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%I #4 Oct 17 2014 23:19:33
%S 7,9,10,12,13,15,17,18,19,21,22,24,25,27,28,29,31,32,34,35,36,38,39,
%T 40,42,43,44,46,47,48,50,51,52,54,55,56,58,59,60,62,63,64,66,67,68,70,
%U 71,72,73,75,76,77,79,80,81,83,84,85,87,88,89,90,92,93,94
%N Least k such that 33/8- sum{(h^3)/3^h, h = 1..k} < 1/4^n.
%C This sequence provides insight into the manner of convergence of sum{(h^3)/3^h, h = 1..k} to 33/8. The difference sequence of A248636 entirely of 1s and 2s, so that A248637 and A248638 partition the positive integers.
%H Clark Kimberling, <a href="/A248636/b248636.txt">Table of n, a(n) for n = 1..1000</a>
%e Let s(n) = 33/8- sum{(h^3)/3^h, h = 1..n}. Approximations follow:
%e n ... s(n) ........ 1/4^n
%e 1 ... 3.79167 ... 0.250000
%e 2 ... 2.90278 ... 0.062500
%e 3 ... 1.90278 ... 0.015625
%e 4 ... 1.11265 ... 0.003906
%e 5 ... 0.59825 ... 0.000976
%e 6 ... 0.30195 ... 0.000244
%e 7 ... 0.14511 ... 0.000061
%e 8 ... 0.06798 ... 0.000015
%e 9 ... 0.03004 ... 0.000004
%e a(2) = 9 because s(9) < 1/16 < s(8).
%t z = 300; p[k_] := p[k] = Sum[(h^3/3^h), {h, 1, k}];
%t d = N[Table[33/8 - p[k], {k, 1, z/5}], 12]
%t f[n_] := f[n] = Select[Range[z], 33/8 - p[#] < 1/4^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248636 *)
%t d = Differences[u]
%t v = Flatten[Position[d, 1]] (* A248637 *)
%t w = Flatten[Position[d, 2]] (* A248638 *)
%Y Cf. A248632, A248630.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Oct 11 2014
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