A239126 -id:A239126 - cp's OEIS frontend

A239127 Rectangular companion array to M(n,k), given in A239126, showing the end numbers N(n, k), k >= 1, for the Collatz operation (ud)^n, n >= 1, ending in an odd number, read by antidiagonals.

5, 11, 17, 17, 35, 53, 23, 53, 107, 161, 29, 71, 161, 323, 485, 35, 89, 215, 485, 971, 1457, 41, 107, 269, 647, 1457, 2915, 4373, 47, 125, 323, 809, 1943, 4373, 8747, 13121, 53, 143, 377, 971, 2429, 5831, 13121, 26243, 39365, 59, 161, 431, 1133, 2915, 7289, 17495, 39365, 78731, 118097

Offset: 1

Wolfdieter Lang, Mar 13 2014

The companion array and triangle for the odd start numbers M(n, k) is given in A239126.
See the comments on A239126 for the Collatz 3x+1 problem and the u and d operations.
This rectangular array is N of the Example 2.2. with x=y = n, n >= 1, of the M. Trümper reference, pp. 7-8, written as a triangle by taking NE-SW diagonals. The Collatz sequence starting  with odd M(n, k) from A239126 and ending in odd N(n, k) has length 2*n+1 for each k.
The first row sequences of the array N (columns of triangle TN) are A016969, A239129, ...

			The rectangular array N(n, k) begins:
n\k      1      2      3      4      5      6     7       8       9      10 ...
1:       5     11     17     23     29     35     41     47      53      59
2:      17     35     53     71     89    107    125    143     161     179
3:      53    107    161    215    269    323    377    431     485     539
4:     161    323    485    647    809    971   1133   1295    1457    1619
5:     485    971   1457   1943   2429   2915   3401   3887    4373    4859
6:    1457   2915   4373   5831   7289   8747  10205  11663   13121   14579
7:    4373   8747  13121  17495  21869  26243  30617  34991   39365   43739
8:   13121  26243  39365  52487  65609  78731  91853 104975  118097  131219
9:   39365  78731 118097 157463 196829 236195 275561 314927  354293  393659
10: 118097 236195 354293 472391 590489 708587 826685 944783 1062881 1180979
...
-------------------------------------------------------------------------------
The triangle TN(m, n) begins (zeros are not shown):
m\n   1   2  3     4    5    6     7     8     9     10 ...
1:    5
2:   11  17
3:   17  35  53
4:   23  53 107  161
5:   29  71 161  323  485
6:   35  89 215  485  971 1457
7:   41 107 269  647 1457 2915  4373
8:   47 125 323  809 1943 4373  8747 13121
9:   53 143 377  971 2429 5831 13121 26243 39365
10:  59 161 431 1133 2915 7289 17495 39365 78731 118097
...
n=1, ud, k=1: M(1, 1) = 3 = TM(1, 1), N(1,1) = 5 with the Collatz sequence  [3, 10, 5] of length 3.
n=1, ud, k=2: M(1, 2) = 7 = TM(2, 1), N(1,2) = 11 with the Collatz sequence  [7, 22, 11] of length 3.
n=4, (ud)^4, k=2: M(4, 2) = 63 = TM(5, 4), N(4,2) = 323 with the Collatz sequence  [63, 190, 95, 286, 143, 430, 215, 646, 323] of length 9.
n=5, (ud)^5, k=1: M(5, 1) = 63 =  TM(5, 5), N(5,1) = 485 with the Collatz sequence  [63, 190, 95, 286, 143, 430, 215, 646, 323, 970, 485]  of length 11.

Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.

Cf. A006577, A139399, A112695, A239126, A016969, A239129.

The array: N(n, k) = 2*3^n*k - 1 for n >= 1 and k >= 1.

The triangle: TN(m, n) = N(n, m-n+1) = 2*3^n*(m-n+1) - 1 for m >= n >= 1 and 0 for m < n.

A239128 a(n) = 32*n - 1, n >= 1. Fourth column of triangle A239126, related to the Collatz problem.

