%I #174 Oct 23 2023 01:43:52

%S 2,2,5,6,15,9,22,11,27,47,16,57,44,20,46,80,78,32,90,66,30,106,75,114,

%T 163,89,42,87,42,100,354,99,165,49,299,58,182,186,128,198,195,76,356,

%U 77,144,75,463,479,168,82,166,270,90,438,275,274,292,91,292,199,99

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

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

%C The lines in the graph correspond to prime gaps of 2, 4, 6, ... . - _T. D. Noe_, Feb 04 2008

%C Lengths of blocks of consecutive primes in A000430 (union of primes and squares of primes). - _Reinhard Zumkeller_, Sep 23 2011

%C 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

%C 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

%C 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

%C 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

%H T. D. Noe, <a href="/A050216/b050216.txt">Table of n, a(n) for n = 0..10000</a>

%H A. G. Shannon and J. V. Leyendekkers, <a href="http://nntdm.net/volume-23-2017/number-2/117-125/">On Legendre's Conjecture</a>, Notes on Number Theory and Discrete Mathematics, Vol. 23, No. 2 (2017): 117-125.

%H Nicolas Vaillant, <a href="/A050216/a050216_9.png">Graph of A050216(n) / (n * A001223(n))</a>

%H Nicolas Vaillant, <a href="/A050216/a050216_10.png">Average density of primes between prime(n)^2 and prime(n+1)^2</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BrocardsConjecture.html">Brocard's Conjecture</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Brocard%27s_conjecture">Brocard's Conjecture</a>

%F For all n >= 1, a(n) = A256468(n) + A256469(n). - _Antti Karttunen_, Mar 30 2015

%F Limit_{N->oo} (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} prime(n)) = 1. - _Alain Rocchelli_, Sep 30 2023

%e 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]

%t PrimePi[ Prime[ n+1 ]^2 ]-PrimePi[ Prime[ n ]^2 ]

%o (Haskell)

%o import Data.List (group)

%o a050216 n = a050216_list !! (n-1)

%o a050216_list =

%o map length $ filter (/= [0]) $ group $ map a010051 a000430_list

%o -- _Reinhard Zumkeller_, Sep 23 2011

%o (PARI) a(n)={n||return(2);primepi(prime(n+1)^2)-primepi(prime(n)^2)} \\ _M. F. Hasler_, Dec 31 2014

%Y First differences of A000879.

%Y One more than A251723.

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

%K nonn,look

%O 0,1

%A _Eric W. Weisstein_

%E Edited by _N. J. A. Sloane_, Nov 15 2009