A160541 Number of odd-then-even runs to reach 1 from n under the modified "3x+1" map: x -> x/2 if x is even, x -> (3x+1)/2 if x is odd.

0, 1, 1, 1, 1, 2, 3, 1, 4, 2, 3, 2, 2, 4, 2, 1, 3, 5, 4, 2, 1, 4, 2, 2, 5, 3, 17, 4, 4, 3, 16, 1, 6, 4, 2, 5, 4, 5, 6, 2, 17, 2, 6, 4, 4, 3, 16, 2, 5, 6, 5, 3, 2, 18, 17, 4, 7, 5, 6, 3, 3, 17, 15, 1, 6, 7, 5, 4, 3, 3, 16, 5, 18, 5, 2, 5, 5, 7, 6, 2, 4, 18, 17

Offset: 1

Author

Brenton Bostick (bostick(AT)gmail.com), May 18 2009

Keywords

• nonn

Comments

The 2->1 step is not counted.

From Dustin Theriault, May 24 2023: (Start)

The ratio of the partial sum of a(n) to the partial sum of A006577(n) appears to approach 1/6 (observation for n = 1..10^10).

The ratio of the partial sum of a(n) to the partial sum of A286380(n) appears to approach 1/2 (observation for n = 1..10^10). (End)

Number of steps x -> A363270(x) to go from n to 1. - Dustin Theriault, Jul 09 2023

Examples

			7->11->17->26->13->20->10->5->8->4->2->1, so the odd-then-even runs are (7->11->17->26) (13->20->10) (5->8->4->2), and a(7) is 3.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Francis Laclé, 2-adic parity explorations of the 3n+1 problem, hal-03201180v2 [cs.DM], 2021.
• Dustin Theriault, Ratio between the partial sum of a(n) to the partial sum of A006577(n), for n = 1..10^10.
• Dustin Theriault, Ratio between the partial sum of a(n) to the partial sum of A286380(n), for n = 1..10^10.
• Dustin Theriault, Histogram of a(n), for n = 1..10^10.
• Dustin Theriault, Combined histograms of A006577, A286380, A160541, for n = 1..10^9.

Crossrefs

Cf. A006577, A286380, A363270.

Programs

C

int a(int n) {
  int steps = 0;
  while (n > 1) {
    while (n & 1) n += (n >> 1) + 1;
    while (!(n & 1)) n >>= 1;
    ++steps;
  }
  return steps;
} /* Dustin Theriault, May 23 2023 */

Mathematica

Array[Length@ Split[Most@ NestWhileList[If[EvenQ@ #, #/2, (3 # + 1)/2] &, #, # > 1 &], Or[OddQ[#1], EvenQ[#2]] &] &, 120] (* Corrected by Michael De Vlieger, Jul 19 2021 *)

Formula

From Alan Michael Gómez Calderón, Mar 19 2025: (Start)

a(n) = A346965(A000265(n)) - (n mod 2) + 1;

a(n) = a(A363270(n)) + 1 for n >= 2. (End)