A337712 -id:A337712 - cp's OEIS frontend

A337936 Irregular triangle read by rows: row n gives the complete system of tripling sequences modulo N = floor((3*n-1)/2), for n >= 1.

1, 1, 1, 3, 1, 3, 4, 2, 1, 3, 2, 6, 4, 5, 1, 3, 5, 7, 1, 3, 9, 7, 1, 3, 9, 5, 4, 2, 6, 7, 10, 8, 1, 3, 9, 2, 6, 5, 4, 12, 10, 7, 8, 11, 1, 3, 9, 13, 11, 5, 1, 3, 9, 11, 5, 15, 13, 7, 1, 3, 9, 10, 13, 5, 15, 11, 16, 14, 8, 7, 4, 12, 2, 6, 1, 3, 9, 8, 5, 15, 7, 2, 6, 18, 16, 10, 11, 14, 4, 12, 17, 13, 1, 3, 9, 7, 11, 13, 19, 17, 1, 3, 9, 5, 15, 7, 21, 19, 13, 17

Offset: 1

Wolfdieter Lang, Oct 22 2020

The length of row n is A053446(n)*A337714(n) = phi(floor((3*n-1)/2)) = A337937(n), for n >= 1.
The tripling sequence modulo N(n), with N(n) = floor((3*n-1)/2) = A001651(n) (i.e., gcd(3, N(n)) = 1), for n >= 1, has entries TS(N, s(N,i), j) = s(N, i) 3^j (mod N), for j >= 0 and with certain positive odd integer seeds s(N, i), for i = 1, 2, ..., S(N(n)) = A337714(n), where gcd(s(N, i), N) = 1 (restricted seeds modulo N).
These tripling sequences are periodic with period length P(N(n)) = A053446(n) (order of 3 modulo N(n)). Only the periods (cycles) {TS(N, s(N, i), j)}_{j=0..P(N)-1}, for i = 1, 2, ..., S(N), are listed.
For n >= 2 the seeds start with s(N, 1) = 1 and if the first cycle does not cover all members of the restricted residue system modulo N = N(n) (RRS(N(n)) then the smallest missing member is chosen as second seed s(N, 2), etc., until all members of RRS(N(n)) have been reached. For N(1) = 1 one uses here RRS(1) = [1] (not [0]).
For the complete system of doubling sequences modulo 2*n + 1, for n >= 0, see A337712.
This entry generalizes A337712, given together with Gary W. Adamson. [added Dec 14 2020]

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

Cf. A001651, A053446, A337712 (doubling), A337714, A337937.

{1}~Join~Array[Block[{a = {}, k = 3, n = Floor[(3 # - 1)/2], m}, m = EulerPhi[n]; While[Length@ Flatten@ a < m, AppendTo[a, Most@ NestWhileList[Mod[3 #, n] &, If[Length@ a == 0, 1, k], UnsameQ, All]];
Set[k, SelectFirst[Complement[Range[n], Union@ Flatten@ a], GCD[#, n] == 1 &] ]]; a] &, 14, 2] // Flatten (* Michael De Vlieger, Nov 06 2020 *)

T(n, k) gives the k-th entry in the complete tripling system modulo N(n), with N(n) = floor((3*n-1)/2), for n >= 1, where the S(N(n)) = A337714(n) cycles of length P(N(n)) = A053446(n) are written in row n. See the comment above for TS(N, s(N,i), j), i = 1, 2, ..., S(N), and j = 0, 1, ..., P(N) - 1.

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

Original entry on oeis.org

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

Wolfdieter Lang, Dec 13 2020

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.

			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.
...

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

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).

cp's OEIS Frontend

A337936 Irregular triangle read by rows: row n gives the complete system of tripling sequences modulo N = floor((3*n-1)/2), for n >= 1.

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