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Godel Escher Bach: Interesting number sequence

  1. Mar 7, 2008 #1

    DaveC426913

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    Gold Member

    I've been re-reading Hofstadter's Godel, Escher and Bach and came across an interesting algorithm.

    Start with any whole number.
    If it is even, halve it.
    If it is odd, triple it and add 1.
    Repeat until the number reaches 1.
    Count # of steps it took.

    Code (Text):

    Number/Count
     1: (1)                  count: 0
     2: (2 1)                count: 1
     3:  (3 10 5 16 8 4 2 1) count: 7
    etc.
     
    This acts very well-behaved for the most part; the count never goes higher than 23


    Until you get to 27...
    Code (Text):

    Number/Count
    25: (25 76 38 19 58 29 88 44 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 ) count: 23
    26: (26 13 40 20 10 5 16 8 4 2 1 ) count: 10
    27:  (27 82 41 124 62 31 94 47 142 71 214 107 322 161 484 242 121 364 182 91 274 137 412 206 103 310 155 466 233 700 350 175 526 263 790 395 1186 593 1780 890 445 1336 668 334 167 502 251 754 377 1132 566 283 850 425 1276 638 319 958 479 1438 719 2158 1079 3238 1619 4858 2429 7288 3644 1822 911 2734 1367 4102 2051 6154 3077 9232 4616 2308 1154 577 1732 866 433 1300 650 325 976 488 244 122 61 184 92 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1 ) count: 111
     
    then it jumps to a whopping 111.

    I've been doing some perfunctory analysis on this (including a little javascript snippet to generate the #s for me (don't get me wrong, the JavaScript came only AFTER I did it all by hand up to 27 (including triple-checking my #s (I was stranded on a plane and had nothing to keep me busy except a pencil and a barf bag with lots of white space on it...(BTW, did I mention that the book is all about recursion?))))) and have found some tantalizing patterns.

    Has anyone seen this algorithm before?
     
    Last edited: Mar 7, 2008
  2. jcsd
  3. Mar 7, 2008 #2
  4. Mar 7, 2008 #3

    DaveC426913

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    Gold Member

    Oooh! Thank you!


    Little did I know, when I was scribbling on the back of that barf bag, that I was generating the first few pixels of a Mandelbrot-esque set!
     
  5. Mar 9, 2008 #4
    If I state the "hailstone function" as,

    f(1) = 1, f(2n) = f(n), f(2n+1)=(3(2n+1)+1)/2

    In the language of recursion theory, this is a partial function. The conjecture is whether this is total, correct? Is this related to the halting problem in some way? Is there a theorem stating whether all recursively defined functions of a certain type are total/partial?
     
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