cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A378845 Smallest starting x which takes n steps to reach the minimum of a cycle in the 3x-1 iteration.

Original entry on oeis.org

1, 2, 4, 7, 3, 6, 11, 19, 21, 13, 26, 9, 18, 35, 37, 73, 25, 49, 98, 33, 66, 131, 45, 90, 175, 127, 117, 85, 149, 57, 113, 199, 209, 133, 265, 89, 177, 65, 119, 237, 87, 159, 165, 329, 231, 225, 439, 309, 293, 585, 377, 391, 273, 261, 521, 1042, 671, 695, 485
Offset: 0

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Author

Kevin Ryde, Dec 09 2024

Keywords

Comments

Each step is x -> 3x-1 if x odd, or x -> x/2 if x even (A001281).
The number of steps is A135730(x) so that a(n) = x is the smallest x for which A135730(x) = n.
a(n) <= 2*a(n-1) since x = 2*a(n-1) is a candidate for a(n) by first step x -> x/2.
Even terms are always a(n) = 2*a(n-1) since any smaller even a(n) would imply a smaller a(n-1) after first step x -> x/2.
No term is of the form 12*k+4, since its first step to 6*k+2 is also where the first step from 2*k+1 goes and the latter is a smaller start.
a(n) >= (a(n-1) + 1)/3 is a lower bound since a(n) = x must at least have a first step 3x-1 which reaches somewhere with n-1 further steps, so 3x-1 >= a(n-1).
Equality a(n) = (a(n-1) + 1)/3 = x occurs iff that x is an odd integer and not a cycle minimum, so its first step is to 3x-1 = a(n-1) (as for example at n=11).

Crossrefs

Cf. A001281 (step), A135730 (number of steps).
Cf. A378846 (with halving steps), A378847 (with tripling steps).
Cf. A033491 (in 3x+1).

Programs

  • C
    /* See links. */

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

Views

Author

Kevin Ryde, Dec 15 2024

Keywords

Comments

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).

Examples

			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.
		

Crossrefs

Cf. A001281 (step), A378833 (number of triplings).
Cf. A378845 (with all steps), A378846 (with halving steps).

Programs

  • C
    /* See links. */
Showing 1-2 of 2 results.