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A337712 Irregular triangle read by rows: row n gives the complete system of cycles of the doubling sequences modulo N = 2*n+1, for n >= 0.

%I A337712 #40 Jun 04 2021 23:22:45
%S A337712 1,2,1,2,4,3,1,2,4,3,6,5,1,2,4,8,7,5,1,2,4,8,5,10,9,7,3,6,1,2,4,8,3,6,
%T A337712 12,11,9,5,10,7,1,2,4,8,7,14,13,11,1,2,4,8,16,15,13,9,3,6,12,7,14,11,
%U A337712 5,10,1,2,4,8,16,13,7,14,9,18,17,15,11,3,6,12,5,10
%N A337712 Irregular triangle read by rows: row n gives the complete system of cycles of the doubling sequences modulo N = 2*n+1, for n >= 0.
%C A337712 The length of row n is A037225(n), for n >= 0.
%C A337712 The doubling sequence modulo N = 2*n+1, for n >= 0, has entries DS(N, s(N,i), j) = s(N,i)*2^j (mod N), with j >= 0, and certain positive odd integer seeds s(N, i), for i = 1, 2, ..., S(N) = A037226((N-1)/2), where gcd(s(N, i), N) = 1 (restricted seeds modulo N). These doubling sequences are periodic with period length P(N) = A002326((N-1)/2) (order of 2 modulo N). Only the periods (cycles) {DS(N, s(N, i), j)}_{j=0..P(N)-1}, for i = 1, 2, ..., S(N), are listed.
%C A337712 N = 1 (n=0) is special: one takes here the restricted residue system modulo N not as [0] but as [1]. The order of 2 modulo 1 is 1, because 2^1 == 1 (mod 1) (== 0 (mod 1)).
%C A337712 In order to obtain the complete system of doubling 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) = phi(N)/P(N) = A037226((N-1)/2)).
%C A337712 The irregular subtriangle where only seed s(N, 1) = 1 has been used is given in A201908. But there 0 (not 1) for N = 1 has been used.
%C A337712 From _Gary W. Adamson_ and _Wolfdieter Lang_, Dec 15 2020: (Start)
%C A337712 The cycles in row n, for N = 2*n + 1, of period length P(N) = A002326((N-1)/2) give the periods of the iterated doubling function D(x) = frac(2*x) with seeds x = s(N, i)/N, for i = 1, 2, ..., S(N) = A037226((N-1)/2), after multiplication with N. This is the doubling function used in the Devaney reference, pp. 24-25, 27, 125. 132, 171,289.
%C A337712 Each cycle in row n can also be used to find from the base 2 version of its first entry (the seed s = s(N, i)) divided by N the other entries by repeated application of a cyclic left shift by one step (called sigma operation) to the period of the base 2 expression of s/N. E.g., n = 7, N = 15, P(N) = 4, s = 1: (1/15)_{10->2} = .repeat(0001), then (.repeat(0010))_{2->10} = 2/10, (.repeat(0100))_{2->10} = 4/10 and (.repeat(1000))_{2->10} = 8/15. Similarly for s = 7: from (7/15)_{10->2} = .repeat(0111) one obtains by repeated sigma operations 14/15, 13/15 and 11/15. The proof uses the elementary formulas for the conversion from base 10 to base 2, and the reverse one, from base 2 to base 10. See also a comment on the period length P(N) given in A002326. (End)
%D A337712 Robert L. Devaney, A First Course in Chaotic Dynamical Systems, Addison-Wesley., 1992. pp. 24-25, 27, 125, 132, 171, 289. Second edition 2020.
%H A337712 Michael De Vlieger, <a href="/A337712/b337712.txt">Table of n, a(n) for n = 0..11921</a> (rows 0 <= n <= 120, flattened)
%F A337712 T(n, k) gives the k-th entry in the complete doubling system modulo N = 2*n+1, for n >= 0, with the S(N) = A037226((N-1)/2) cycles of length A002326((N-1)/2) written in row n. See the comment above for DS(N,s(N,i)), i = 1, 2, ..., S(N).
%e A337712 The irregular triangle T(n, k) begins (cycles are separated by a vertical bar)
%e A337712 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 ...
%e A337712 0,  1:    1
%e A337712 1,  3:    1 2
%e A337712 2,  5:    1 2 4 3
%e A337712 3,  7:    1 2 4|3  6  5
%e A337712 4,  9:    1 2 4 8  7  5
%e A337712 5,  11:   1 2 4 8  5 10  9  7  3  6
%e A337712 6,  13:   1 2 4 8  3  6 12 11  9  5 10  7
%e A337712 7,  15:   1 2 4 8| 7 14 13 11
%e A337712 8,  17:   1 2 4 8 16 15 13  9| 7 14 11  5 10  3  6 12
%e A337712 9,  19:   1 2 4 8 16 13  7 14  9 18 17 15 11  3  6 12 5 10
%e A337712 10, 21:   1 2 4 8 16 11| 5 10 20 19 17 13
%e A337712 11, 23:   1 2 4 8 16  9 18 13  3  6 12| 5 10 20 17 11 22 21 19 15  7 14
%e A337712 12, 25:   1 2 4 8 16  7 14  3  6 12 24 23 21 17  9 18 11 22 19 13
%e A337712 13, 27:   1 2 4 8 16  5 10 20 13 26 25 23 19 11 22 17  7 14
%e A337712 ...
%e A337712 n = 14, N = 29:  1 2 4 8 16  3  6 12 24 19  9 18  7 14 28 27 25 21 13 26 23 17  5 10 20 11 22 15,
%e A337712 n = 15, N = 31: 1 2 4 8 16|3 6 12 24 17|5 10 20 9 18|7 14 28 25 19|11 22 13 26 21|15 30 29 27 23.
%t A337712 Array[Block[{a = {}, k = 2, n = 2 # + 1, m}, m = EulerPhi[n]; While[Length@ Flatten@ a < m, AppendTo[a, Most@ NestWhileList[Mod[2 #, n] &, If[Length@ a == 0, 1, k], UnsameQ, All]]; Set[k, SelectFirst[Complement[Range[n], Union@ Flatten@ a], GCD[#, n] == 1 &] ]]; a] &, 9] // Flatten (* _Michael De Vlieger_, Nov 06 2020 *)
%Y A337712 Cf. A000010, A002326, A037225, A037226, A201908, A038566, A334430 (modified doubling), A337936 (tripling), A339046 (quadrupling).
%K A337712 nonn,tabf,easy
%O A337712 0,2
%A A337712 _Gary W. Adamson_ and _Wolfdieter Lang_, Oct 14 2020