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Search: id:A069323
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| A069323 |
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Triangle in which n-th row gives ascending sequence of numbers derived from the (3x+1) problem, beginning with n. Numbers in one row share the same number of iteration steps required to reach the value of '1' when applying the (3x+1) algorithm. Each row terminates with a power of 2. |
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1
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| 1, 2, 3, 20, 128, 4, 5, 32, 6, 40, 256, 7, 44, 272, 1664, 10240, 65536, 8, 9, 56, 352, 2176, 13312, 81920, 524288, 10, 64, 11, 68, 416, 2560, 16384, 12, 80, 512, 13, 80, 512, 14, 88, 544, 3328, 20480, 131072, 15, 92, 560, 3392, 20480, 131072, 16, 17, 104, 640
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Provided that a number m will iterate to the number 1 by the (3x+1) algorithm, taking s steps, and also provided that m is not a power of 2, then the sequence beginning with the number m will terminate at 2^s
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LINKS
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Jeffrey C. Lagarias, The 3x + 1 Problem and its Generalizations
Index entries for sequences related to 3x+1 (or Collatz) problem
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FORMULA
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A sequence begins with any positive integer. If a(n) = 2^k then the sequence terminates. a(n+1)=6.a(n) + 2^(k+1) where 2^k is derived from the formula a(n)=m.2^k, with m odd.
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EXAMPLE
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If a(1)=7 then a(2) = 6.7 + 2 =44 44 can be expressed as 11.2^2, therefore a(3) = 6.44 + 2^3 =272
2; 3,20,128; 4; 5,32; 6,40,256; 7,44,272,1664,10240,65536; ...
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CROSSREFS
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Cf. A006577.
Sequence in context: A125763 A042441 A128977 this_sequence A009721 A013340 A012416
Adjacent sequences: A069320 A069321 A069322 this_sequence A069324 A069325 A069326
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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John Hulbert (john.hulbert(AT)velnet.co.uk), Apr 15 2002
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EXTENSIONS
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More terms from David Wasserman (wasserma(AT)spawar.navy.mil), Apr 07 2003
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