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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. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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)).
LINKS
FORMULA
a(n+1) = ceiling(a(n)*log_2(10)) = ceiling(a(n)*A020862). - Conjectured by R. J. Mathar, proved by Max Alekseyev
EXAMPLE
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.
CROSSREFS
Sequence in context: A082574 A289780 A320404 * A365987 A228178 A000908
KEYWORD
nonn,base
AUTHOR
EXTENSIONS
Edited and extended by Max Alekseyev, May 13 2009
STATUS
approved

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Last modified April 16 17:08 EDT 2024. Contains 371749 sequences. (Running on oeis4.)