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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].
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%I #9 Jul 05 2015 20:11:18

%S 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,

%T 10,-2,7,-15,-4,-8,0,-14,-8,-4,-2,10,19,11,-1,10,12,-39,-27,-28,-20,

%U -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

%N 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].

%C 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.

%H Daniel Forgues, <a href="/A163160/b163160.txt">Table of n, a(n) for n=1..1000</a>

%Y Cf. A162996 (Round(kn * (log(kn)+1)), with k = 2.216 as an approximation of R_n = n-th Ramanujan Prime).

%Y 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).

%K sign

%O 1,1

%A _Daniel Forgues_, Jul 21 2009, Jul 29 2009