A135282 -id:A135282 - cp's OEIS frontend

A231610 The least k such that the Collatz (3x+1) iteration of k contains 2^n as the largest power of 2.

1, 2, 4, 8, 3, 32, 21, 128, 75, 512, 151, 2048, 1365, 8192, 5461, 32768, 14563, 131072, 87381, 524288, 184111, 2097152, 932067, 8388608, 5592405, 33554432, 13256071, 134217728, 26512143, 536870912, 357913941, 2147483648, 1431655765, 8589934592, 3817748707

Offset: 0

T. D. Noe, Dec 02 2013

nonn

Very similar to A225124, where 2^n is the largest number in the Collatz iteration of A225124(n). The only difference appears to be a(8), which is 75 here and 85 in A225124. The Collatz iteration of 75 is {75, 226, 113, 340, 170, 85, 256, 128, 64, 32, 16, 8, 4, 2, 1}.

			The iteration for 21 is {21, 64, 32, 16, 8, 4, 2, 1}, which shows that 64 = 2^6 is a term. However, 32 is not the first power of two. We have to wait until the iteration for 32, which is {32, 16, 8, 4, 2, 1}, to see 32 = 2^5 as the first power of two.

Cf. A010120, A054646 (similar sequences).
Cf. A135282, A232503 (largest power of 2 in the Collatz iteration of n).
Cf. A225124.

Collatz[n_?OddQ] := 3*n + 1; Collatz[n_?EvenQ] := n/2; nn = 21; t = Table[-1, {nn}]; n = 0; cnt = 0; While[cnt < nn, n++; q = Log[2, NestWhile[Collatz, n, Not[IntegerQ[Log[2, #]]] &]]; If[q < nn && t[[q + 1]] == -1, t[[q + 1]] = n; cnt++]]; t

a(n) = 2^n for odd n.

A355187 Number of Collatz trajectories (A070165) for all positive integers <= 10^n that contain 2^4 as the greatest power of 2 within its trajectory.

Original entry on oeis.org

6, 89, 933, 9401, 93744, 937712, 9379078, 93773848

Offset: 1

Frank M Jackson, Jun 23 2022

It is conjectured that lim_{n->infinity} a(n)/10^n = 15/16. Empirically, 93.75% of all trajectories have 2^4 as the greatest power of 2 within its trajectory. Sequence A135282(n) is the maximum power of 2 reached in the Collatz trajectory for integer n.

			a(1)=6 because the first 10 positive integers have trajectories, of which 6 have 2^4 as the greatest power of 2 in their trajectory.
These integers are 3, 5, 6, 7, 9, 10. See trajectory tables below.
  1:    1
  2:    2  1
  3:    3 10  5 16  8   4  2   1
  4:    4  2  1
  5:    5 16  8  4  2   1
  6:    6  3 10  5 16   8  4   2  1
  7:    7 22 11 34 17  52 26  13 40 20 10  5 16  8  4  2  1
  8:    8  4  2  1
  9:    9 28 14  7 22  11 34  17 52 26 13 40 20 10  5 16  8  4  2  1
  10:  10  5 16  8  4   2  1

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

Cf. A070165, A135282.

collatz[n_] := Module[{}, If[OddQ[n], 3n+1, n/2]]; step[n_] := Module[{p=0, m=n, q}, While[!IntegerQ[q=Log[2, m]], m=collatz[m]; p++]; {p, q}]; Counts[Table[Last@step[n], {n, 1, 10^5}]][[Key[4]]]

cp's OEIS Frontend

A231610 The least k such that the Collatz (3x+1) iteration of k contains 2^n as the largest power of 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula