A163160 a(n) = A162996(n) - R_n = round(kn * (log(kn)+1)) - R_n, with k = 2.216 and R_n = n-th Ramanujan Prime A104272(n) and where Abs(a(n)) < 2 * sqrt(A162996(n)) for n in [1..1000].

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

Offset: 1

Author

Daniel Forgues, Jul 21 2009, Jul 29 2009

Keywords

• sign

Comments

A162996(n) approximates the {kn}-th prime number, which in turn approximates the n-th 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 = n-th 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).