A248186 Least k such that 1/18 - Sum_{h=1..k} 1/(h*(h+1)*(h+2)*(h+3)) < 1/n^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

Offset

1,5

Comments

This sequence gives a measure of the convergence rate of Sum_{h=1..k} 1/(h*(h+1)*(h+2)*(h+3)) 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

Formula

Conjecture: a(n) = floor(r*n + r^2/n - 1), where r = 3^(-1/3), for n > 2. - Ridouane Oudra, Oct 06 2023

Example

Let s(k) = Sum_{h = 1..k} 1/(h*(h+1)*(h+2)*(h+3)).
For each k in 1..5, the table below shows the value of 1/18 - s(k), the largest value of n for which a(n) = k, and the corresponding value of 1/n^3.
.
  k  1/18 - s(k)   n  1/n^3
  -  -----------  --  --------------
  1  0.013888...   4  0.015625
  2  0.005555...   5  0.008
  3  0.002777...   7  0.002915451895
  4  0.001587...   8  0.001953125
  5  0.000992...  10  0.001
.
a(5) = 2 because 1/18 - s(2) = 0.005555... < 0.008 (= 1/n^3)
             but 1/18 - s(1) = 0.013888... > 0.008.

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}]]   (* this sequence *)
Flatten[Position[Differences[u], 0]]  (* A248187 *)
Flatten[Position[Differences[u], 1]]  (* A248188 *)

Crossrefs

Cf. A248187, A248188, A248183.

Sequence in context: A136746 A003003 A356988 * A248183 A049474 A076874

Adjacent sequences: A248183 A248184 A248185 * A248187 A248188 A248189

Keyword

nonn,easy

Author

Clark Kimberling, Oct 04 2014

Extensions

Example edited by Jon E. Schoenfield, Oct 06 2023

Status

approved