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A137284
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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 tailing zeros in its decimal representation.
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1
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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; internal format)
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OFFSET
| 0,2
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COMMENTS
| First and last nonzero decimal digits of 2^(-m) appear respectively at the ceil(m/log2(10))-th and m-th positions after the point. Hence a(n+1) equals the minimum solution to ceil(x/log2(10))=a(n)+1, which is x = ceil(a(n)*log2(10)).
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FORMULA
| a(n+1) = ceil(a(n)*log2(10)) = ceil(a(n)*A020862). - Conjectured by R. J. Mathar, proved by Max Alekseyev.
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EXAMPLE
| Start from 0.
0 + 2^(-1)= 0.5
0.5 + 2^(-4)= 0.5625 (first digit "5" is equal to the decimal of prevoius number)
0.5625 + 2^(-14)=0.56256103515625 (first digits "5625" are equal to the decimals of prevoius number)
etc.
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CROSSREFS
| Sequence in context: A104487 A094789 A082574 * A121095 A000908 A015651
Adjacent sequences: A137281 A137282 A137283 * A137285 A137286 A137287
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KEYWORD
| nonn
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AUTHOR
| Paolo P. Lava & Giorgio Balzarotti (paoloplava(AT)gmail.com), Mar 14 2008
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EXTENSIONS
| Edited and extended by Max Alekseyev (maxale(AT)gmail.com), May 13 2009
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