OFFSET
1,4
COMMENTS
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 04 2014
STATUS
approved