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Floor(1/s(n)), where s(n) = n*log(1+1/n) - (2n-1)/(2n).
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%I #6 Jul 24 2013 12:50:15

%S 5,16,33,56,86,121,162,209,263,322,387,458,536,619,708,803,905,1012,

%T 1125,1244,1370,1501,1638,1781,1931,2086,2247,2414,2588,2767,2952,

%U 3143,3341,3544,3753,3968,4190,4417,4650,4889,5135,5386,5643,5906,6176,6451,6732

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

%C 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.

%H Clark Kimberling, <a href="/A227719/b227719.txt">Table of n, a(n) for n = 1..1000</a>

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

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

%t s[n_] := n*Log[1 + 1/n] - (2 n - 1)/(2 n);

%t Table[Floor[1/s[n]], {n, 1, 100}] (* A227719 *)

%t Table[Round[1/s[n]], {n, 1, 100}] (* A227720 *)

%Y Cf. A227720, A227721, A094159.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jul 22 2013