A324246 Irregular triangle T read by rows: T(n, k) = (A324038(n, k) - 1)/2.

0, 2, 1, 10, 6, 42, 8, 26, 56, 170, 5, 34, 17, 106, 37, 226, 113, 682, 3, 22, 138, 11, 70, 426, 150, 906, 75, 454, 2730, 4, 14, 90, 184, 554, 7, 46, 282, 568, 1706, 200, 602, 1208, 3626, 100, 302, 1818, 3640, 10922, 18, 9, 58, 120, 362, 738, 369, 2218, 30, 186, 376, 1130, 2274, 1137, 6826, 133, 802, 401, 2410, 805, 4834, 2417, 14506, 402, 201, 1210, 2424, 7274, 14562, 7281, 43690

Offset: 0

Author

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

Keywords

• nonn
• tabf
• easy

Comments

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

This is the incomplete binary tree corresponding to the modified Collatz map f (from the Vaillant and Delarue link) given in A324245.

The branches of this tree, called CfTree, give the iterations under the Vaillant and Delarue map f of the vertex labels of level n until label 0 on level n = 0 is reached.

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

The number of vertex labels on level n which are 1 (mod 3) is given in A324040.

The corresponding tree CfsTree with only odd vertex labels t(n, k) = 2*T(n,k) + 1 is given in A324038.

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

For this tree see Figure 2 in the Vaillant-Delarue link. It is also shown in the W. Lang link given in A324038.

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:    0
1:    2
2:   (1 10)
3:    6 42
4:   (8 26) (56 170)
5:   (5 34) (17 106)  (37  226) (113 682)
6:   (3 22) 138 (11    70) 426   150 906 (75 454) 2730
...
Row n = 7: (4 14) 90 (184 554)  (7 46) 282 (568 1706) (200 602) (1208 3626) (100 302) 1818 (3640 10922);
Row n = 8: 18 (9 58) (120 362) 738 (369 2218) 30 186 (376 1130) 2274 (1137 6826) (133 802) (401 2410) (805 4834) (2417 14506) 402 (201 1210) (2424 7274) 14562 (7281 43690).
...
The successors of T(1,1) = 2 == 2 (mod 3) are (-1 + 2*2 )/3 = 1 and 2*(1 + 2*2) = 10. The successor of T(2, 1) = 1  == 1 (mod 3) is 2*(1 + 2*1) = 6. The successors of T(3, 1) = 6  == 0 (mod 3) are 4*6/3 = 8 and 2*(1 + 2*6) = 26.
		

Links

• Nicolas Vaillant and Philippe Delarue, The hidden face of the 3x+1 problem. Part I: Intrinsic algorithm, April 26 2019.

Crossrefs

Cf. A248573 (Collatz-Terras tree), A324038 (CfsTree), A324039, A324040, A324245.

Formula

Recurrence for the set of vertex labels CfTree(n) = {T(n, k), k = 1..A324039(n)} on level (row) n:

This set is obtained, with the map f from A324245, from CfTree(0) = {0}, CfTree(1) = {2}, and for n >= 2 CfTree(n) = {m >= 0: f(m) = T(n-1, k), for k = 1.. A324039(n-1)}.

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

a label with T(n, k) == 1 (mod 3) produces the label 2*(1 + 2*T(n, k)) on row n+1; label T(n, k) == 0 (mod 3) produces the two labels 4*T(n, k)/3 and 2*(1 + 2*T(n, k)); label T(n, k) == 2 (mod 3) produces the two labels (-1 + 2*T(n, k))/3 and 2*(1 + 2*T(n, k)).