

A163160


a(n) = A162996(n)  R_n = round(kn * (log(kn)+1))  R_n, with k = 2.216 and R_n = nth Ramanujan Prime A104272(n) and where Abs(a(n)) < 2 * sqrt(A162996(n)) for n in [1..1000].


3



2, 0, 2, 1, 3, 1, 1, 2, 9, 6, 1, 7, 1, 11, 1, 5, 5, 6, 27, 17, 8, 1, 10, 2, 9, 10, 2, 7, 15, 4, 8, 0, 14, 8, 4, 2, 10, 19, 11, 1, 10, 12, 39, 27, 28, 20, 11, 2, 11, 9, 4, 15, 24, 33, 30, 3, 11, 14, 17, 14, 11, 7, 6, 18, 7, 18, 10, 31, 19, 9, 14
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OFFSET

1,1


COMMENTS

A162996(n) approximates the {kn}th prime number, which in turn approximates the nth Ramanujan prime, with k = 2.216 nearly optimal for n in [1..1000] since a(n)  2*sqrt(a(n)) < R_n < a(n) + 2*sqrt(a(n)) in that range. Since R_n ~ Prime(2n) ~ 2n * (log(2n)+1) ~ 2n * log(2n), whereas A162996(n) ~ Prime(kn) ~ kn * (log(kn)+1) ~ kn * log(kn), giving A162996(n) / R_n ~ k/2 = 2.216/2 = 1.108 which implies an asymptotic overestimate of about 10.8% (a better approximation would need k to depend on n and be asymptotic to 2). Consequently, a(n)  2*sqrt(a(n)) < R_n < a(n) + 2*sqrt(a(n)) will fail pretty early (R_n falling below the lower bound) as n grows beyond 1000.


LINKS

Daniel Forgues, Table of n, a(n) for n=1..1000


CROSSREFS

Cf. A162996 (Round(kn * (log(kn)+1)), with k = 2.216 as an approximation of R_n = nth Ramanujan Prime).
Cf. A104272 (Ramanujan primes: a(n) is the smallest number such that if x >= a(n), then pi(x)  pi(x/2) >= n, where pi(x) is the number of primes <= x).
Sequence in context: A308263 A028932 A076473 * A306696 A237184 A029240
Adjacent sequences: A163157 A163158 A163159 * A163161 A163162 A163163


KEYWORD

sign


AUTHOR

Daniel Forgues, Jul 21 2009, Jul 29 2009


STATUS

approved



