A176999 An encoding of the Collatz iteration of n.

1, 1111010, 11, 11110, 11110101, 1111011101101010, 111, 1111011101101010110, 111101, 11110111011010, 111101011, 111101110, 11110111011010101, 11110111110101010, 1111, 111101110110, 11110111011010101101, 11110111011010111010, 1111011, 1111110, 111101110110101

Offset: 2

Author

T. D. Noe, Apr 30 2010

Keywords

• nonn
• base

Comments

Working from right to left, the sequence of 0's and 1's in a(n) encode, respectively, the sequence of 3x+1 and x/2 steps in the Collatz iteration of n. This is reverse one's complement of Garner's parity vector. Criswell mentions this encoding.

The length of a(n) is A006577(n). The number of 1's in a(n) is A006666(n). The number of 0's in a(n) is A006667(n). The number of terms having length k is A005186(k).

Examples

			a(5)=11110 because the Collatz iteration for 5 is a 3x+1 step (0) followed by 4 x/2 steps (four 1's).
		

Links

• T. D. Noe, Table of n, a(n) for n = 2..10000
• Evans A. Criswell, The Collatz Problem (3x+1)
• Lynn E. Garner, On heights in the Collatz 3n+1 problem, Discrete Math, 55 (1985), 57-64.

Crossrefs

Cf. A006577, A006666, A006667, A005186.

Cf. A220145, A260592.

Programs

Mathematica

encode[n_]:=Module[{m=n,p,lst={}}, While[m>1, p=Mod[m,2]; AppendTo[lst,1-p]; If[p==0, m=m/2, m=3m+1]]; FromDigits[Reverse[lst]]]; Table[encode[n], {n,2,26}]