Original entry on oeis.org

31, 63, 95, 127, 159, 191, 223, 255, 287, 319, 351, 383, 415, 447, 479, 511, 543, 575, 607, 639, 671, 703, 735, 767, 799, 831, 863, 895, 927, 959, 991, 1023, 1055, 1087, 1119, 1151, 1183, 1215, 1247, 1279, 1311, 1343, 1375, 1407, 1439, 1471, 1503, 1535, 1567, 1599

Offset: 1

Wolfdieter Lang, Mar 13 2014

This sequence gives all starting values a(n) (in increasing order) of Collatz sequences of length 9 following the pattern (ud)^4, with u (for `up'), mapping an odd number m to 3*m+1, and d (for `down'), mapping an even number m to m/2. The last entry of this sequence is required to be odd and it is given by 162*n-1.

This appears in Example 2.2. for x=y = 4 in the M. Trümper paper on p. 7, given as a link below.

			a(1) = 31 because the Collatz sequence following the pattern udududud is [31, 94, 47, 142, 71, 214, 107, 322, 161], with length 9, ending in the odd number N(4,1) = 161 = 162*1 - 1 from the array A239127, and 31 is the smallest positive number whose Collatz sequence follows this pattern and ends in an odd number.
a(4) = 127 with the Collatz sequence [127, 382, 191, 574, 287, 862, 431, 1294, 647] ending in N(4,4) = 647 = 32*4 - 1. 127 is the fourth smallest positive number following this pattern with odd end number.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A125169 (third column), A239126, A239127.

CoefficientList[Series[(31 + x)/(1 - x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 16 2014 *)
32*Range[50]-1 (* Harvey P. Dale, Jan 25 2021 *)

O.g.f.: x*(31+x)/(1-x)^2.
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: exp(x)*(32*x - 1) + 1.
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)

A004767 a(n) = 4*n + 3.

Original entry on oeis.org

3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 123, 127, 131, 135, 139, 143, 147, 151, 155, 159, 163, 167, 171, 175, 179, 183, 187, 191, 195, 199, 203, 207, 211, 215, 219, 223

Offset: 0

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(12).
Binary expansion ends 11.
These the numbers for which zeta(2*x+1) needs just 2 terms to be evaluated. - Jorge Coveiro, Dec 16 2004 [This comment needs clarification]
a(n) is the smallest k such that for every r from 0 to 2n - 1 there exist j and i, k >= j > i > 2n - 1, such that j - i == r (mod (2n - 1)), with (k, (2n - 1)) = (j,(2n - 1)) = (i, (2n - 1)) = 1. - Amarnath Murthy, Sep 24 2003
Complement of A004773. - Reinhard Zumkeller, Aug 29 2005
Any (4n+3)-dimensional manifold endowed with a mixed 3-Sasakian structure is an Einstein space with Einstein constant lambda = 4n + 2 [Theorem 3, p. 10 of Ianus et al.]. - Jonathan Vos Post, Nov 24 2008
Solutions to the equation x^(2*x) = 3*x (mod 4*x). - Farideh Firoozbakht, May 02 2010
Subsequence of A022544. - Vincenzo Librandi, Nov 20 2010
First differences of A084849. - Reinhard Zumkeller, Apr 02 2011
Numbers n such that {1, 2, 3, ..., n} is a losing position in the game of Nim. - Franklin T. Adams-Watters, Jul 16 2011
Numbers n such that there are no primes p that satisfy the relationship p XOR n = p + n. - Brad Clardy, Jul 22 2012
The XOR of all numbers from 1 to a(n) is 0. - David W. Wilson, Apr 21 2013
A089911(4*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
First differences of A014105. - Ivan N. Ianakiev, Sep 21 2013
All triangular numbers in the sequence are congruent to {3, 7} mod 8. - Ivan N. Ianakiev, Nov 12 2013
Apart from the initial term, length of minimal path on an n-dimensional cubic lattice (n > 1) of side length 2, until a self-avoiding walk gets stuck. Construct a path connecting all 2n points orthogonally adjacent from the center, ending at the center. Starting at any point adjacent to the center, there are 2 steps to reach each of the remaining 2n - 1 points, resulting in path length 4n - 2 with a final step connecting the center, for a total path length of 4n - 1, comprising 4n points. - Matthew Lehman, Dec 10 2013
a(n-1), n >= 1, appears as first column in the triangles A238476 and A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014
For the Collatz Conjecture, we identify two types of odd numbers. This sequence contains all the ascenders: where (3*a(n) + 1) / 2 is odd and greater than a(n). See A016813 for the descenders. - Jaroslav Krizek, Jul 29 2016

			G.f. = 3 + 7*x + 11*x^2 + 15*x^3 + 19*x^4 + 23*x^5 + 27*x^6 + 31*x^7 + ...

Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 85.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999. See Theorem 8.1 on page 240.

Ivan Panchenko, Table of n, a(n) for n = 0..200
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Stere Ianus, Mihai Visinescu and Gabriel-Eduard Vilcu, Hidden symmetries and Killing tensors on curved spaces, arXiv:0811.3478 [math-ph], 2008. - _Jonathan Vos Post_, Nov 24 2008
Tanya Khovanova, Recursive Sequences
Juan F. Pulido, José L. Ramírez, and Andrés R. Vindas-Meléndez, Generating Trees and Fibonacci Polyominoes, arXiv:2411.17812 [math.CO], 2024. See p. 8.
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N))
William A. Stein, The modular forms database
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A004773, A005408, A008545 (partial products), A008586, A014105, A016813, A016825, A017137, A017629, A022544, A084849, A181049, A238476, A239126.
Cf. A017101 and A004771 (bisection: 3 and 7 mod 8).
Cf. A016838 (square).

a004767 = (+ 3) . (* 4)
a004767_list = [3, 7 ..]  -- Reinhard Zumkeller, Oct 03 2012

[4*n+3: n in [0..50]]; // Wesley Ivan Hurt, Jul 09 2014

Maple
```
seq( 3+4*n, n=0..100 );
```

4 Range[50] - 1 (* Wesley Ivan Hurt, Jul 09 2014 *)

a(n)=4*n+3 \\ Charles R Greathouse IV, Jul 28 2015

Vec((3+x)/(1-x)^2 + O(x^200)) \\ Altug Alkan, Jan 15 2016

for n in range(0,50): print(4*n+3, end=', ') # Stefano Spezia, Dec 12 2018

[4*n+3 for n in range(50)] # G. C. Greubel, Dec 09 2018

(0 to 59).map(4 *  + 3) // _Alonso del Arte, Dec 12 2018

G.f.: (3+x)/(1-x)^2. - Paul Barry, Feb 27 2003
a(n) = 2*a(n-1) - a(n-2) for n > 1, a(0) = 3, a(1) = 7. - Philippe Deléham, Nov 03 2008
a(n) = A017137(n)/2. - Reinhard Zumkeller, Jul 13 2010
a(n) = 8*n - a(n-1) + 2 for n > 0, a(0) = 3. - Vincenzo Librandi, Nov 20 2010
a(n) = A005408(A005408(n)). - Reinhard Zumkeller, Jun 27 2011
a(n) = 3 + A008586(n). - Omar E. Pol, Jul 27 2012
a(n) = A014105(n+1) - A014105(n). - Michel Marcus, Sep 21 2013
a(n) = A016813(n) + 2. - Jean-Bernard François, Sep 27 2013
a(n) = 4*n - 1, with offset 1. - Wesley Ivan Hurt, Mar 12 2014
From Ilya Gutkovskiy, Jul 29 2016: (Start)
E.g.f.: (3 + 4*x)*exp(x).
Sum_{n >= 0} (-1)^n/a(n) = (Pi + 2*log(sqrt(2) - 1))/(4*sqrt(2)) = A181049. (End)

A004771 a(n) = 8*n + 7. Or, numbers whose binary expansion ends in 111.

