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%I #20 Dec 18 2023 01:27:31
%S 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,
%T 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,
%U 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
%N Irregular triangle read by rows: r-tuples (lengths) of the complete coach system Sigma(2*n+1), for n >= 1.
%C The length of row n of this irregular triangle is A135303(n). The row sums are given in A332435, where more details are found.
%C 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.
%C The parity of the entries of each row is identical ([HP], p. 261).
%C 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), ...
%D Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, (3rd printing 2012) pp. 261-281.
%H Wolfdieter Lang, <a href="https://arxiv.org/abs/2008.04300">On the Equivalence of Three Complete Cyclic Systems of Integers</a>, arXiv:2008.04300 [math.NT], 2020.
%F 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).
%e The irregular triangle T(n, j) begins:
%e n, b \ j 1 2 3 ... | A135303(n) A332435(n)
%e 1, 3: 1 1 1
%e 2, 5: 1 1 1
%e 3, 7: 2 1 2
%e 4, 9: 1 1 1
%e 5, 11: 3 1 3
%e 6, 13: 3 1 3
%e 7, 15: 2 1 2
%e 8, 17: 1 3 2 4
%e 9, 19: 5 1 5
%e 10, 21: 2 1 2
%e 11, 23: 6 1 6
%e 12, 25: 5 1 5
%e 13, 27: 5 1 5
%e 14, 29: 7 1 7
%e 15, 31: 2 2 4 3 8
%e 16, 33: 1 3 2 4
%e 17, 35: 6 1 6
%e 18, 37: 9 1 9
%e 19, 39: 6 1 6
%e 20, 41: 3 7 2 10
%e ...
%e 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.
%e n = 5, b = 11: Sigma(11) = [[1, 5, 3; 1, 1, 3]], hence T(5, 1) = 3 or R(11) = (r(11,1)) = (3).
%e n = 8, b = 17: Sigma(17) = [[1; 4], [3, 7, 5; 1, 1, 2]], hence T(8, 1) = 1, T(8, 2) = 3.
%e n = 16, b = 33: Sigma(33) = [[1; 5], [5, 7, 13; 2, 1, 2]], hence T(16, 1) = 1, T(16, 2) = 3.
%Y Cf. A003558, A135303, A332435.
%K nonn,tabf,easy
%O 1,3
%A _Wolfdieter Lang_, Feb 26 2020
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