A247983 Least number k such that log(2) - sum{1/(h*2^h), h=1..k} < 1/2^n.

1, 1, 2, 3, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 15, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63

Offset: 1

Author

Clark Kimberling, Sep 28 2014

Keywords

• nonn
• easy

Comments

It appears that a(n+1) - a(n) if and only if n is in A083706(n), for n >= 1.

Examples

			Let w(n) = log(2) - sum{1/(h*2^h), h=1..n}.  Approximations are shown here:
n .... w(n)  ...... 1/2^n
1 ... 0.193147 .... 0.5
2 ... 0.0681472 ... 0.25
3 ... 0.0264805 ... 0.125
4 ... 0.0108555 ... 0.0625
5 ... 0.0046055 ... 0. 03125
a(4) = 3 because w(3) < 1/2^4 < w(2).
		

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. A083706, A247968.

Programs

Mathematica

z = 200; s[k_] := s[k] = Sum[1/(h*2^h), {h, 1, k}];
N[Table[Log[2] - s[n], {n, 1, z/8}]]
f[n_] := f[n] = Select[Range[z], Log[2] - s[#] < 1/2^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]   (* A247983 *)