. Hasler, Dec 31 2014

Legendre's conjecture (see A014085) would imply that a(n) >= 2 for all n and that sequences A054272, A250473 and A250474 were thus strictly increasing (see the Wikipedia article about Brocard's conjecture). - Antti Karttunen, Jan 01 2015

a(n) >= 4 up to at least n = 4*10^5. - Eric W. Weisstein, Jan 13 2025

Examples

			There are 2 primes less than 2^2, there are 2 primes between 2^2 and 3^2, 5 primes between 3^2 and 5^2, etc. [corrected by Jonathan Sperry, Aug 30 2013]
		

References

• Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 183.

Links

• T. D. Noe, Table of n, a(n) for n = 0..10000
• Joel E. Cohen, Conjectures about Primes and Cyclic Numbers, arXiv:2508.08335 [math.NT], 2025. See p. 16.
• Anthony G. Shannon and Jean V. Leyendekkers, On Legendre's Conjecture, Notes on Number Theory and Discrete Mathematics, Vol. 23, No. 2 (2017): 117-125.
• Nicolas Vaillant, Graph of A050216(n) / (n * A001223(n))
• Nicolas Vaillant, Average density of primes between prime(n)^2 and prime(n+1)^2
• Eric Weisstein's World of Mathematics, Brocard's Conjecture.
• Wikipedia, Brocard's Conjecture

Crossrefs

First differences of A000879.

One more than A251723.

Cf. A010051, A001248, A014085, A089609, A251719, A256468, A256469, A256470, A029707, A137859.

Cf. A380135 (High water marks for number of primes between prime(n)^2 and prime(n+1)^2).

Cf. A380136 (Positions of the high water marks for number of primes between prime(n)^2 and prime(n+1)^2).

Programs

Haskell

import Data.List (group)
a050216 n = a050216_list !! (n-1)
a050216_list =
   map length $ filter (/= [0]) $ group $ map a010051 a000430_list
-- Reinhard Zumkeller, Sep 23 2011

Maple

A050216 := proc(n)
    local p,pn ;
    if n = 0 then
        2;
    else
        p := ithprime(n) ;
        pn := nextprime(p) ;
        numtheory[pi](pn^2)-numtheory[pi](p^2) ;
    end if;
end proc:
seq(A050216(n),n=0..40) ; # R. J. Mathar, Jan 27 2025

Mathematica

-Subtract @@@ Partition[PrimePi[Prime[Range[20]]^2], 2, 1] (* Eric W. Weisstein, Jan 10 2025 *)

PARI

a(n)={n||return(2);primepi(prime(n+1)^2)-primepi(prime(n)^2)} \\ M. F. Hasler, Dec 31 2014

Formula

For all n >= 1, a(n) = A256468(n) + A256469(n). - Antti Karttunen, Mar 30 2015

Limit_{N->oo} (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} prime(n)) = 1. - Alain Rocchelli, Sep 30 2023

Extensions

Edited by N. J. A. Sloane, Nov 15 2009