A143224 Numbers n such that (number of primes between n^2 and (n+1)^2) = (number of primes between n and 2n).

0, 9, 36, 37, 46, 49, 85, 102, 107, 118, 122, 127, 129, 140, 157, 184, 194, 216, 228, 360, 365, 377, 378, 406, 416, 487, 511, 571, 609, 614, 672, 733, 767, 806, 813, 863, 869, 916, 923, 950, 978, 988, 1249, 1279, 1280, 1385, 1427, 1437, 1483, 1539, 1551, 1690

Offset: 1

Author

Jonathan Sondow, Jul 31 2008

Keywords

• nonn

Comments

The sequence gives the positions of zeros in A143223. The number of primes in question is A143225(n).

Legendre's conjecture (still open) says there is always a prime between n^2 and (n+1)^2. Bertrand's postulate (actually a theorem due to Chebyshev) says there is always a prime between n and 2n.

Examples

			There is the same number of primes (namely 3) between 9^2 and 10^2 as between 9 and 2*9, so 9 is a term.
		

References

• M. Aigner and C. M. Ziegler, Proofs from The Book, Chapter 2, Springer, NY, 2001.
• G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 5th ed., Oxford Univ. Press, 1989, p. 19.
• S. Ramanujan, Collected Papers of Srinivasa Ramanujan (G. H. Hardy, S. Aiyar, P. Venkatesvara and B. M. Wilson, eds.), Amer. Math. Soc., Providence, 2000, pp. 208-209. [Jonathan Sondow, Aug 03 2008]

Links

• T. D. Noe, Table of n, a(n) for n=1..97 (no other n < 10^6)
• T. Hashimoto, On a certain relation between Legendre's conjecture and Bertrand's postulate, arXiv:0807.3690 [math.GM], 2008.
• M. Hassani, Counting primes in the interval (n^2,(n+1)^2), arXiv:math/0607096 [math.NT], 2006.
• J. Pintz, Landau's problems on primes
• S. Ramanujan, A proof of Bertrand's postulate, J. Indian Math. Soc., 11 (1919), 181-182.
• J. Sondow, Ramanujan Prime in MathWorld.
• J. Sondow and E. W. Weisstein, Bertrand's Postulate in MathWorld.
• Eric Weisstein's World of Mathematics, Legendre's Conjecture.

Crossrefs

See A000720, A014085, A060715, A143223, A143225, A143226.

Cf. A104272, A143227. [Jonathan Sondow, Aug 03 2008]

Programs

Maple

with(numtheory): A143224:=n->`if`(pi((n+1)^2)-pi(n^2) = pi(2*n)-pi(n), n, NULL): seq(A143224(n), n=0..2000); # Wesley Ivan Hurt, Jul 25 2017

Mathematica

L={}; Do[If[PrimePi[(n+1)^2]-PrimePi[n^2] == PrimePi[2n]-PrimePi[n], L=Append[L,n]], {n,0,2000}]; L
(* Second program *)
With[{nn = 2000}, {0}~Join~Position[#, {0}][[All, 1]] &@ Map[Differences, Transpose@ {Differences@ Array[PrimePi[#^2] &, nn], Array[PrimePi[2 #] - PrimePi[#] &, nn - 1]}]] (* Michael De Vlieger, Jul 25 2017 *)

PARI

is(n) = primepi((n+1)^2)-primepi(n^2)==primepi(2*n)-primepi(n) \\ Felix Fröhlich, Jul 25 2017

Formula

A143223(a(n)) = 0.