A347834 An array A of the positive odd numbers, read by antidiagonals upwards, giving the present triangle T.

1, 3, 5, 7, 13, 21, 9, 29, 53, 85, 11, 37, 117, 213, 341, 15, 45, 149, 469, 853, 1365, 17, 61, 181, 597, 1877, 3413, 5461, 19, 69, 245, 725, 2389, 7509, 13653, 21845, 23, 77, 277, 981, 2901, 9557, 30037, 54613, 87381, 25, 93, 309, 1109, 3925, 11605, 38229, 120149, 218453, 349525

Offset: 1

Author

Wolfdieter Lang, Sep 20 2021

Keywords

• nonn
• tabl
• easy

Comments

For the definition of this array A see the formula section.

The rows of A appear in a draft by Immmo O. Kerner in eqs. (1) and (2) as so-called horizontal sequences (horizontale Folgen). Thanks to Dr. A. Eckert for sending me this paper.

This array with entry A(k, n) becomes equal to the array T with T(n, k) given in A178415 by using a permutation of the rows, and changing the offset: A(k, n) = T(pe(k), n+1), with pe(3*(L+1)) = 4*(L+1), pe(1+3*L) = 1 + 2*L, pe(2+3*L) = 2*(1 + 2*L), for L >= 0. This permutation appears in A265667.

A proper sub-array is A238475(n, k) = A(1 + 3*(k-1), n-1), for k >= 1 and n >= 1.

In the directed Collatz tree with nodes labeled with only positive odd numbers (see A256598 for the paths), here called CTodd, the level L = 0 (on the top) has the node with label 1 as root. Because 1 -> 1 there is an arrow (a 1-cycle or loop) at the root. The level L = 1 consists of the nodes with labels A(1, n), for n >= 1, and each node is connected to 1 by a downwards directed arrow. The next levels for L >= 2 are obtained using the successor rule (used also by Kerner): S(u) = (4*u - 1)/3 if u == 1 (mod 3), (2*u - 1)/3 if u == 5 (mod 3), and there is no successor S(u) = empty if u = 3 (mod 6), that is, this node is a leaf.

However, each node with label u on level L >= 1, except a leaf, has as successors at level L + 1 not only the node with S(u) but all the nodes with labels A(S(u), n), for n >= 0.

In this way each node (also the root) of this CTodd has in-degree 1 and infinite out-degree (for L >= 2 there are infinitely many infinite outgoing arrows). All nodes with label A(k, n) with n >= 1, have the same precursor as the node A(k,0) in this tree for each k >= 1.

Except for the loop (1-cycle) for the root 1 there are no cycles in this directed tree CTodd.

That each number N = 5 + 8*K, for K >= 0 appears in array A for some column n >= 1 uniquely can be proved, using the fact of strictly increasing rows and columns, by showing that the columns n = 1, 2, ..., c contain all positive integers congruent to 5 modulo 8 except those of the positive congruence class A(1, c+1) modulo 2^(2*c+3) by induction on c. [added Dec 05 2021]

Row index k for numbers congruent to 5 modulo 8: Each number N = 5 + 8*K, for K >= 0, from A004770 is a member of row k of the array A starting with element A(k, 0) = (2*A065883(2 + 3*N) - 1)/3. For this surjective map see A347840. [simplified Dec 05 2021]

The Collatz conjecture can be reduced to the conjecture that in this rooted and directed tree CTodd each positive odd number appears as a label once, that is, all entries of the array A appear.

