A281665 -id:A281665 - cp's OEIS frontend

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

Michel Lagneau, Nov 16 2019

nonn

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.

			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.

Cf. A006666, A006667.

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

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

cp's OEIS Frontend

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

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