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Least k such that 20/27- sum{(h^2)/4^h, h = 1..k} < 1/8^n.
3

%I #4 Oct 17 2014 23:18:43

%S 3,5,7,9,11,12,14,16,17,19,21,22,24,25,27,29,30,32,33,35,36,38,40,41,

%T 43,44,46,47,49,50,52,53,55,57,58,60,61,63,64,66,67,69,70,72,73,75,76,

%U 78,80,81,83,84,86,87,89,90,92,93,95,96,98,99,101,102,104

%N Least k such that 20/27- sum{(h^2)/4^h, h = 1..k} < 1/8^n.

%C This sequence provides insight into the manner of convergence of sum{(h^2)/4^h, h = 1..k} to 20/27. The difference sequence of A248633 entirely of 1s and 2s, so that A248634 and A248635 partition the positive integers.

%H Clark Kimberling, <a href="/A248633/b248633.txt">Table of n, a(n) for n = 1..1000</a>

%e Let s(n) = 20/27 - sum{(h^2)/4^h, h = 1..n}. Approximations follow:

%e n ... s(n) ........ 1/8^n

%e 1 ... 0.49074 ... 0.125000

%e 2 ... 0.24074 ... 0.015625

%e 3 ... 0.10011 ... 0.001953

%e 4 ... 0.03761 ... 0.000244

%e 5 ... 0.01320 ... 0.000030

%e a(2) = 5 because s(5) < 1/8^2 < s(2).

%t z= 300; p[k_] := p[k] = Sum[(h^2/4^h), {h, 1, k}];

%t d = N[Table[20/27 - p[k], {k, 1, z/5}], 12];

%t f[n_] := f[n] = Select[Range[z], 20/27 - p[#] < 1/8^n &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248633 *)

%t d = Differences[u]

%t Flatten[Position[d, 1]] (* A248634 *)

%t Flatten[Position[d, 2]] (* A248635 *)

%Y Cf. A248632, A248630.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 11 2014