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Search: id:A160348
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| A160348 |
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Minimal recursive sequence such that if a(n)>0 then always a(n)>a((f(2n+1)-1)/2), where f is defined as in A159885 |
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3
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| 0, 2, 1, 6, 7, 5, 3, 11, 4, 13, 14, 10, 15, 52, 12, 50, 53, 9, 54, 59, 51, 62, 63, 49, 60, 65, 8, 68, 69, 58, 16, 75, 61, 56, 76, 48, 77, 80, 64, 84, 85, 67, 78, 88, 57, 44
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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If (3x+1)-Collatz conjecture is true, then this sequence is a permutation of nonnegative integers.
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EXAMPLE
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Put a(0)=0. Let m=3. Then f(m)=5, f^2(m)=1. The corresponding numbers n=(m-1)/2 are: 1,2,0. By the condition, a(1)>a(2)>a(0)=0. Therefore put a(2)=1, a(1)=2. Furthermore, consider m=7. Then f(m)=11, f^2(m)=17, f^3(m)=13, f^4(m)=5. The corresponding numbers n=(m-1)/2 are: 3,5,8,6,2 and, by the condition, a(3)>a(5)>a(8)>a(6)>a(2)=1.Therefore put a(6)=3 (the minimal value which yet did not appear), a(8)=4, a(5)=5, a(3)=6, etc.
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CROSSREFS
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A159985 A159945 A160198 A006519 A122458
Sequence in context: A084312 A066752 A059364 this_sequence A047708 A110608 A112007
Adjacent sequences: A160345 A160346 A160347 this_sequence A160349 A160350 A160351
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KEYWORD
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nonn,uned
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AUTHOR
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Vladimir Shevelev (shevelev(AT)bgu.ac.il), May 10 2009; corrected May 13 2009, May 19 2009
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