|
Search: id:A005186
|
|
|
| A005186 |
|
a(n) = number of m which take n steps to reach 1 in `3x+1' problem. (Formerly M0305)
|
|
+0 3
|
|
| 1, 1, 1, 1, 1, 2, 2, 4, 4, 6, 6, 8, 10, 14, 18, 24, 29, 36, 44, 58, 72, 91, 113, 143, 179, 227, 287, 366, 460, 578, 732, 926, 1174, 1489, 1879, 2365, 2988, 3780, 4788, 6049, 7628, 9635, 12190, 15409, 19452, 24561, 31025, 39229, 49580, 62680, 79255, 100144
(list; graph; listen)
|
|
|
OFFSET
|
0,6
|
|
|
REFERENCES
|
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
S. N. Anderson, Struggling with the 3x+1 problem, Math. Gazette, 71 (1987), 271-274.
R. K. Guy, personal communication.
|
|
LINKS
|
T. D. Noe, Table of n, a(n) for n=0..70
Index entries for sequences related to 3x+1 (or Collatz) problem
|
|
FORMULA
|
Appears to settle into approximately exponential growth after about 25 terms or so, with a ratio between adjacent terms of roughly 1.264. - Howard A. Landman (howard(AT)riverrock.org), May 24 2003
David W. Wilson (Jun 10, 2003) gives a heuristic argument that the constant should be the largest eigenvalue of the matrix [ 1 0 0 1 0 0 / 0 0 0 0 1/3 0 / 0 1 0 0 1 0 / 0 0 0 0 1/3 0 / 0 0 1 0 0 1 / 0 0 0 0 1/3 0 ], which is (3 + sqrt(21))/6 = 1.2637626...
|
|
PROGRAM
|
#!/usr/bin/perl @old = ( 1 ); while (1) { print scalar(@old), " "; @new = ( ); foreach $n (@old) { $used{$n} = 1; if (($n % 6) == 4) { $m = ($n-1)/3; push(@new, $m) unless ($used{$m}); } $m = $n + $n; push(@new, $m) unless ($used{$m}); } @old = @new; }
|
|
CROSSREFS
|
Cf. A088975
Sequence in context: A131055 A052928 A137501 this_sequence A008642 A001364 A029010
Adjacent sequences: A005183 A005184 A005185 this_sequence A005187 A005188 A005189
|
|
KEYWORD
|
nonn,easy,nice
|
|
AUTHOR
|
N. J. A. Sloane (njas(AT)research.att.com), R. K. Guy
|
|
EXTENSIONS
|
More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001
|
|
|
Search completed in 0.002 seconds
|