login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Least k such that 6 - sum{(h^2)/2^h, h = 1..k} < 1/3^n.
4

%I #5 Oct 15 2014 21:02:25

%S 9,11,13,15,17,19,21,22,24,26,28,29,31,33,35,36,38,40,41,43,45,47,48,

%T 50,52,53,55,57,58,60,62,63,65,67,68,70,72,73,75,76,78,80,81,83,85,86,

%U 88,90,91,93,95,96,98,99,101,103,104,106,108,109,111,112,114

%N Least k such that 6 - sum{(h^2)/2^h, h = 1..k} < 1/3^n.

%C This sequence provides insight into the manner of convergence of sum{(h^2)/2^h, h = 1..k} to 6.

%H Clark Kimberling, <a href="/A248629/b248629.txt">Table of n, a(n) for n = 1..1000</a>

%e Let s(n) = 6 - sum{(h^2)/2^h, h = 1..n}. Approximations follow:

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

%e 1 ... 5.50000 ... 0.333333

%e 2 ... 4.50000 ... 0.111111

%e 3 ... 3.37500 ... 0.037037

%e 4 ... 2.37500 ... 0.012345

%e 5 ... 1.59375 ... 0.004115

%e 6 ... 1.03125 ... 0.001371

%e 7 ... 0.64843 ... 0.000457

%e 8 ... 0.39843 ... 0.000152

%e 9 ... 0.24023 ... 0.000050

%e 10 .. 0.14257 ... 0.000018

%e 11 .. 0.08349 ... 0.000006

%e a(2) = 11 because s(11) < 1/9 < s(10).

%t z = 300; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}]

%t d = N[Table[6 - p[k], {k, 1, z/5}], 12]

%t f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1]

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248629 *)

%t d = Differences[u]

%t v = Flatten[Position[d, 1]] (* A248630 *)

%Y Cf. A248630, A248631.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 10 2014