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%I #15 Sep 07 2023 12:50:35
%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_{h = 1..k} 1/(h*(h+1)*(h+2)) < 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>
%F Conjecture: a(n) = floor(sqrt(n^2/2 + 1) - 1/2), for n>1. - _Ridouane Oudra_, Sep 06 2023
%e Let s(n) = Sum_{h = 1..k} 1/(h*(h+1)*(h+2)).
%e 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 *)
%o (PARI) a(n) = my(k=1); while (1/4 - sum(h = 1, k, 1/(h*(h+1)*(h+2))) >= 1/n^2, k++); k; \\ _Michel Marcus_, Sep 06 2023
%Y Cf. A248184, A248185.
%K nonn,easy
%O 1,4
%A _Clark Kimberling_, Oct 04 2014
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