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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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%I A332434 #26 Dec 20 2024 09:26:35
%S A332434 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 A332434 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 A332434 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 A332434 Irregular triangle read by rows: r-tuples (lengths) of the complete coach system Sigma(2*n+1), for n >= 1.
%C A332434 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 A332434 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 A332434 The parity of the entries of each row is identical ([HP], p. 261).
%C A332434 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 A332434 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 A332434 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 A332434 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 A332434 The irregular triangle T(n, j) begins:
%e A332434   n,   b \ j   1  2  3 ... | A135303(n)   A332435(n)
%e A332434   1,   3:      1                1            1
%e A332434   2,   5:      1                1            1
%e A332434   3,   7:      2                1            2
%e A332434   4,   9:      1                1            1
%e A332434   5,  11:      3                1            3
%e A332434   6,  13:      3                1            3
%e A332434   7,  15:      2                1            2
%e A332434   8,  17:      1  3             2            4
%e A332434   9,  19:      5                1            5
%e A332434  10,  21:      2                1            2
%e A332434  11,  23:      6                1            6
%e A332434  12,  25:      5                1            5
%e A332434  13,  27:      5                1            5
%e A332434  14,  29:      7                1            7
%e A332434  15,  31:      2  2  4          3            8
%e A332434  16,  33:      1  3             2            4
%e A332434  17,  35:      6                1            6
%e A332434  18,  37:      9                1            9
%e A332434  19,  39:      6                1            6
%e A332434  20,  41:      3  7             2           10
%e A332434  ...
%e A332434 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 A332434 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 A332434 n = 8, b = 17: Sigma(17) = [[1; 4], [3, 7, 5; 1, 1, 2]], hence T(8, 1) = 1, T(8, 2) = 3.
%e A332434 n = 16, b = 33:  Sigma(33) = [[1; 5], [5, 7, 13; 2, 1, 2]], hence T(16, 1) = 1, T(16, 2) = 3.
%Y A332434 Cf. A003558, A135303, A332435.
%K A332434 nonn,tabf,easy
%O A332434 1,3
%A A332434 _Wolfdieter Lang_, Feb 26 2020