A324039 -id:A324039 - cp's OEIS frontend

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

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

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.

			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.

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.

Cf. A324036, A324039, A324040, A324246.

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.

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

Original entry on oeis.org

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

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

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.

			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.

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

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

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)).

A324040 Number of vertex labels congruent to 1 modulo 3 of level n of the irregular triangle A324246.

Original entry on oeis.org

0, 0, 2, 0, 0, 5, 3, 7, 12, 12, 30, 51, 75, 139, 232, 365, 640, 1029, 1717, 2872, 4789, 7996, 13338, 22288, 36896, 61942, 102746, 170993, 286029, 476053, 793800

Offset: 0

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

a(n) is also the number of vertex labels congruent to 3 modulo 6 of row n of the irregular triangle A324038.

This entry is interesting because it determines the number of vertices with out-degree 1 of level n, for n >= 1, of the modified reduced Collatz trees A324038 and A324246. All other vertices have out-degree 2. Hence this sequence determines recursively the number A324039(n) of vertices of label n of these two trees.

Cf. A324038, A324039, A324246.

a(n) = 2*A324039(n) - A324039(n-1), for n >= 1, and a(0) = 0. Implied by the definition of a(n) given in the name.

cp's OEIS Frontend

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.

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

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A324040 Number of vertex labels congruent to 1 modulo 3 of level n of the irregular triangle A324246.

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