A185050 Least k such that G(k) > 3 - 1/2^n, where G(k) is the sum of the first k terms of the geometric series 1 + 2/3 + (2/3)^2 + ....

3, 5, 7, 8, 10, 12, 13, 15, 17, 19, 20, 22, 24, 25, 27, 29, 31, 32, 34, 36, 37, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 58, 60, 61, 63, 65, 66, 68, 70, 72, 73, 75, 77, 78, 80, 82, 84, 85, 87, 89, 90, 92, 94, 96, 97, 99, 101, 102, 104, 106, 107, 109, 111, 113

Offset: 0

Author

Arkadiusz Wesolowski, Dec 25 2012

Keywords

• nonn

Comments

Many of terms in this sequence are that same as A186219(n+2) but not all.

Examples

			a(1) = 5 because 1 + 2/3 + (2/3)^2 + (2/3)^3 + (2/3)^4 > 3 - 1/2.
		

References

• Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Eric Weisstein's World of Mathematics, Geometric Series

Programs

Mathematica

lst = {}; n = s = 0; Do[s = s + (2/3)^k; If[s > 3 - 1/2^n, AppendTo[lst, k + 1]; n++], {k, 0, 112}]; lst