Revision History for A075485
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Showing entries 1-10
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#12 by Alois P. Heinz at Fri Dec 15 08:01:04 EST 2023
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#11 by Joerg Arndt at Fri Dec 15 06:22:08 EST 2023
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#10 by Michel Marcus at Fri Dec 15 03:09:40 EST 2023
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#9 by Michel Marcus at Fri Dec 15 03:09:19 EST 2023
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#8 by Kevin Ryde at Fri Dec 15 02:50:22 EST 2023
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#7 by Kevin Ryde at Fri Dec 15 02:48:36 EST 2023
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| LINKS
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<a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
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| STATUS
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approved
editing
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#6 by N. J. A. Sloane at Tue Oct 15 22:31:31 EDT 2013
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| AUTHOR
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_Labos E. (labos(AT)ana.sote.hu), Elemer_, Sep 26 2002
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Discussion
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Tue Oct 15
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| OEIS Server: https://oeis.org/edit/global/2029
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#5 by T. D. Noe at Thu Jan 17 19:30:14 EST 2013
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#4 by T. D. Noe at Thu Jan 17 19:29:04 EST 2013
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| COMMENTS
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n=4, 2^n - 1 = 15, list = {15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1}, so a(4) = 18.
Somewhat surprisingly, these iterations take almost twice as long as the iterations for 2^n + 1. See A075486. - T. D. Noe, Jan 17 2013
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| LINKS
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T. D. Noe, <a href="/A075485/b075485.txt">Table of n, a(n) for n = 1..10000</a>
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| FORMULA
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a(n)=) = A008908(2^n-1)).
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| EXAMPLE
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n=4, 2^n - 1 = 15, list = {15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1}, so a(4) = 18.
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#3 by T. D. Noe at Thu Jan 17 16:44:39 EST 2013
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| NAME
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Length of iteration list when Collatz-function is iterated with initial value -1+ 2^n - 1.
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| COMMENTS
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n=4,-1+, 2^n= - 1 = 15,, list={ = {15,, 46,, 23,, 70,, 35,, 106,, 53,, 160,, 80,, 40,, 20, 10,, 5,, 16,, 8,, 4,, 2,, 1}, so a(4)=) = 18.
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| MATHEMATICA
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Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Table[Length[Collatz[2^n - 1]], {n, 100}] (* T. D. Noe, Jan 17 2013 *)
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approved
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