Examples

			The array A(k, n) begins:
k\n  0   1   2    3    4     5      6      7       8       9       10 ...
-------------------------------------------------------------------------
1:   1   5  21   85  341  1365   5461  21845   87381  349525  1398101
2:   3  13  53  213  853  3413  13653  54613  218453  873813  3495253
3:   7  29 117  469 1877  7509  30037 120149  480597 1922389  7689557
4:   9  37 149  597 2389  9557  38229 152917  611669 2446677  9786709
5:  11  45 181  725 2901 11605  46421 185685  742741 2970965 11883861
6:  15  61 245  981 3925 15701  62805 251221 1004885 4019541 16078165
7:  17  69 277 1109 4437 17749  70997 283989 1135957 4543829 18175317
8:  19  77 309 1237 4949 19797  79189 316757 1267029 5068117 20272469
9:  23  93 373 1493 5973 23893  95573 382293 1529173 6116693 24466773
10: 25 101 405 1621 6485 25941 103765 415061 1660245 6640981 26563925
...
--------------------------------------------------------------------
The triangle T(k, n) begins:
k\n  0  1   2    3    4     5     6      7      8      9 ...
------------------------------------------------------------
1:   1
2:   3  5
3:   7 13  21
4:   9 29  53   85
5:  11 37 117  213  341
6:  15 45 149  469 853   1365
7:  17 61 181  597 1877  3413  5461
8:  19 69 245  725 2389  7509 13653  21845
9:  23 77 277  981 2901  9557 30037  54613  87381
10: 25 93 309 1109 3925 11605 38229 120149 218453 349525
...
-------------------------------------------------------------
Row index k of array A, for entries 5 (mod 8).
213 = 5 + 8*26. K = 28 is even, (3*231+1)/16 = 40, A065883(40) = 10, hence A(k, 0) = N' = (10-1)/3 = 3, and k = 2. Moreover, n = log_4((3*213 + 1)/(3*A(2,0) + 1)) = log_4(64) = 3. 213 = A(2, 3).
85 = 5 + 8*10. K = 10 is even, (3*85 + 1)/16 = 16, A065883(16) = 1, N' = (1-1)/3 = 0 is even, hence A(k, 0) = 4*0 + 1 = 1, k = 1. 85 = A(1, 3).
61 = 5 + 8*7, K = 7 is odd, k = (7+1)/2 + ceiling((7+1)/4) = 6, and n = log_4((3*61 + 1)/(3*A(6,0) + 1)) = 1. 61 = A(6, 1).
----------------------------------------------------------------------------
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..11325 (first 150 antidiagonals, flattened).
• Immo O. Kerner, Elementarer Lösungsansatz zum Collatz-Ulam-Problem CUP oder das "3a + 1" Problem, Version Nov 26 2000.
• Eric Weisstein's World of Mathematics, Collatz Problem
• Wikipedia, Collatz conjecture
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Row sequences of the array A, also diagonal sequences of the triangle T: -A007583 (k=0), A002450(n+1), A072197, A072261(n+1), A206374(n+1), A072262(n+1), A072262(n+1), A072201(n+1), A330246(n+1), ...

Column sequences of the array A, also of the triangle T (shifted): A047529, A347836, A347837, ...

Cf. A004770, A007494, A016777, A065883, A047395, A047529, A178415, A238475, A256598, A265667, A319451, A347839, A347840.

Programs

Maple

# Seen as an array:
A := (n, k) -> ((3*(n + floor(n/3)) - 1)*4^(k+1) - 2)/6:
for n from 1 to 6 do seq(A(n, k), k = 0..9) od;
# Seen as a triangle:
T := (n, k) -> 2^(2*k + 1)*(floor((n - k)/3) - k + n - 1/3) - 1/3:
for n from 1 to 9 do seq(T(n, k), k = 0..n-1) od;
# Using row expansion:
gf_row := k -> (1 / (x - 1) - A047395(k)) / (4*x - 1):
for k from 1 to 10 do seq(coeff(series(gf_row(k), x, 11), x, n), n = 0..10) od;
# Peter Luschny, Oct 09 2021

Mathematica

A347834[k_, n_] := (4^n*(6*(Floor[k/3] + k) - 2) - 1)/3;
Table[A347834[k - n, n], {k, 10}, {n, 0, k - 1}] (* Paolo Xausa, Jun 26 2025 *)

Formula

Array A:

A(k, 0) = A047529(k) (the positive odd numbers {1, 3, 7} (mod 8));

A(k, n) = ((3* A(k, 0) + 1)*4^n - 1)/3, for k >= 1 and n >= 0.

Recurrence for rows k >= 1: A(k, n) = 4*A(k, n-1) + 1, for n >= 1, with A(k, 0) = 2*(k + floor(k/3)) - 1 = A047529(k).

Explicit form: A(k, n) = ((3*(k + floor(k/3)) - 1)*4^(n+1) - 2)/6, k >= 1, n >= 0. Here 3*(k + floor(k/3)) = A319451(k).

Hence A(k, n) = 5 + 8*(2*A(k, n-2)), for n >= 1, with A(k, 0) = 2*(k + floor(k/3)) - 1 = A047529(k), and 2*A(k, -1) = (A(k, 1) - 5)/8 = k - 1 + floor(k/3) (equals index n of A(k, 1) in the sequence (A004770(n+1))_{n >= 0}). A(k, -1) is half-integer if k = A007494(m) = m + ceiling(m/2), for m >= 1, and A(k, -1) = 2*K if k = 1 + 3*K = A016777(K), for K >= 0.

O.g.f.: expansion in z gives o.g.f.s for rows k, also for k = 0: -A007583; expansion in x gives o.g.f.s for columns n.

G(z, x) = (2*(-1 + 3*z + 3*z^2 + 7*z^3)*(1-x) - (1-4*x)*(1-z^3)) / (3*(1-x)*(1-4*x)*(1-z)*(1-z^3)).

Triangle T:

T(k, n) = A(k - n, n), for k >= 1 and n = 0..k-1.

A(k, n) = [x^n] (1/(x - 1) - A047395(k)) / (4*x - 1). - Peter Luschny, Oct 09 2021