Original entry on oeis.org

7, 15, 23, 31, 39, 47, 55, 63, 71, 79, 87, 95, 103, 111, 119, 127, 135, 143, 151, 159, 167, 175, 183, 191, 199, 207, 215, 223, 231, 239, 247, 255, 263, 271, 279, 287, 295, 303, 311, 319, 327, 335, 343, 351, 359, 367, 375, 383, 391, 399, 407, 415, 423, 431

Offset: 0

N. J. A. Sloane

These numbers cannot be expressed as the sum of 3 squares. - Artur Jasinski, Nov 22 2006
These numbers cannot be perfect squares. - Cino Hilliard, Sep 03 2006
a(n-2), n >= 2, appears in the second column of triangle A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014
The initial terms 7, 15, 23, 31 are the generating set for the rest of the sequence in the sense that, by Lagrange's Four Square Theorem, any number n of the form 8*k+7 can always be written as a sum of no fewer than four squares, and if n = a^2 + b^2 + c^2 + d^2, then (a mod 4)^2 + (b mod 4)^2 + (c mod 4)^2 + (d mod 4)^2 must be one of 7, 15, 23, 31. - Walter Kehowski, Jul 07 2014
Define a set of consecutive positive odd numbers {1, 3, 5, ..., 12*n + 9} and skip the number 6*n + 5. Then the contraharmonic mean of that set gives this sequence. For example, ContraharmonicMean[{1, 3, 7, 9}] = 7. - Hilko Koning, Aug 27 2018
Jacobi symbol (2, a(n)) = Kronecker symbol (a(n), 2) = 1. - Jianing Song, Aug 28 2018

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 246.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Tanya Khovanova, Recursive Sequences
Leo Tavares, Illustration: Twin Square Frames
Leo Tavares, Illustration: Mid-line Hexagons
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 962
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A004767, A008590, A017077, A017101, A017137, A033954.
Cf. A007522 (primes), subsequence of A047522.
Cf. A056753, A227144, A227146, A239126.

List([0..60],n->8*n+7); # Muniru A Asiru, Aug 28 2018

a004771 = (+ 7) . (* 8)
a004771_list = [7, 15 ..]  -- Reinhard Zumkeller, Jan 29 2013

[8*n+7: n in [0..60]]; // Vincenzo Librandi, May 28 2011

A004771:=n->8*n+7; seq(A004771(n), n=0..100); # Wesley Ivan Hurt, Dec 22 2013

8 Range[0, 60] + 7 (* or *) Range[7, 500, 8] (* or *) Table[8 n + 7, {n, 0, 60}] (* Bruno Berselli, Dec 28 2016 *)

a(n)=8*n+7 \\ Charles R Greathouse IV, Sep 23 2012

O.g.f: (7 + x)/(1 - x)^2 = 8/(1 - x)^2 - 1/(1 - x). - R. J. Mathar, Nov 30 2007
a(n) = 2*a(n-1) - a(n-2) for n >= 2.  - Vincenzo Librandi, May 28 2011
A056753(a(n)) = 7. - Reinhard Zumkeller, Aug 23 2009
a(n) = t(t(t(n))), where t(i) = 2*i + 1.
a(n) = A004767(2*n+1), for n >= 0. See also A004767(2*n) = A017101(n). - Wolfdieter Lang, Feb 03 2022
From Elmo R. Oliveira, Apr 11 2024: (Start)
E.g.f.: exp(x)*(7 + 8*x).
a(n) = A033954(n+1) - A033954(n). (End)

A125169 a(n) = 16*n + 15.

Original entry on oeis.org

15, 31, 47, 63, 79, 95, 111, 127, 143, 159, 175, 191, 207, 223, 239, 255, 271, 287, 303, 319, 335, 351, 367, 383, 399, 415, 431, 447, 463, 479, 495, 511, 527, 543, 559, 575, 591, 607, 623, 639, 655, 671, 687, 703, 719, 735, 751, 767, 783, 799, 815, 831, 847

Offset: 0

Artur Jasinski, Nov 22 2006

The identity (16*n + 15)^2 - (16*(n+1)^2 - 2*(n+1))*4^2 = 1 can be written as a(n)^2 - A158058(n+1)*4^2 = 1. - Vincenzo Librandi, Feb 01 2012

a(n-3), n >= 3, appears in the third column of triangle A239126 related to the Collatz problem. - Wolfdieter Lang, Mar 14 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Tanya Khovanova, Recursive Sequences.
Edward J. Barbeau, Polynomial Excursions, Chapter 10: Diophantine equations (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(4^2*t-2)).
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A004771, A158058, A239126.

