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A056735
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Numbers n such that 2^n in base 3 has same number of 1's as 2^(n+1) in base 3 and 2^n and 2^(n+1) have the same number of digits in base 3.
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1
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5, 27, 32, 40, 54, 92, 135, 138, 151, 159, 167, 176, 189, 281, 284, 319, 401, 503, 718, 723, 734, 820, 929, 1035, 1086, 1127, 1311, 1341, 1371, 1693, 1785, 1869, 1948, 2010, 2181, 2408, 2563, 2771, 2923, 2983, 3004, 3007, 3210, 3213, 3479, 3527, 4037
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Using empirical data for 1 <= n <= 10000, it has been found that the distribution of these terms correlates well (R^2 = 0.9798) to g(n) = b*n^(1/2) with 'b' a constant approximately 0.70. In addition, g'(n) approximates the probability that any particular n has this property. Any terms in sequence A056154 must also satisfy this sequence.
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EXAMPLE
| First term: 2^5 = 1012, 2^6 = 2101, both with 2 ones and both of length 4. Second term: 2^27 = 100100112222002222, 2^28 = 200201002221012221, both with 4 ones and both of length 18.
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CROSSREFS
| A056154.
Sequence in context: A091721 A039283 A045162 * A056154 A058490 A156215
Adjacent sequences: A056732 A056733 A056734 * A056736 A056737 A056738
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KEYWORD
| easy,nonn,base
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AUTHOR
| Russell Harper (rharper(AT)intouchsurvey.com), Aug 13 2000
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