A227721 Floor(1/s(n)), where s(n) = (2n+1)/(2n+2) - n*log((n+1)/n).

17, 44, 83, 134, 197, 272, 359, 458, 569, 692, 827, 974, 1133, 1304, 1487, 1682, 1889, 2108, 2339, 2582, 2837, 3104, 3383, 3674, 3977, 4292, 4619, 4958, 5309, 5672, 6047, 6434, 6833, 7244, 7667, 8102, 8549, 9008, 9479, 9962, 10457, 10964, 11483, 12014, 12557

Offset: 1

Author

Clark Kimberling, Jul 22 2013

Keywords

• nonn
• easy

Comments

That s(n) > 0 for n >=1 follows from the chain 1 < log 2 < 3/4 < 2 log 3/2 < 5/6 < 3 log 4/3 < 7/8 < 4 log 5/4 < ... ; i.e., n log((n+1)/n) - (2n-1)/(2n) > 0 and (2n+1)/(2n+2) - n log((n+1)/n) > 0. For the first, closeness to 0 is indicated by A227719 and A227720, and for the second, by A227721 and a sequence which possibly equals A094159. Conjecture: the four sequences are linearly recurrent.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A227719, A227720, A094159.

Programs

Mathematica

s[n_] := s[n] = (2 n + 1)/(2 n + 2) - n*Log[1 + 1/n]
Table[Floor[1/s[n]], {n, 1, 100}] (* A227721 *)
Table[Round[1/s[n]], {n, 1, 100}] (* conjecture: A094159 *)

Formula

a(n) = 2 + 9*n + 6*n^2 (conjectured).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) (conjectured).

G.f.: (-17 + 7 x - 2 x^2)/(-1 + x)^3 (conjectured).