A332434 Irregular triangle read by rows: r-tuples (lengths) of the complete coach system Sigma(2*n+1), for n >= 1.

1, 1, 2, 1, 3, 3, 2, 1, 3, 5, 2, 6, 5, 5, 7, 2, 2, 4, 1, 3, 6, 9, 6, 3, 7, 3, 3, 5, 6, 12, 10, 4, 4, 13, 10, 3, 5, 15, 15, 2, 4, 4, 1, 3, 3, 5, 17, 10, 18, 2, 6, 4, 6, 10, 14, 20, 13, 21, 2, 4, 6, 4, 14, 4, 6, 4, 8, 6, 4, 8, 4, 4, 6, 18, 11, 13, 9, 7, 25, 26, 4, 8, 27, 9, 7, 11, 18, 5, 7, 9, 7, 22, 4, 6, 8

Offset: 1

Author

Wolfdieter Lang, Feb 26 2020

Keywords

• nonn
• tabf
• easy

Comments

The length of row n of this irregular triangle is A135303(n). The row sums are given in A332435, where more details are found.

For the complete coach system Sigma(b), with b = 2*n+1, for n >= 1, see the Hilton and Pedersen [HP] reference. The coach numbers are c(b) = c(2*n+1) = A135303(n), and the quasi-order of 2 modulo b is k(n) = A003558(n). The number of entries (length) of a specific coach of Sigma(b), say C(b; j), for j from {1, 2, ..., c(b)} is r(b;j), and the present array lists the r-tuples R(b) = (r(b; 1), ..., (b; c(b)). These R(b) numbers give the length of the (primitive) periods of the cycles of the first rows of each coach.

The parity of the entries of each row is identical ([HP], p. 261).

This table and a computation shows that part two of item (2) of 'Some open questions' of [HP], p. 281, namely 'Is it the case that the smallest r always occurs in the first coach (where a_1 =1)?' has the answer no. For the first counterexamples see: b = 46, 99, 109, 155, 157, 189, ..., with the r-tuples (6,4,8), (9,7), (9,7,11), (10,8,12), (13,11,15) (10,8,8), ...

Examples

			The irregular triangle T(n, j) begins:
  n,   b \ j   1  2  3 ... | A135303(n)   A332435(n)
  1,   3:      1                1            1
  2,   5:      1                1            1
  3,   7:      2                1            2
  4,   9:      1                1            1
  5,  11:      3                1            3
  6,  13:      3                1            3
  7,  15:      2                1            2
  8,  17:      1  3             2            4
  9,  19:      5                1            5
 10,  21:      2                1            2
 11,  23:      6                1            6
 12,  25:      5                1            5
 13,  27:      5                1            5
 14,  29:      7                1            7
 15,  31:      2  2  4          3            8
 16,  33:      1  3             2            4
 17,  35:      6                1            6
 18,  37:      9                1            9
 19,  39:      6                1            6
 20,  41:      3  7             2           10
 ...
In the following the complete coach is written as a list of list of coaches, and the first and second rows (the a- and k-numbers) of a coach are separated by a semicolon. Here only the first part of a coach list (the top row of a coach) is of interest.
n = 5, b = 11: Sigma(11) = [[1, 5, 3;  1, 1, 3]], hence T(5, 1) = 3 or R(11) = (r(11,1)) = (3).
n = 8, b = 17: Sigma(17) = [[1; 4], [3, 7, 5; 1, 1, 2]], hence T(8, 1) = 1, T(8, 2) = 3.
n = 16, b = 33:  Sigma(33) = [[1; 5], [5, 7, 13; 2, 1, 2]], hence T(16, 1) = 1, T(16, 2) = 3.
		

References

• Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, (3rd printing 2012) pp. 261-281.

Links

• Wolfdieter Lang, On the Equivalence of Three Complete Cyclic Systems of Integers, arXiv:2008.04300 [math.NT], 2020.

Crossrefs

Cf. A003558, A135303, A332435.

Formula

T(n, j) gives the length of the j-th coach of the complete coach system Sigma(b), with b = 2*n+1, for n >= 1, and j = 1, 2, ..., A135303(n).