A248121 a(n) = floor(1 / (1/n - Pi^2/6 + Sum_{h=1..n} 1/h^2)).

2, 9, 20, 34, 53, 76, 102, 133, 168, 206, 249, 296, 346, 401, 460, 522, 589, 660, 734, 813, 896, 982, 1073, 1168, 1266, 1369, 1476, 1586, 1701, 1820, 1942, 2069, 2200, 2334, 2473, 2616, 2762, 2913, 3068, 3226, 3389, 3556, 3726, 3901, 4080, 4262, 4449, 4640

Offset: 1

Author

Clark Kimberling, Oct 02 2014

Keywords

• nonn
• easy

Comments

It is well known that Sum_{h>=1} 1/h^2 = Pi^2/6; this sequence provides insight into the manner of convergence.

Examples

			Let d(n) = Pi^2/6 - Sum_{h=1..n} 1/h^2. Approximations are shown here:
n ... 1/n .... d(n) ....... 1/n - d(n) ... a(n)
1 ... 1 ...... 0.644934 ... 0.355066 ..... 2
2 ... 0.5 .... 0.394934 ... 0.105066 ..... 9
3 ... 0.33 ... 0.283823 ... 0.04951 ...... 20
4 ... 0.25 ... 0.221323 ... 0.028677 ..... 34
		

References

• Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 20.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000027, A264938 (second conjecture).

Programs

Mathematica

z = 200; p[k_] := p[k] = Sum[1/h^2, {h, 1, k}];
N[Table[Pi^2/6 - p[n], {n, 1, z/4}]]
f[n_] := f[n] = Select[Range[z], Pi^2/6 - p[#] < 1/n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A000027 *)
v = Floor[Table[1/(1/n - (Pi^2/6 - p[n])), {n, 1, z}]]  (* A248121 *)

Formula

a(n) ~ 2*n^2. - Vaclav Kotesovec, Oct 09 2014

Conjectures from Chai Wah Wu, Aug 03 2022: (Start)

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5) for n > 5.

G.f.: -x*(x + 1)^2*(x + 2)/((x - 1)^3*(x^2 + x + 1)). (End)

Extensions

Typo in name corrected by Vaclav Kotesovec, Oct 09 2014