A248559 Least k such that log(2) - sum{1/(h*2^h), h = 1..k} < 1/3^n.

1, 2, 3, 4, 6, 7, 8, 10, 11, 13, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 40, 41, 43, 44, 46, 47, 49, 50, 52, 53, 55, 56, 58, 60, 61, 63, 64, 66, 67, 69, 70, 72, 74, 75, 77, 78, 80, 81, 83, 84, 86, 88, 89, 91, 92, 94, 95, 97, 98

Offset: 1

Author

Clark Kimberling, Oct 09 2014

Keywords

• nonn
• easy

Comments

This sequence provides insight into the manner of convergence of sum{1/(h*2^h), h = 1..k} to log 2. Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248560 and A248561 partition the positive integers.

Examples

			Let s(n) = log(2) - sum{1/(h*2^h), h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/3^n
1 ... 0.193147 .... 0.33333
2 ... 0.0681472 ... 0.11111
3 ... 0.0264805 ... 0.037037
4 ... 0.0108555 ... 0.0123457
5 ... 0.0046066 ... 0.004115
6 ... 0.0020013 ... 0.00137174
a(5) = 6 because s(6) < 1/3^5 < s(5).
		

References

• Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 15.

Links

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

Crossrefs

Cf. A002162 (log(2)), A248560, A248561.

Programs

Mathematica

z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]
N[Table[Log[2] - p[n], {n, 1, z/5}]]
f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]    (* A248559 *)
Flatten[Position[Differences[u], 1]]   (* A248560 *)
Flatten[Position[Differences[u], 2]]   (* A248561 *)