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Search: id:A100961
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| A100961 |
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For a decimal string s, let f(s) = decimal string ijk, where i = number of even digits in s, j = number of odd digits in s, k=i+j. Start with s = decimal expansion of n; a(n) = number of applications of f needed to reach the string 123. |
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+0
3
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| 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 1, 2, 1, 2
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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Obviously if the digits of m and n have the same parity then a(m) = a(n). E.g. a(334) = a(110). In other words, a(n) = a(A065031(n)).
It is easy to show that (i) the trajectory of every number under f eventually reaches 123 (if s has more than three digits then f(s) has fewer digits than s) and (ii) since each string ijk has only finitely many preimages, a(n) is unbounded.
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EXAMPLE
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n=0: s=0 -> f(s) = 101 -> f(f(s)) = 123, stop, a(0) = 2.
n=1: s=1 => f(s) = 011 -> f(f(s)) = 123, stop, f(1) = 2.
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CROSSREFS
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A073054 gives another version. f(n) is (essentially) A073053. Cf. A065031.
Sequence in context: A098708 A067394 A076925 this_sequence A064458 A125891 A153675
Adjacent sequences: A100958 A100959 A100960 this_sequence A100962 A100963 A100964
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KEYWORD
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nonn,easy,base
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Jun 17 2005
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EXTENSIONS
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More terms from Zak Seidov, Jun 18 2005
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