I:=[15, 31]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..60]]; // Vincenzo Librandi, Jan 04 2012

Table[16n + 15, {n, 0, 100}]
LinearRecurrence[{2,-1},{15,31},100] (* or *) Range[15,1620,16] (* Harvey P. Dale, Jan 03 2012 *)

a(n) = 16*n + 15 \\ Vincenzo Librandi, Jan 04 2012

a(n) = 2*a(n-1) - a(n-2); a(0)=15, a(1)=31. - Harvey P. Dale, Jan 03 2012
O.g.f.: (15 + x)/(1 - x)^2. - Wolfdieter Lang, Mar 14 2014
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: exp(x)*(15 + 16*x).
a(n) = A004771(2*n+1). (End)

A239129 a(n) = 18*n - 1, n >= 1, the second column of triangle A239127 related to the Collatz problem.

Original entry on oeis.org

17, 35, 53, 71, 89, 107, 125, 143, 161, 179, 197, 215, 233, 251, 269, 287, 305, 323, 341, 359, 377, 395, 413, 431, 449, 467, 485, 503, 521, 539, 557, 575, 593, 611, 629, 647, 665, 683, 701, 719, 737, 755, 773, 791, 809, 827, 845, 863, 881, 899, 917, 935, 953, 971

Offset: 1

Wolfdieter Lang, Mar 13 2014

This sequence gives all ending values a(n) (in increasing order) of Collatz sequences of length 5 following the pattern (ud)^2, with u (for `up'), mapping an odd number m to 3*m+1, and d (for `down'), mapping an even number m to m/2. The last entry of this sequence is required to be odd. The first entry is also odd and is given by M(2,n) = 8*n-1 from the array A239126.

This appears as N in Example 2.2. for x=y = 2 in the M. Trümper paper on p. 7, given as a link below.

			a(1) = 17 because the Collatz sequence for M(2,1) = 8*1 - 1 = 7 from A239126 is [7, 22, 11, 34, 17] ending in the odd number 17.
a(4) = 71 with the Collatz sequence of length 5 starting with M(2,4) = 31 given by [31, 94, 47, 142, 71], ending in a(4).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Wolfdieter Lang, On Collatz' Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A016969 (first column), A239126, A239127.

CoefficientList[Series[(x + 17)/(1 - x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 16 2014 *)

a(n) = 18*n - 1 for n >= 1.
O.g.f.: x*(x+17)/(1-x)^2.
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: exp(x)*(18*x - 1) + 1.
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)

A240222 Rectangular array giving all start values M(n, k), k >= 1, for Collatz sequences following the pattern (udd)^(n-1) ud, n >= 1, read by antidiagonals.

Original entry on oeis.org

1, 3, 1, 5, 9, 1, 7, 17, 33, 1, 9, 25, 65, 129, 1, 11, 33, 97, 257, 513, 1, 13, 41, 129, 385, 1025, 2049, 1, 15, 49, 161, 513, 1537, 4097, 8193, 1, 17, 57, 193, 641, 2049, 6145, 16385, 32769, 1, 19, 65, 225, 769, 2561, 8193, 24577, 65537, 131073, 1, 21, 73, 257, 897, 3073, 10241, 32769, 98305

Offset: 1

Wolfdieter Lang, Apr 02 2014

The companion array and triangle for the end numbers N(n, k) is given in A240223.

The two operations on natural numbers m used in the Collatz 3x+1 conjecture are here (following the M. Trümper paper given in the link) denoted by u for 'up' and d for 'down': u m = 3*m+1, if m is odd, and d m = m/2 if m is even. The present array gives all start numbers M(n, k) for Collatz sequences realizing the Collatz word (udd)^n ud = (sd)^n s (s = ud is useful because, except for the one letter word u, at least one d follows a letter u), with n >= 1, and k >= 1. The length of these Collatz sequences 3*n. For these Collatz sequences M(n, 0) = M(1, 0) = 1 and N(n, 0) = N(1, 0) = 2.

