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A050216 Number of primes between (prime(n))^2 and (prime(n+1))^2, with a(0) = 2 by convention. 15
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

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

LINKS

T. D. Noe, Table of n, a(n) for n=0..10000

Eric Weisstein's World of Mathematics, Brocard's Conjecture

Wikipedia, Brocard's Conjecture

FORMULA

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

EXAMPLE

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.

MATHEMATICA

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

PROG

(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

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

CROSSREFS

First differences of A000879.

One more than A251723.

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

Sequence in context: A147766 A034420 A028410 * A080880 A120843 A205674

Adjacent sequences:  A050213 A050214 A050215 * A050217 A050218 A050219

KEYWORD

nonn,look

AUTHOR

Eric W. Weisstein

EXTENSIONS

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

Example corrected by Jonathan Sperry, Aug 30 2013

STATUS

approved

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Last modified August 2 06:21 EDT 2015. Contains 260193 sequences.