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 A248183 Least k such that 1/4 - sum{1/(h*(h+1)*(h+2))}, h = 1..k} < 1/n^2. 6

%I

%S 1,1,1,2,3,3,4,5,5,6,7,8,8,9,10,10,11,12,12,13,14,15,15,16,17,17,18,

%T 19,20,20,21,22,22,23,24,24,25,26,27,27,28,29,29,30,31,32,32,33,34,34,

%U 35,36,36,37,38,39,39,40,41,41,42,43,44,44,45,46,46,47

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

%C This sequence gives a measure of the convergence rate of sum{1/(h*(h+1)*(h+2))}, h = 1..k} to 1/4. Since a(n+1) - a(n) is in {0,1} for n >= 0, the sequences A248184 and A248185 partition the positive integers.

%H Clark Kimberling, <a href="/A248183/b248183.txt">Table of n, a(n) for n = 1..2000</a>

%e Let s(n) = sum{1/(h*(h+1)*(h+2))}, h = 1..k}. Approximations are shown here:

%e n ... 1/4 - s(n) ... 1/n^2

%e 1 ... 0.08333 ...... 1

%e 2 ... 0.04166 ...... 0.25

%e 3 ... 0.025 ........ 0.111

%e 4 ... 0.01666 ...... 0.0625

%e 5 ... 0.01190 ...... 0.004

%e 6 ... 0.00893 ...... 0.02777

%e a(4) = 2 because 1/4 - s(2) < 1/16 < 1/4 - s(1).

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

%t N[Table[1/4 - p[n], {n, 1, z/10}]]

%t f[n_] := f[n] = Select[Range[z], 1/4 - p[#] < 1/n^2 &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248183 *)

%t Flatten[Position[Differences[u], 0]] (* A248184 *)

%t Flatten[Position[Differences[u], 1]] (* A248185 *)

%Y Cf. A248184, A248185.

%K nonn,easy

%O 1,4

%A _Clark Kimberling_, Oct 04 2014

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Last modified May 30 15:21 EDT 2020. Contains 334726 sequences. (Running on oeis4.)