

A143227


(Number of primes between n and 2n)  (number of primes between n^2 and (n+1)^2), if > 0.


11



1, 2, 1, 1, 1, 1, 2, 2, 3, 3, 3, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 2, 1, 1, 2, 2, 6, 3, 3, 1, 1, 1, 2, 1, 1, 1, 1, 6, 3, 8, 3, 2, 3, 2, 3, 1, 1, 4, 3, 10, 2, 1, 1, 2, 3, 1, 3, 4, 2, 2, 9, 7, 2, 2, 4, 3, 3, 1, 2, 3, 5, 1, 2, 3, 2, 11, 3, 1, 2, 4, 7, 1, 1, 1, 1, 1, 5, 1, 2, 3, 3, 4, 2, 2, 9, 5, 1, 4, 2, 2
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OFFSET

1,2


COMMENTS

If the sequence is bounded (e.g., if it is finite), then Legendre's conjecture is true: there is always a prime between n^2 and (n+1)^2, at least for all sufficiently large n. This follows from the strong form of Bertrand's postulate proved by Ramanujan (see A104272 Ramanujan primes).


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. 208209.


LINKS



FORMULA



EXAMPLE

The first positive value of ((pi(2n)  pi(n))  (pi((n+1)^2)  pi(n^2))) is 1 (at n = 42), the 2nd is 2 (at n = 55) and the 3rd is 1 (at n = 56), so a(1) = 1, a(2) = 2, a(3) = 1.


MATHEMATICA

L={}; Do[ With[ {d=(PrimePi[2n]PrimePi[n])(PrimePi[(n+1)^2]PrimePi[n^2])}, If[d>0, L=Append[L, d]]], {n, 0, 1000}]; L
Select[Table[(PrimePi[2n]PrimePi[n])(PrimePi[(n+1)^2]PrimePi[n^2]), {n, 1000}], #>0&] (* Harvey P. Dale, Jun 19 2019 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



