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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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%I #23 Feb 06 2022 21:23:23

%S 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,

%T 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,

%U 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

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

%C The length of row n is A053446(n)*A337714(n) = phi(floor((3*n-1)/2)) = A337937(n), for n >= 1.

%C 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).

%C 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.

%C 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]).

%C For the complete system of doubling sequences modulo 2*n + 1, for n >= 0, see A337712.

%C This entry generalizes A337712, given together with _Gary W. Adamson_. [added Dec 14 2020]

%F 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.

%e The irregular triangle T(n, k) begins (cycles are separated by a vertical bar)

%e 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 ...

%e 1, 1: 1

%e 2, 2: 1

%e 3, 4: 1 3

%e 4, 5: 1 3 4 2

%e 5, 7: 1 3 2 6 4 5

%e 6, 8: 1 3|5 7

%e 7, 10: 1 3 9 7

%e 8, 11: 1 3 9 5 4| 2 6 7 10 8

%e 9, 13: 1 3 9| 2 6 5| 4 12 10| 7 8 11

%e 10, 14: 1 3 9 13 11 5

%e 11, 16: 1 3 9 11| 5 15 13 7

%e 12, 17: 1 3 9 10 13 5 15 11 16 14 8 7 4 12 2 6

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

%e 14, 20: 1 3 9 7| 11 13 19 17

%e 15, 22: 1 3 9 5 15| 7 21 19 13 17

%e 16, 23: 1 3 9 4 12 13 16 2 6 18 8| 5 15 22 20 14 19 11 10 7 21 17

%e 17, 25: 1 3 9 2 6 18 4 12 11 8 24 22 16 23 19 7 21 13 14 17

%e 18, 26: 1 3 9| 5 15 19| 7 21 11|17 25 23

%e 19, 28: 1 3 9 27 25 19| 5 15 17 23 13 11

%e ...

%e 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.

%e ...

%t {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]];

%t Set[k, SelectFirst[Complement[Range[n], Union@ Flatten@ a], GCD[#, n] == 1 &] ]]; a] &, 14, 2] // Flatten (* _Michael De Vlieger_, Nov 06 2020 *)

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

%K nonn,tabf,easy

%O 1,4

%A _Wolfdieter Lang_, Oct 22 2020