A248629 Least k such that 6 - sum{(h^2)/2^h, h = 1..k} < 1/3^n.

9, 11, 13, 15, 17, 19, 21, 22, 24, 26, 28, 29, 31, 33, 35, 36, 38, 40, 41, 43, 45, 47, 48, 50, 52, 53, 55, 57, 58, 60, 62, 63, 65, 67, 68, 70, 72, 73, 75, 76, 78, 80, 81, 83, 85, 86, 88, 90, 91, 93, 95, 96, 98, 99, 101, 103, 104, 106, 108, 109, 111, 112, 114

Offset: 1

Author

Clark Kimberling, Oct 10 2014

Keywords

• nonn
• easy

Comments

This sequence provides insight into the manner of convergence of sum{(h^2)/2^h, h = 1..k} to 6.

Examples

			Let s(n) = 6 - sum{(h^2)/2^h, h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/3^n
1 ... 5.50000 ... 0.333333
2 ... 4.50000 ... 0.111111
3 ... 3.37500 ... 0.037037
4 ... 2.37500 ... 0.012345
5 ... 1.59375 ... 0.004115
6 ... 1.03125 ... 0.001371
7 ... 0.64843 ... 0.000457
8 ... 0.39843 ... 0.000152
9 ... 0.24023 ... 0.000050
10 .. 0.14257 ... 0.000018
11 .. 0.08349 ... 0.000006
a(2) = 11 because s(11) < 1/9 < s(10).
		

Links

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

Crossrefs

Cf. A248630, A248631.

Programs

Mathematica

z = 300; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}]
d = N[Table[6 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248629 *)
d = Differences[u]
v = Flatten[Position[d, 1]]  (* A248630 *)