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

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

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

Formula

Conjecture: a(n) = floor(sqrt(n^2/2 + 1) - 1/2), for n>1. - Ridouane Oudra, Sep 06 2023

Example

Let s(n) = Sum_{h = 1..k} 1/(h*(h+1)*(h+2)).
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 *)

Prog

(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

Crossrefs

Cf. A248184, A248185.

Sequence in context: A003003 A356988 A248186 * A049474 A076874 A127041

Adjacent sequences: A248180 A248181 A248182 * A248184 A248185 A248186

Keyword

nonn,easy

Author

Clark Kimberling, Oct 04 2014

Status

approved