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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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%I #5 Mar 31 2012 13:20:45

%S 1,2,3,20,128,4,5,32,6,40,256,7,44,272,1664,10240,65536,8,9,56,352,

%T 2176,13312,81920,524288,10,64,11,68,416,2560,16384,12,80,512,13,80,

%U 512,14,88,544,3328,20480,131072,15,92,560,3392,20480,131072,16,17,104,640

%N 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.

%C 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

%H Jeffrey C. Lagarias, <a href="http://www.cecm.sfu.ca/organics/papers/lagarias/paper/html/paper.html">The 3x + 1 Problem and its Generalizations </a>

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

%F 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.

%e 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

%e 2; 3,20,128; 4; 5,32; 6,40,256; 7,44,272,1664,10240,65536; ...

%Y Cf. A006577.

%K nonn,tabf,easy

%O 1,2

%A John Hulbert (john.hulbert(AT)velnet.co.uk), Apr 15 2002

%E More terms from _David Wasserman_, Apr 07 2003