cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

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Author

Wolfdieter Lang, Feb 26 2020

Keywords

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.

Crossrefs

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).

A332436 The number of even numbers <= n of the smallest nonnegative reduced residue system modulo 2*n + 1, for n >= 0.

Original entry on oeis.org

1, 0, 1, 1, 2, 2, 3, 2, 4, 4, 4, 5, 5, 4, 7, 7, 6, 6, 9, 6, 10, 10, 6, 11, 11, 8, 13, 10, 10, 14, 15, 8, 12, 16, 12, 17, 18, 10, 16, 19, 14, 20, 16, 14, 22, 18, 16, 18, 24, 14, 25, 25, 12, 26, 27, 18, 28, 22, 18, 24, 28, 20, 25, 31, 22, 32, 28, 18, 34, 34, 24
Offset: 0

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Author

Wolfdieter Lang, Feb 29 2020

Keywords

Comments

For the smallest positive reduced residue system modulo N see the array A038566. Here the nonnegative residue system [0, 1, ..., N-1] is considered, differing only for N = 1 from A038566, with [0] (instead of [1]).
This sequence gives the complement of A332435 (with 0 for n = 0 included) relative to the number of positive numbers <= n of the smallest nonnegative reduced residue system modulo (2*n+1). Thus a(n) + A332435(n) = phi(n)/2, for n >= 1, with phi = A000010. For n = 0 one has 1 + 0 = 1.
a(n) gives also the number of even numbers appearing in the complete modified doubling sequence system (name it MDS(b)), for b = 2*n + 1, with n >= 1, proposed in a comment from Gary W. Adamson, Aug 24 2019, in the example section of A135303 for prime b.

Examples

			n = 4, b = 9: the even numbers <= 4 in RRS(9) := [1, 2, 4, 5, 7, 8] are {2, 4}, hence a(4) = 2.
The complete MDS(9) system has one cycles of length 3: Cy*(9, 1) = (2, 4, 1), with the even numbers {2, 4}.
n = 8, b = 17: the even numbers <= 8 in RRS(17) := [1, 2, ..., 16] are {2, 4, 6 ,8}, hence a(8) = 4.
The complete MDS(17) system has two cycles of length 4: Cy*(17, 1) = (2, 4, 8, 1) and Cy*(17, 2) = (6, 5, 7, 3) and the even numbers are {2, 4, 6 ,8}.
		

Crossrefs

Formula

a(n) = A000010(n)/2 - A332435(n), for n >= 1, and a(0) = 1.

A334430 Irregular triangle read by rows. Row n gives the cycles of positive integers of the complete modified doubling sequence MDS(b) for b = 2*n + 1, for n >= 1.

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 4, 1, 2, 4, 3, 5, 1, 2, 4, 5, 3, 6, 1, 2, 4, 7, 1, 2, 4, 8, 1, 6, 5, 7, 3, 2, 4, 8, 3, 6, 7, 5, 9, 1, 2, 4, 8, 5, 10, 1, 2, 4, 8, 7, 9, 5, 10, 3, 6, 11, 1, 2, 4, 8, 9, 7, 11, 3, 6, 12, 1, 2, 4, 8, 11, 5, 10, 7, 13, 1, 2, 4, 8, 13, 3, 6, 12, 5, 10, 9, 11, 7, 14, 1
Offset: 1

