A137284 a(0)=1 and a(n) for n > 0 equals the minimal positive integer such that addition of 2^(-a(n)) to Sum_{k = 0,1,...,n-1} 2^(-a(k)) changes only trailing zeros in its decimal representation.

1, 4, 14, 47, 157, 522, 1735, 5764, 19148, 63609, 211305, 701941, 2331798, 7746066, 25731875, 85479439, 283956550, 943283242, 3133519104, 10409325148, 34579029658, 114869050115, 381586724811, 1267603661786, 4210888217270, 13988267873380, 46468020047392

Offset: 0

Author

Paolo P. Lava and Giorgio Balzarotti, Mar 14 2008

Keywords

• nonn
• base

Comments

First and last nonzero decimal digits of 2^(-m) appear respectively at the ceiling(m/log_2(10))-th and m-th positions after the point. Hence a(n+1) equals the minimum solution to ceiling(x/log_2(10)) = a(n) + 1, which is x = ceiling(a(n)*log_2(10)).

Examples

			Start from 0;
0 + 2^(-1) = 0.5;
0.5 + 2^(-4) = 0.5625 (first digit "5" is equal to the decimal of previous number);
0.5625 + 2^(-14) = 0.56256103515625 (first digits "5625" are equal to the decimals of previous number);
etc.
		

Formula

a(n+1) = ceiling(a(n)*log_2(10)) = ceiling(a(n)*A020862). - Conjectured by R. J. Mathar, proved by Max Alekseyev

Extensions

Edited and extended by Max Alekseyev, May 13 2009