A227512 - OEIS

%I #16 Nov 16 2014 06:43:21

%S 10,92,318,760,1490,2581,4103,6129,8731,11981,15952,20714,26340,32902,

%T 40472,49123,58925,69951,82273,95963,111094,127736,145962,165844,

%U 187454,210865,236147,263373,292615,323945,357436,393158,431184,471586,514436,559807

%N Floor(-1/n + 1/log((2n+1)/(2n-1))).

%C log(u/v), where u = n + 1/2 and v = n - 1/2, is the area under the curve y = 1/x that matches the rectangle of width 1 and height 1/n with base centered at (1/n,0); a(n) -> oo since -1/n + log(u/v) -> 0.

%H Clark Kimberling, <a href="/A227512/b227512.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = 12*n^3 - floor(9*n/5) - 1. (conjectured, based on computations by _Peter J. C. Moses_, Jul 14 2013)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-5) - 3*a(n-6) + 3*a(n-7) - a(n-8) (conjectured; verified up to n = 100000 ).

%F G.f.: (10 + 62 x + 72 x^2 + 72 x^3 + 72 x^4 + 63 x^5 + 8 x^6 + x^7)/((-1 + x)^4 (1 + x + x^2 + x^3 + x^4)) (conjectured).

%e -1/4 + log(9/7) = 0.0013144..., so 1/u = 760.78..., so a(4) = 760.

%t z = 120; a[n_] := Floor[1/(Log[(2 n + 1)/(2 n - 1)] - 1/n)]; t = Table[a[n], {n, 1, z}]

%Y Cf. A227513.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jul 14 2013