A248565 Least k such that log(4/3) - sum{1/(h*4^h), h = 1..k} < 1/8^n.

1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 16, 17, 19, 20, 21, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 43, 45, 46, 48, 49, 51, 52, 54, 55, 57, 58, 60, 61, 63, 64, 66, 67, 69, 70, 72, 73, 75, 76, 78, 79, 81, 82, 84, 85, 86, 88, 89, 91, 92, 94, 95

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*4^h), h = 1..k} to log(4/3). Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248566 and A248567 partition the positive integers.

Examples

			Let s(n) = log(4/3) - sum{1/(h*4^h), h = 1..n}.  Approximations follow:
n ... s(n) ........ 1/8^n
1 ... 0.037682 ... 0.125
2 ... 0.006432 ... 0.015625
3 ... 0.001223 ... 0.001953
4 ... 0.000247 ... 0.000244
5 ... 0.000051 ... 0.000030
a(4) = 5 because s(5) < 1/8^4 < s(4).
		

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. A083679 (log(4/3)), A248566, A248567, A248559, A248565.

Programs

Mathematica

z = 2500; p[k_] := p[k] = Sum[1/(h*4^h), {h, 1, k}];
N[Table[p[k], {k, 1, z/5}], 12];
N[Table[Log[4/3] - p[n], {n, 1, z/5}]];
f[n_] := f[n] = Select[Range[z], Log[4/3] - p[#] < 1/8^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]] ; (* A248565 *)
Flatten[Position[Differences[u], 1]]; (* A248566 *)
Flatten[Position[Differences[u], 2]]; (* A248567 *)