A134191 Impure numbers in the Collatz (3x+1) iteration.

2, 4, 5, 8, 10, 11, 13, 14, 16, 17, 20, 22, 23, 26, 28, 29, 31, 32, 34, 35, 38, 40, 41, 44, 46, 47, 49, 50, 52, 53, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 74, 76, 77, 80, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 98, 100, 101, 103, 104, 106, 107, 110, 112, 113, 116, 118

Offset: 1

Author

T. D. Noe, Oct 12 2007

Keywords

• nonn

Comments

Let f(k) be the trajectory of the Collatz iteration of the number k. Then Shaw calls a number n impure if n is in f(k) for some k < n. Shaw has an algorithm for finding congruences that the impure numbers satisfy.

Examples

			The Collatz trajectory of 3 is (3,10,5,16,8,4,2,1), showing that the numbers 4,5,8,10,16 are impure.
		

Links

• T. D. Noe, Table of n, a(n) for n=1..10000
• Douglas J. Shaw, The Pure Numbers Generated by the Collatz Sequence, The Fibonacci Quarterly, Vol. 44, Number 3, August 2006, pp. 194-201.

Programs

Mathematica

c[n_] := If[EvenQ[n], n/2, 3n + 1]; nn=1000; t=Table[0,{nn}]; Do[If[t[[n]]==0, m=n; While[m=c[m]; If[nn>=m>n && t[[m]]==0, t[[m]]=n]; m>nn || t[[m]]>0]], {n,nn}]; Flatten[Position[t,_?(#>0&)]]

Formula

Complement of A061641.