A092893 Smallest starting value in a Collatz '3x+1' sequence such that the sequence contains exactly n tripling steps.

1, 5, 3, 17, 11, 7, 9, 25, 33, 43, 57, 39, 105, 135, 185, 123, 169, 219, 159, 379, 283, 377, 251, 167, 111, 297, 395, 263, 175, 233, 155, 103, 137, 91, 121, 161, 107, 71, 47, 31, 41, 27, 73, 97, 129, 171, 231, 313, 411, 543, 731, 487, 327, 859, 1145, 763, 1017, 1351

Offset: 0

Author

Hugo Pfoertner, Mar 11 2004

Keywords

• nonn

Comments

First occurrence of n in A006667.

These are the odd (primitive) terms in A129304. - T. D. Noe, Apr 09 2007

Except for a(1) = 5, all values are congruent {1, 3, 7} (mod 8). Reason: If n is 5 (mod 8) then the Collatz trajectory starting with m = (n - 1)/4 contains the same number of tripling steps, because n = 4m + 1 and the Collatz 3x + 1 step results in 3*(4m + 1) + 1 = 12m + 4 which gets reduced by halving to 3m + 1, without changing the number of tripling steps. - Ralf Stephan, Jun 19 2025

Examples

			a(4)=11 because the Collatz sequence 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 is the first sequence containing 4 tripling steps.
		

Links

• T. D. Noe, Table of n, a(n) for n=0..300
• Jeffrey R. Goodwin, The 3x+1 Problem and Integer Representations, Arxiv preprint arXiv:1504.03040 [math.NT], 2015.
• Eric Weisstein's World of Mathematics, Collatz Problem
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006667, A033491, A092892, A087228.

Row n=1 of A354236.

Programs

Mathematica

a[n_]:=Length[Select[NestWhileList[If[EvenQ[#],#/2,3#+1] &,n,#>1 &],OddQ]]; Table[i=1; While[a[i]!=n,i=i+2]; i,{n,58}] (* Jayanta Basu, May 27 2013 *)