			The rectangular array M(n, k) begins:
n\k 0       1       2       3       4       5 ...
1:  1       3       5       7       9      11
2:  1       9      17      25      33      41
3:  1      33      65      97     129     161
4:  1     129     257     385     513     641
5:  1     513    1025    1537    2049    2561
6:  1    2049    4097    6145    8193   10241
7:  1    8193   16385   24577   32769   40961
8:  1   32769   65537   98305  131073  163841
9:  1  131073  262145  393217  524289  655361
10: 1  524289 1048577 1572865 2097153 2621441
...
For more columns see the link.
The triangle TM(m, n) begins (zeros are not shown):
k\n  1  2   3   4    5     6      7 ...
0:   1
1:   3  1
2:   5  9   1
3:   7 17  33   1
4:   9 25  65 129    1
5:  11 33  97 257  513     1
6:  13 41 129 385 1025  2049      1
...
For more rows see the link.
n=1, ud, k=0: M(1, 0) = 1 = TM(0, 1), N(1, 0) = 2 with the Collatz sequence [1, 4, 2] of
length 3.
n=1, ud, k=2: M(1, 2) = 5 = TM(2, 1), N(1,  2) = 8 with the Collatz sequence [5, 16, 8] of length 3.
n=2, uddud, k=0: M(2, 0) = 1 = TM(1, 2), Ne(2, 0) = 2 with the Collatz sequence [1, 4, 2, 1, 4, 2, 1, 4, 2] of length 9.

Wolfdieter Lang, Rectangular array and triangle.
Wolfdieter Lang, On Collatz' Words, Sequences and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.
Eric Weisstein's World of Mathematics, Collatz Problem.
Wikipedia, Collatz Conjecture

Cf. A238475, A238476, A239126, A239127.

The array: M(n, k) = 1 + 2^(2*n-1)*k for n >= 1 and k >= 0.

The triangle: TM(m, n) = M(n,m-n+1) = 1 + 2^(2*n-1)*(m-n+1) for m+1 >= n >= 1 and 0 for m+1 < n.

A240223 Rectangular companion array to M(n,k), given in A240222, showing the end numbers N(n, k), k >= 1, for the Collatz operation (udd)^(n-1) ud, n >= 1, read by antidiagonals.

Original entry on oeis.org

2, 5, 2, 8, 11, 2, 11, 20, 29, 2, 14, 29, 56, 83, 2, 17, 38, 83, 164, 245, 2, 20, 47, 110, 245, 488, 731, 2, 23, 56, 137, 326, 731, 1460, 2189, 2, 26, 65, 164, 407, 974, 2189, 4376, 6563, 2, 29, 74, 191, 488, 1217, 2918, 6563, 13124, 19685, 2, 32, 83, 218, 569, 1460, 3647, 8750, 19685, 39368, 59051, 2

Offset: 0

Wolfdieter Lang, Apr 04 2014

The companion array and triangle for the start numbers M(n, k) is given in A240222.

For the Collatz operations u (for 'up') and d (for 'down') see the comment on A240222, also for links, especially for the M. Trümper paper.

			The rectangular array N(n, k) begins
  n\k 0      1       2       3       4       5 ...
  1:  2      5       8      11      14      17
  2:  2     11      20      29      38      47
  3:  2     29      56      83     110     137
  4:  2     83     164     245     326     407
  5:  2    245     488     731     974    1217
  6:  2    731    1460    2189    2918    3647
  7:  2   2189    4376    6563    8750   10937
  8:  2   6563   13124   19685   26246   32807
  9:  2  19685   39368   59051   78734   98417
  10: 2  59051  118100  177149  236198  295247
  ...
For more columns see the link.
The triangle TN(m, n) begins (zeros are not shown):
  m\n  1  2   3   4    5    6    7 ...
  0:   2
  1:   5  2
  2:   8 11   2
  3:  11 20  29   2
  4:  14 29  56  83    2
  5:  17 38  83 164  245    2
  6:  20 47 110 245  488  731    2
  ...
For more rows see the link.
n=1, ud, k=0: M(1, 0) = 1, N(1, 0) = TN(0, 1) = 2 with the Collatz sequence [1, 4, 2] of length 3.
n=1, ud, k=2: M(1, 2) = 5, N(1, 2) = TN(2, 1) = 8 with the Collatz sequence [5, 16, 8] of length 3.
n=2, uddud, k=0: M(2, 0) = 1, Ne(2, 0) = TN(1, 2) = 2 with the Collatz sequence [1, 4, 2, 1, 4, 2, 1, 4, 2] of length 9.

