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A248186 Least k such that 1/18 - sum{1/(h*(h+1)*(h+2)*(h+3))}, h = 1..k} < 1/n^3. 3
1, 1, 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 10, 11, 12, 12, 13, 14, 14, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 32, 32, 33, 34, 35, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 44, 44, 45, 46 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

COMMENTS

This sequence gives a measure of the convergence rate of  sum{1/(h*(h+1)*(h+2)*(h+3))}, h = 1..k} to 1/18.  Since a(n+1) - a(n) is in {0,1} for n >= 0, the sequences A248187 and A248188 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+3))}, h = 1..k}.  Approximations are shown here:

n ... 1/18 - s(n) ... 1/n^2

1 ... 0.0138889 ..... 1

2 ... 0.00555556 .... 0.125

3 ... 0.00277778 .... 0.037037

4 ... 0.0015873 ..... 0.015625

5 ... 0.0009920 ..... 0.008

a(5) = 2 because 1/18 - s(2) < 1/125 < 1/4 - s(1).

MATHEMATICA

$MaxExtraPrecision = Infinity;

z = 800; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)*(h + 3)), {h, 1, k}];

N[Table[1/18 - p[n], {n, 1, z/10}]]

f[n_] := f[n] = Select[Range[z], 1/18 - p[#] < 1/n^3 &, 1]

u = Flatten[Table[f[n], {n, 1, z}]]   (* A248186 *)

Flatten[Position[Differences[u], 0]]  (* A248187 *)

Flatten[Position[Differences[u], 1]]  (* A248188 *)

CROSSREFS

Cf. A248187, A248188, A248183.

Sequence in context: A113818 A136746 A003003 * A248183 A049474 A076874

Adjacent sequences:  A248183 A248184 A248185 * A248187 A248188 A248189

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Oct 04 2014

STATUS

approved

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Last modified August 23 11:41 EDT 2017. Contains 290995 sequences.