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A248186
Least k such that 1/18 - Sum_{h=1..k} 1/(h*(h+1)*(h+2)*(h+3)) < 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
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
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 04 2014
EXTENSIONS
Example edited by Jon E. Schoenfield, Oct 06 2023
STATUS
approved