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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
 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, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 29, 30, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 38, 39, 39, 40, 41, 41, 42, 43, 44, 44, 45, 46, 46, 47 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS 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. LINKS Clark Kimberling, Table of n, a(n) for n = 1..2000 EXAMPLE Let s(n) = sum{1/(h*(h+1)*(h+2))}, h = 1..k}.  Approximations are shown here: n ... 1/4 - s(n) ... 1/n^2 1 ... 0.08333 ...... 1 2 ... 0.04166 ...... 0.25 3 ... 0.025 ........ 0.111 4 ... 0.01666 ...... 0.0625 5 ... 0.01190 ...... 0.004 6 ... 0.00893 ...... 0.02777 a(4) = 2 because 1/4 - s(2) < 1/16 < 1/4 - s(1). MATHEMATICA z = 200; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)), {h, 1, k}] ; N[Table[1/4 - p[n], {n, 1, z/10}]] f[n_] := f[n] = Select[Range[z], 1/4 - p[#] < 1/n^2 &, 1]; u = Flatten[Table[f[n], {n, 1, z}]]   (* A248183 *) Flatten[Position[Differences[u], 0]]  (* A248184 *) Flatten[Position[Differences[u], 1]]  (* A248185 *) CROSSREFS Cf. A248184, A248185. Sequence in context: A136746 A003003 A248186 * A049474 A076874 A127041 Adjacent sequences:  A248180 A248181 A248182 * A248184 A248185 A248186 KEYWORD nonn,easy AUTHOR Clark Kimberling, Oct 04 2014 STATUS approved

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Last modified March 31 19:37 EDT 2020. Contains 333151 sequences. (Running on oeis4.)