A329535 Numbers with twice as many halving steps before reaching 1 in the 3x + 1 problem as tripling steps.

1, 159, 283, 377, 502, 503, 603, 615, 668, 669, 670, 799, 807, 888, 890, 892, 893, 1063, 1065, 1095, 1186, 1187, 1188, 1189, 1190, 1417, 1435, 1580, 1581, 1582, 1585, 1586, 1587, 1889, 1913, 1947, 1959, 1963, 2104, 2106, 2108, 2109, 2113, 2114, 2115, 2119, 2518

Offset

1,2

Comments

Essentially the same as A281665. - R. J. Mathar, Feb 07 2020

Numbers m such that A006666(m) = 2 * A006667(m).

Steps after reaching 1 the first time are ignored. For example, for 5, 16, 8, 4, 2, 1, 4, 2, 1, ..., only 8, 4, 2, 1 are counted for halving steps, the subsequent 4, 2, 1 subcycles are ignored.

Links

Gus Michel, Table of n, a(n) for n = 1..2000

Example

159 is in the sequence because its trajectory, 159, 478, 239, 718, ..., has 36 halving steps and 18 tripling steps.
160 is not in the sequence because its trajectory, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1, has nine even terms but only two odd terms.

Mathematica

collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 50; t = {}; n = 0; While[Length[t] < nn, n++; c = collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; If[ev == 2 * od, AppendTo[t, n]]]; t

Prog

(Scala) def halfTripleCompare(n: Int): Int = {
  var curr = n
  var htc = 0
  while (curr > 1) {
    curr = (curr % 2) match {
      case 0 => htc = htc + 1
        curr / 2
      case 1 => htc = htc - 2
        3 * curr + 1
    }
  }
  htc
}
(1 to 1000).filter(halfTripleCompare(_) == 0) // Alonso del Arte, Nov 18 2019

Crossrefs

Cf. A006666, A006667.

Sequence in context: A093472 A270304 A281665 * A250659 A280971 A090948

Adjacent sequences: A329532 A329533 A329534 * A329536 A329537 A329538

Keyword

nonn

Author

Michel Lagneau, Nov 16 2019

Status

approved