A248559 - OEIS

%I #9 Oct 15 2014 20:56:30

%S 1,2,3,4,6,7,8,10,11,13,14,15,17,18,20,21,23,24,26,27,29,30,32,33,35,

%T 36,38,40,41,43,44,46,47,49,50,52,53,55,56,58,60,61,63,64,66,67,69,70,

%U 72,74,75,77,78,80,81,83,84,86,88,89,91,92,94,95,97,98

%N Least k such that log(2) - sum{1/(h*2^h), h = 1..k} < 1/3^n.

%C This sequence provides insight into the manner of convergence of  sum{1/(h*2^h), h = 1..k} to log 2.  Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248560 and A248561 partition the positive integers.

%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 15.

%H Clark Kimberling, <a href="/A248559/b248559.txt">Table of n, a(n) for n = 1..1000</a>

%e Let s(n) = log(2) - sum{1/(h*2^h), h = 1..n}.  Approximations follow:

%e n ... s(n) ........ 1/3^n

%e 1 ... 0.193147 .... 0.33333

%e 2 ... 0.0681472 ... 0.11111

%e 3 ... 0.0264805 ... 0.037037

%e 4 ... 0.0108555 ... 0.0123457

%e 5 ... 0.0046066 ... 0.004115

%e 6 ... 0.0020013 ... 0.00137174

%e a(5) = 6 because s(6) < 1/3^5 < s(5).

%t z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]

%t N[Table[Log[2] - p[n], {n, 1, z/5}]]

%t f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]

%t u = Flatten[Table[f[n], {n, 1, z}]]    (* A248559 *)

%t Flatten[Position[Differences[u], 1]]   (* A248560 *)

%t Flatten[Position[Differences[u], 2]]   (* A248561 *)

%Y Cf. A002162 (log(2)), A248560, A248561.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Oct 09 2014