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Search: id:A163160
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| A163160 |
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a(n) = A162996(n) - R_n = round(kn * (ln(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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3
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| 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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 * (ln(2n)+1) ~ 2n * ln(2n), whereas A162996(n) ~ Prime(kn) ~ kn * (ln(kn)+1) ~ kn * ln(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.
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LINKS
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Daniel Forgues, Table of n, a(n) for n=1..1000
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CROSSREFS
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Cf. A162996 Round(kn * (ln(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.
Sequence in context: A078152 A028932 A076473 this_sequence A029240 A025803 A029185
Adjacent sequences: A163157 A163158 A163159 this_sequence A163161 A163162 A163163
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KEYWORD
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sign
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AUTHOR
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Daniel Forgues (squid(AT)zensearch.com), Jul 21 2009, Jul 29 2009
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