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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.

%I A137284 #26 Jul 27 2017 03:41:36
%S A137284 1,4,14,47,157,522,1735,5764,19148,63609,211305,701941,2331798,
%T A137284 7746066,25731875,85479439,283956550,943283242,3133519104,10409325148,
%U A137284 34579029658,114869050115,381586724811,1267603661786,4210888217270,13988267873380,46468020047392
%N 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.
%C A137284 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)).
%F A137284 a(n+1) = ceiling(a(n)*log_2(10)) = ceiling(a(n)*A020862). - Conjectured by _R. J. Mathar_, proved by _Max Alekseyev_
%e A137284 Start from 0;
%e A137284 0 + 2^(-1) = 0.5;
%e A137284 0.5 + 2^(-4) = 0.5625 (first digit "5" is equal to the decimal of previous number);
%e A137284 0.5625 + 2^(-14) = 0.56256103515625 (first digits "5625" are equal to the decimals of previous number);
%e A137284 etc.
%K A137284 nonn,base
%O A137284 0,2
%A A137284 _Paolo P. Lava_ and _Giorgio Balzarotti_, Mar 14 2008
%E A137284 Edited and extended by _Max Alekseyev_, May 13 2009