A339046 Irregular triangle read by rows: row n gives the complete quadrupling system modulo N = 2*n + 1, for n >= 0.

1, 1, 2, 1, 4, 2, 3, 1, 4, 2, 3, 5, 6, 1, 4, 7, 2, 8, 5, 1, 4, 5, 9, 3, 2, 8, 10, 7, 6, 1, 4, 3, 12, 9, 10, 2, 8, 6, 11, 5, 7, 1, 4, 2, 8, 7, 13, 11, 14, 1, 4, 16, 13, 2, 8, 15, 9, 3, 12, 14, 5, 6, 7, 11, 10, 1, 4, 16, 2, 8, 11, 5, 20, 17, 10, 19, 13, 1, 4, 16, 18, 3, 12, 2, 8, 9, 13, 6, 2, 8, 9, 13, 6, 1, 4, 16, 18, 3, 12

Offset

0,3

Comments

The length of row n is given by phi(2*n + 1), with phi = A000010, for n >= 0.

The quadrupling sequence modulo N = 2*n + 1, for n >= 0, has entries QS(N, s(N,i), j) = s(N,i)*4^j (mod N), with j >= 0, and certain positive integer seeds s(N, i), for i = 1, 2, ..., S(N) = A339049((N-1)/2), where gcd(s(N, i), N) = 1 (restricted seeds modulo N). These quadrupling sequences are periodic with period length P(N) = A053447((N-1)/2) (order of 4 modulo N). Only the periods (cycles) QS(N, s(N,i)) = {QS(N, s(N, i), j)}_{j=0..P(N)-1}, for i = 1, 2, ..., S(N), are listed.

N = 1 (n = 0) is special: one takes here the restricted residue system modulo N not as [0] but as [1]. The order of 4 modulo 1 is 1, because 4^1 == 1 (mod 1) (== 0 (mod 1)).

In order to obtain the complete system of quadrupling sequences one starts with seed s(N, 1) = 1, and if all numbers from the smallest positive reduced residue system modulo N (called RRS(N), given in row N of A038566) are obtained, i.e., if P(N) = #RRS(N) = phi(N) = A000010(N), then the system is complete. Otherwise the smallest missing number from RRS(N) is taken as new seed s(N, 2), etc. until the system is complete. This means that the number of seeds needed is S(N) given above.

This entry generalizes A337712, given together with Gary W. Adamson. See also A337936.

Links

Table of n, a(n) for n=0..98.

Formula

T(n, k) gives the k-th entry in the complete quadrupling system modulo N = 2*n + 1, for n >= 0, with the S(N) = A339049((N-1)/2) cycles of length A053447((N-1)/2) written in row n. See the comment above for QS(N,s(N,i)), i = 1, 2, ..., S(N).

Example

The irregular triangle begins (the vertical bar separates the cycles):
n,  N \ k  1 2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 ...
0,  1:     1
1,  3:     1|2
2,  5:     1 4| 2  3
3,  7:     1 4  2| 3  5  6
4,  9:     1 4  7| 2  8  5
5,  11:    1 4  5  9  3| 2  8 10  7  6
6,  13:    1 4  3 12  9 10| 2  8  6 11  5  7
7,  15:    1 4| 2  8| 7 13|11 14
8,  17:    1 4 16 13| 2  8 15  9| 3 12 14  5| 6  7 11 10
9,  19:    1 4 16  7  9 17 11  6  5| 2  8 13 14 18 15  3 12 10
10, 21:    1 4 16| 2  8 11| 5 20 17|10 19 13
11, 23:    1 4 16 18  3 12  2  8  9 13  6| 2  8  9 13  6  1  4 16 18  3 12
12, 25:    1 4 16 14  6 24 21  9 11 19| 2  8  7  3 12 23 17 18 22 13
13, 27:    1 4 16 10 13 25 19 22  7| 2  8  5 20 26 23 11 17 14
...
n = 14, N = 29: 1 4 16 6 24 9 7 28 25 13 23 5 20 22 | 2 8 3 12 19 18 14 27 21 26 17 10 11 15,
n = 15, N = 31: 1 4 16 2 8 | 3 12 17 6 24 | 5 20 18 10 9 | 7 28 19 14 25 | 11 13 21 22 26 | 15 29 23 30 27.
...

Crossrefs

Cf. A000010, A053447, A337712 (doubling), A337936 (tripling), A339049.

Sequence in context: A153281 A338654 A130584 * A265911 A363893 A078458

Adjacent sequences: A339043 A339044 A339045 * A339047 A339048 A339049

Keyword

nonn,tabf,easy

Author

Wolfdieter Lang, Dec 13 2020

Status

approved