A378845 -id:A378845 - cp's OEIS frontend

A378846 Smallest starting x which takes n halving steps to reach the minimum of a cycle in the 3x-1 iteration.

1, 2, 4, 3, 6, 11, 13, 9, 18, 35, 25, 47, 33, 63, 45, 81, 95, 117, 127, 85, 57, 113, 133, 89, 97, 65, 129, 87, 173, 225, 231, 293, 309, 377, 261, 273, 545, 671, 465, 485, 597, 647, 741, 529, 353, 705, 471, 941, 1029, 1241, 837, 577, 385, 257, 513, 343, 229, 153

Offset: 0

Kevin Ryde, Dec 15 2024

nonn

Each step is x -> 3x-1 if x odd, or x -> x/2 if x even (A001281) and here only the halving steps x/2 are counted.
The number of halving steps is A377524(x) so that a(n) = x is the smallest x for which A377524(x) = n.
a(n) <= 2*a(n-1) is an upper bound since x = 2*a(n-1) is a candidate for a(n) by first step x -> x/2.
All even terms are a(n) = 2*a(n-1), since any smaller even a(n) would imply a smaller a(n-1) by first step x -> x/2.
No term is of the form y = 6*k + 2, apart from a(1)=2, since odd x = 2*k+1 takes a tripling step to 3*x-1 = y and x is a smaller start with the same number of halvings as y.

Kevin Ryde, Table of n, a(n) for n = 0..932
Kevin Ryde, C Code (set count type HALF)

Cf. A001281 (step), A377524 (number of halving steps).

Cf. A378845 (with all steps), A378847 (with tripling steps).

C
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A378847 Smallest starting x which takes n tripling steps to reach the minimum of a cycle in the 3x-1 iteration.

Original entry on oeis.org

1, 3, 15, 13, 9, 37, 25, 33, 45, 57, 145, 97, 65, 87, 159, 165, 225, 273, 391, 261, 647, 465, 741, 529, 353, 471, 921, 837, 865, 577, 385, 257, 343, 229, 153, 407, 543, 721, 481, 321, 855, 1141, 761, 1015, 677, 903, 1209, 1605, 2149, 1433, 1911, 2529, 3397, 2265

Offset: 0

Kevin Ryde, Dec 15 2024

nonn

Each step is x -> 3x-1 if x odd, or x -> x/2 if x even (A001281) and here only the tripling steps 3x-1 are counted.
The number of tripling steps is A378833(x) so that a(n) = x is the smallest x for which A378833(x) = n.
All terms are odd since any even x takes a first step to x/2 which is a smaller start for the same number of tripling steps.
a(n) >= L(n) = (2*a(n-1) + 1)/3 is a lower bound since a(n) = x must at least have a first step 3x-1 and halve to (3x-1)/2, then n-1 further tripling steps, so (3x-1)/2 >= a(n-1).
Equality a(n) = L(n) occurs iff L(n) is an integer and not a cycle minimum.
A large upper bound for n>=1, showing a(n) always exists, is a(n) <= U(n) = (4^(3^n) - 1)*2^n/3^n + 1, since U(n) is a candidate for a(n) by taking n steps of (3x-1)/2 to reach 4^(3^n) which is a power of 2.
Tighter upper bounds on a(n) can be found by taking predecessor steps back from a(n-c) seeking c tripling steps to reach a(n-c) if that's possible (which for instance it's not if a(n-c) == 0 (mod 3)).
Such predecessors are candidates for a(n), but the actual a(n) might have a trajectory which does not go through any previous a(n-c).

			For n=4, a(4) = 9 has 4 tripling steps on its way to 5 which is the minimum of a cycle:
  9 -> 26 -> 13 -> 38 -> 19 -> 56 -> 28 -> 14 -> 7 -> 20 -> 10 -> 5
    ^            ^           ^                      ^
This a(4) = 9 is an example where a(n) is at its lower bound L(n), in this case a(3) = 13 has L(4) = (2*a(3)+1)/3 = 9 which is an integer and not a cycle minimum.

Kevin Ryde, Table of n, a(n) for n = 0..625
Kevin Ryde, C Code (set count type TRIPLE)

Cf. A001281 (step), A378833 (number of triplings).

Cf. A378845 (with all steps), A378846 (with halving steps).

C
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cp's OEIS Frontend

A378846 Smallest starting x which takes n halving steps to reach the minimum of a cycle in the 3x-1 iteration.

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