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A332434 Irregular triangle read by rows: r-tuples (lengths) of the complete coach system Sigma(2*n+1), for n >= 1. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
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), ...
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.
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).
EXAMPLE
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.
CROSSREFS
Sequence in context: A302555 A355255 A336692 * A262209 A324338 A047679
KEYWORD
nonn,tabf,easy
AUTHOR
Wolfdieter Lang, Feb 26 2020
STATUS
approved

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Last modified May 13 21:17 EDT 2024. Contains 372523 sequences. (Running on oeis4.)