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A143224
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Numbers n such that (number of primes between n^2 and (n+1)^2) = (number of primes between n and 2n).
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10
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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
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OFFSET
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1,2
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COMMENTS
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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.
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REFERENCES
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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]
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LINKS
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FORMULA
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EXAMPLE
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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.
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MAPLE
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MATHEMATICA
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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 *)
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PROG
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(PARI) is(n) = primepi((n+1)^2)-primepi(n^2)==primepi(2*n)-primepi(n) \\ Felix Fröhlich, Jul 25 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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