The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A324246 Irregular triangle T read by rows: T(n, k) = (A324038(n, k) - 1)/2. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 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. LINKS Nicolas Vaillant and Philippe Delarue, The hidden face of the 3x+1 problem. Part I: Intrinsic algorithm, April 26 2019. 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)). EXAMPLE 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. CROSSREFS Cf. A248573 (Collatz-Terras tree), A324038 (CfsTree), A324039, A324040, A324245. Sequence in context: A143172 A004747 A155810 * A225470 A081099 A213252 Adjacent sequences:  A324243 A324244 A324245 * A324247 A324248 A324249 KEYWORD nonn,tabf,easy AUTHOR Nicolas Vaillant, Philippe Delarue, Wolfdieter Lang, May 09 2019 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified April 23 04:58 EDT 2021. Contains 343199 sequences. (Running on oeis4.)