OFFSET
1,5
COMMENTS
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 04 2014
EXTENSIONS
Example edited by Jon E. Schoenfield, Oct 06 2023
STATUS
approved