Views

Author

Gary W. Adamson and Wolfdieter Lang, Aug 11 2020

Keywords

Comments

The length of row n is phi(2*n+1)/2 with phi = A000010, for n > = 1.
There are c(2*n+1) = A135303(n) cycles in row n, and the length of each cycle is k(2*n+1) = A003558(n). c(2*n+1)*k(2*n+1) = phi(n)/2 (in Hilton and Pedersen [HP], the coach theorem, p. 262).
In the construction of the cycles a modified modular equivalence relation, called mod* b, for b = 2*n + 1, with n >= 1, proposed in the Brändli and Beyne [BB] paper, is used. It is defined as mod*(a, b) = mod(a, b) (with values in I(b) = {0, 1, ..., b-1}) for integer numbers a, if mod(a, b) <= (b-1)/2 and mod*(a, b) = mod(-a, b) if mod(a, b) > (b-1)/2. The equivalence relation k ~ m if mod*(k, b) = mod*(m,b) is multiplicative (not additive).
The MDS(b) cycles are obtained from the doubling sequence {a(b,i)*2^j}A003558(n)%20for%20each%20odd%20a(b,%20i)%20with%20gcd(a(b,%20i),%20b)%20=%201%20and%20a(b,%20i)%20%3C=%20(b-1)/2.%20The%20first%20cycle%20is%20Cy*(b,1)%20obtained%20for%20a(b,%201)%20=%201.%20If%20not%20all%20odd%20numbers%20relatively%20prime%20to%20b%20and%20%3C=%20(b-1)/2%20are%20present,%20the%20next%20cycle%20Cy*(b,2)%20uses%20the%20smallest%20missing%20reduced%20odd%20number,%20etc.%20The%20number%20of%20such%20inputs%20a(b,%20i),%20hence%20cycles,%20is%20c*(b),%20coinciding%20with%20c(2*n+1)%20=%20A135303(n),%20and%20all%20odd%20and%20even%20positive%20numbers%20with%20gcd(m,%20b)%20=%201%20and%20m%20%3C=%20(b-1)/2%20appear%20just%20once%20in%20the%20complete%20cycle%20system%20for%20b,%20called%20Cy*(b)%20=%20%7BCy*(b,i)%7D">{j=1..P(b)} evaluated mod*b: Cy*(b,i)_j = mod*(a(b,i)*2^j, b), with certain inputs (seeds) a(b, i). P(b), the length of the period independent, coincides with k(2*n+1) = A003558(n) for each odd a(b, i) with gcd(a(b, i), b) = 1 and a(b, i) <= (b-1)/2. The first cycle is Cy*(b,1) obtained for a(b, 1) = 1. If not all odd numbers relatively prime to b and <= (b-1)/2 are present, the next cycle Cy*(b,2) uses the smallest missing reduced odd number, etc. The number of such inputs a(b, i), hence cycles, is c*(b), coinciding with c(2*n+1) = A135303(n), and all odd and even positive numbers with gcd(m, b) = 1 and m <= (b-1)/2 appear just once in the complete cycle system for b, called Cy*(b) = {Cy*(b,i)}{i=1..c(b)}.
Such modified doubling sequences have been considered in the Kappraff-Adamson paper using iterations of x^2 - 2, and also in comments and examples by Adamson like the Aug 25 2019 comment in A065941, where it is named "r-t table" (for roots trajectory).
The recurrence relation for the cycle Cy*(b,i) with elements {c_j(b,i)} is
The complete cycle system Cy*(b) is equivalent to the complete coach system of Hilton and Pedersen [HP], with the coach number c(b) = c*(b) and k(b) = P(b). The odd numbers of each cycle Cy*(b,i), read backwards (anticyclically), give the first row of the coach; and the number of steps to go from one odd number to the next odd number, ending with the starting number, gives the second row of the coach with sum k(b). See [HP] pp. 102-103 for the modified coach system and the extended list, eq. (7. 12), in the proof of their quasi-order theorem.
The complete cycle system of Schick (unsigned) and Brändli-Beyne SBB(b) is also equivalent to Cy*(b). Each cycle Cy*(b,i) gives the SBB(b,i) cycle (q(b, i)j) with elements obtained from q(b, i)_j = b - 2*c(b,i){P(b)-1+j}, for j = 0, 1, 2, ..., P(b)-1. Cyclicity is used: c(b,i){P(b)+k} = c(b,i){k}.
For more details and proofs see the arxiv paper by W. Lang.

Examples

			The irregular triangle a(n,m) begins (cycles are enclosed by brackets; blanks are used to fit the index m):
n,   b \ m 1 2 3  4  5   6  7  8  9 10  11 12 13 14 15 16 17 18 19 20 ...
1,   3:   (1)
2,   5:   (2 1)
3,   7:   (2 3 1)
4,   9:   (2 4 1)
5,  11:   (2 4 3  5  1)
6,  13:   (2 4 5  3  6   1)
7,  15:   (2 4 7  1)
8,  17:   (2 4 8  1)(6   5  7  3)
9,  19:   (2 4 8  3  6   7  5  9  1)
10, 21:   (2 4 8  5 10   1)
11, 23:   (2 4 8  7  9   5 10  3  6 11  1)
12, 25:   (2 4 8  9  7  11  3  6 12  1)
13, 27:   (2 4 8 11  5  10  7 13  1)
14, 29:   (2 4 8 13  3   6 12  5 10  9  11  7 14  1)
15, 31:   (2 4 8 15  1) (6 12  7 14  3)(10 11  9 13  5)
16, 33:   (2 4 8 16  1)(10 13  7 14  5)
17, 35:   (2 4 8 16  3   6 12 11 13  9  17  1)
18, 37:   (2 4 8 16  5  10 17  3  6 12  13 11 15  7 14  9 18  1)
19, 39:   (2 4 8 16  7  14 11 17  5 10  19  1)
20, 41:   (2 4 8 16  9  18  5 10 20  1) (6 12 17  7 14 13 15 11 19  3)
...
-------------------------------------------------------------------------
n = 8, b = 17: (2, 4, 8, 16) = (2, 4, 8, 16) (mod(17)) = (2, 4, 8, 1) (mod*(17), and the first odd (reduced) number missing is 3, with cycle (3*2, 3*4, 3*8, 3*16) = (6, 12, 7, 14) (mod(17)) = (6, 5, 7, 3) (mod*(17).
From this the complete coach system Sigma(17) becomes {[[1], [4]]; [[3 7 5], [1 1 2]]]}.
The complete SBB(17) cycle systen is (unsigned, offset 0) {(1 15 13 9); (3 11 5 7)}, from  17 - 2*8 = 1, 17 - 2*1 = 15, 17 - 2*2 = 13,  17 - 2*4 = 9, and from  17 - 2*7 = 3, 17 - 2*3 = 11, 17 - 2*6 = 5, 17 - 2*5 = 7.
		

References

  • Peter Hilton and Jean Pederson, A Mathematical Tapestry, 2010 (3rd printing 2012), Cambridge University Press, pp. 102-103, 260-264.
  • Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, Bokos Druck, Zürich, 2003 (ISBN 3-9522917-0-6). Tables 3.1 to 3.10, for odd p = 3 to 113 (with some gaps), pp. 158-166.

Crossrefs

Formula

See the above given recurrence relation for the cycles Cy*(b, i), i = 1, 2, ..., c(b) = A135303((b-1)/2), for b = 2*n + 1 >= 3, and the procedure to choose the inputs a(b, i).
Showing 1-3 of 3 results.