A069323 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.

1, 2, 3, 20, 128, 4, 5, 32, 6, 40, 256, 7, 44, 272, 1664, 10240, 65536, 8, 9, 56, 352, 2176, 13312, 81920, 524288, 10, 64, 11, 68, 416, 2560, 16384, 12, 80, 512, 13, 80, 512, 14, 88, 544, 3328, 20480, 131072, 15, 92, 560, 3392, 20480, 131072, 16, 17, 104, 640

Offset

1,2

Comments

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

Links

Table of n, a(n) for n=1..54.

Jeffrey C. Lagarias, The 3x + 1 Problem and its Generalizations

Index entries for sequences related to 3x+1 (or Collatz) problem

Formula

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.

Example

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

Crossrefs

Cf. A006577.

Sequence in context: A042441 A128977 A375683 * A356884 A009721 A013340

Adjacent sequences: A069320 A069321 A069322 * A069324 A069325 A069326

Keyword

nonn,tabf,easy

Author

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

Extensions

More terms from David Wasserman, Apr 07 2003

Status

approved