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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.

%I A337936 #23 Feb 06 2022 21:23:23
%S A337936 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 A337936 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 A337936 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 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.
%C A337936 The length of row n is A053446(n)*A337714(n) = phi(floor((3*n-1)/2)) = A337937(n), for n >= 1.
%C A337936 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 A337936 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 A337936 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 A337936 For the complete system of doubling sequences modulo 2*n + 1, for n >= 0, see A337712.
%C A337936 This entry generalizes A337712, given together with _Gary W. Adamson_. [added Dec 14 2020]
%F A337936 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 A337936 The irregular triangle T(n, k) begins (cycles are separated by a vertical bar)
%e A337936 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 A337936 1,  1:   1
%e A337936 2,  2:   1
%e A337936 3,  4:   1 3
%e A337936 4,  5:   1 3 4  2
%e A337936 5,  7:   1 3 2  6   4  5
%e A337936 6,  8:   1 3|5  7
%e A337936 7,  10:  1 3 9  7
%e A337936 8,  11:  1 3 9  5   4| 2  6  7 10  8
%e A337936 9,  13:  1 3 9| 2   6  5| 4 12 10| 7  8 11
%e A337936 10, 14:  1 3 9 13  11  5
%e A337936 11, 16:  1 3 9 11|  5 15 13  7
%e A337936 12, 17:  1 3 9 10  13  5 15 11 16 14  8  7  4 12  2  6
%e A337936 13, 19:  1 3 9  8   5 15  7  2  6 18 16 10 11 14  4 12 17 13
%e A337936 14, 20:  1 3 9  7| 11 13 19 17
%e A337936 15, 22:  1 3 9  5  15| 7 21 19 13 17
%e A337936 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 A337936 17, 25:  1 3 9  2   6 18  4 12 11  8 24 22 16 23 19  7 21 13 14 17
%e A337936 18, 26:  1 3 9| 5  15 19| 7 21 11|17 25 23
%e A337936 19, 28:  1 3 9 27  25 19| 5 15 17 23 13 11
%e A337936 ...
%e A337936 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 A337936 ...
%t A337936 {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 A337936 Set[k, SelectFirst[Complement[Range[n], Union@ Flatten@ a], GCD[#, n] == 1 &] ]]; a] &, 14, 2] // Flatten (* _Michael De Vlieger_, Nov 06 2020 *)
%Y A337936 Cf. A001651, A053446, A337712 (doubling), A337714, A337937.
%K A337936 nonn,tabf,easy
%O A337936 1,4
%A A337936 _Wolfdieter Lang_, Oct 22 2020