A227720 Round(1/s(n)), where s(n) = n*log(1+1/n) - (2n-1)/(2n).

5, 16, 34, 57, 86, 121, 163, 210, 263, 322, 388, 459, 536, 619, 709, 804, 905, 1012, 1126, 1245, 1370, 1501, 1639, 1782, 1931, 2086, 2248, 2415, 2588, 2767, 2953, 3144, 3341, 3544, 3754, 3969, 4190, 4417, 4651, 4890, 5135, 5386, 5644, 5907, 6176, 6451, 6733

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, A227721, A094159.

Programs

Mathematica

s[n_] := n*Log[1 + 1/n] - (2 n - 1)/(2 n);
Table[Floor[1/s[n]], {n, 1, 100}]  (* A227719 *)
Table[Round[1/s[n]], {n, 1, 100}]  (* A227720 *)

Formula

a(n) = -2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) (conjectured).

G.f.: (-5 - 6 x - 7 x^2 - 5 x^3 - x^4)/((-1 + x)^3 (1 + x + x^2 + x^3)) (conjectured).