Wolfdieter Lang, Rectangular array and triangle.
Wolfdieter Lang, On Collatz Words, Sequences, and Trees, J. of Integer Sequences, Vol. 17 (2014), Article 14.11.7.
Manfred Trümper, The Collatz Problem in the Light of an Infinite Free Semigroup, Chinese Journal of Mathematics, Vol. 2014, Article ID 756917, 21 pages.

Cf. A238475, A238476, A239126, A239127, A240222, A016789 (first row of N), A017185 (second row of N).

The array: N(n, k) = 2 + 3^n*k for n >= 1 and k >= 0.

The triangle: TN(m, n) = N(n,m-n+1) = 2 + 3^n*(m-n+1) for m+1 >= n >= 1 and 0 for m+1 < n.

A243115 Starting values of the reduced Collatz function (A014682) where 2 to the power of the "dropping time" is greater than the starting value.

Original entry on oeis.org

3, 7, 11, 15, 23, 27, 31, 39, 47, 59, 63, 71, 79, 91, 95, 103, 111, 123, 127, 155, 159, 167, 175, 191, 199, 207, 219, 223, 231, 239, 251, 255, 283, 287, 303, 319, 327, 347, 359, 367, 383, 411, 415, 423, 447, 463, 479, 487, 495, 507, 511, 539, 543, 559, 575

Offset: 1

K. Spage, Aug 20 2014

nonn

a(n) is the lowest positive starting value of the reduced Collatz function such that all starting values (>1) that are congruent to a(n) (mod 2^d) have the same dropping time (d). The dropping time here counts the (3x+1)/2 and the x/2 steps as listed in A126241. A number is included in this sequence if 2^A126241(a(n)) > a(n).
Starting values that produce new record dropping times as listed in A060412 are necessarily a subset of this sequence.
If at least one iteration is carried out before checking that the absolute iterated value has become less than or equal to the absolute starting value, then a(n) is the lowest positive starting value such that all starting values (positive, zero or negative) that are congruent to a(n) (mod 2^d) have the same dropping time (d). Defined like this, the sequence would start with 0, 1, 3, 7.
For k>0, A076227(k) is the number of terms between 2^k and 2^(k+1)-1. - Ruud H.G. van Tol, Dec 18 2022
All terms are congruent to 3 (mod 4) since any 1 (mod 4) has dropping time A126241(4k+1) = 2, for k>=1. - Ruud H.G. van Tol, Jan 11 2023

			3 is in this sequence because the dropping time starting with 3 is A126241(3) = 4 and 2^4 > 3.

Cf. A014682, A060412, A076227, A126241, A177789, A239126.

is(t)= if(t<3||3!=t%4,0,my(x=t, d=0); until(x<=t, if(x%2, x=(x*3+1)/2, x/=2); d++); 2^d>t); \\ updated by Ruud H.G. van Tol, Jan 10 2023

Offset 1 from Ruud H.G. van Tol, Jan 10 2023

cp's OEIS Frontend

A239127 Rectangular companion array to M(n,k), given in A239126, showing the end numbers N(n, k), k >= 1, for the Collatz operation (ud)^n, n >= 1, ending in an odd number, read by antidiagonals.

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A239128 a(n) = 32*n - 1, n >= 1. Fourth column of triangle A239126, related to the Collatz problem.

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A004767 a(n) = 4*n + 3.

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A004771 a(n) = 8*n + 7. Or, numbers whose binary expansion ends in 111.

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A125169 a(n) = 16*n + 15.

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A239129 a(n) = 18*n - 1, n >= 1, the second column of triangle A239127 related to the Collatz problem.

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