A324038 Irregular triangle T read by rows: Row n gives the vertex labels of level n of the tree related to the modified reduced Collatz map A324036.

1, 5, 3, 21, 13, 85, 17, 53, 113, 341, 11, 69, 35, 213, 75, 453, 227, 1365, 7, 45, 277, 23, 141, 853, 301, 1813, 151, 909, 5461, 9, 29, 181, 369, 1109, 15, 93, 565, 1137, 3413, 401, 1205, 2417, 7253, 201, 605, 3637, 7281, 21845, 37, 19, 117, 241, 725, 1477, 739, 4437, 61, 373, 753, 2261, 4549, 2275, 13653, 267, 1605, 803, 4821, 1611, 9669, 4835, 29013, 805, 403, 2421, 4849, 14549, 29125, 14563, 87381

Offset: 0

Author

Nicolas Vaillant, Philippe Delarue, Wolfdieter Lang, May 09 2019

Keywords

• nonn
• easy
• tabf

Comments

The length of row n is A324039, for n >= 0.

The branches of this incomplete binary tree, called CfsTree, give the iterations of the vertex labels of level n of the modified reduced Collatz map Cfs defined by Cfs(2*k+1) = A324036(k), for k >= 0, until at level n = 0 the label 1 is reached for the first time.

The out-degree of a vertex label T(n, k), for n >= 1, is 1 if T(n, k) == 3 (mod 6) and 2 for all other vertices. For level n = 0 with vertex label 1 this rule does not hold, it has out-degree 1, not 2.

The number of vertex labels on level n which are 3 (mod 6) is given by A324040(n).

The corresponding tree with nonnegative vertex labels t(n, k) = (T(n,k) - 1)/2 is given in A324246.

The Collatz conjecture is that all positive odd integers appear in this CfsTree. Because the sets of labels on the levels are pairwise disjoint these odd numbers will then appear just once.

For this tree see Figure 1 in the Vaillant-Delarue link. It is also shown in the W. Lang link.

Examples

			The irregular triangle T begins (the brackets combine pairs coming from out-degree 2 vertices of the preceding level):
n/k   1  2     3   4     5    6     7    8    9  10    11 ...
-------------------------------------------------------------
0:   1
1:   5
2:  (3  21)
3:  13  85
4:  (17 53) (113 341)
5:  (11 69)  (35 213)  (75  453) (227 1365)
6:  ( 7 45)  277 (23   141) 853   301 1813 (151 909) 5461
...
Row n = 7: (9 29) 181 (369 1109) (15 93)  565 (1137  3413) (401 1205) (2417 7253) (201 605) 3637 (7281 21845);
Row n = 8: 37 (19 117) (241 725) 1477 (739 4437) 61 373 (753 2261) 4549 (2275 13653) (267 1605) (803 4821) (1611 9669) (4835 29013) 805 (403 2421) (4849 14549) 29125 (14563 87381).
...
The successors of T(1,1) = 5 == 5 (mod 6) are (2*5 - 1)/3 = 3 and 4*5 + 1 = 21. The successor of T(2, 1) = 3 == 3 (mod 6) is 4*3 + 1 = 13. The successors of T(3, 1) = 13 == 1 (mod 6) are (4*13 - 1)/3 = 17 and 4*13 + 1 = 53.
		

Links

• Wolfdieter Lang, Collatz Trees from Vaillant-Delarue Maps
• Nicolas Vaillant and Philippe Delarue, The hidden face of the 3x+1 problem. Part I: Intrinsic algorithm, April 26 2019.

Crossrefs

Cf. A324036, A324039, A324040, A324246.

Formula

Recurrence: CfsTree(n), the list of vertex labels {T(n, k), for k = 1..A324038(n)} of level n, is obtained from: CfsTree(0) = {1}, CfsTree(1) = {5}, and for n >= 2, CfsTree(n) = {2*m + 1 >= 1: fs(2*m+1) = T(n-1, k)), for k = 1..A324038(n-1)}, with fs from A324036.

Explicit form for the successor(s) of T(n, k) on level n+1, for n >= 1:

a vertex label with T(n, k) == 3 (mod 6) produces the label 4*T(n, k) + 1 on level n+1; label T(n, k) == 1 (mod 6) produces the two labels (4*T(n, k) - 1)/3 and 4*T(n, k) + 1; label T(n, k) == 5 (mod 6) produces the two labels (2*T(n, k) - 1)/3 and 4*T(n, k) + 1.