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%I #36 Oct 16 2019 13:32:18
%S 0,0,0,0,0,0,0,0,2,0,0,0,1,0,0,2,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,2,
%T 0,0,0,0,1,0,2,0,0,0,0,0,0,0,0,0,0,3,0,1,0,2,0,0,0,1,5,0,0,2,6,20,15,
%U 5,17,3,9,29,2,8,24,74,27,5,15,47,17,53
%N Irregular triangle T(n,k) read by rows: "quotient trajectories" in reduced Collatz sequences; i.e., T(n,k) = q-value(A256598(n,k)) where q-value(z) = (z - A259663(m,j))/2^(m+j) and (m,j) is the unique pair such that z == A259663(m,j) (mod 2^(m+j)). (See Comments for definitions.)
%C Coefficients T(m,j) in the array A259663 are least residues in congruence classes T(m,j) mod 2^(m+j). T"(m,j) denotes all members of that class.
%C Reduced Collatz sequences (i.e., reduced sequences) are standard Collatz sequences excluding even terms. Row n in triangle A256598 shows the reduced sequence starting with 2n+1.
%C Let every positive odd number z = T"(m,j)_q, where q is the quotient of z in T"(m,j). For example, T(2,3) = 7 in A259663, so T"(2,3) contains all numbers == 7 (mod 32). So z = T"(2,3)_0 = 7, z = T"(2,3)_1 = 39, z = T"(2,3)_2 = 71, etc.
%C T(n,k) is the q-value of A256598(n,k) (see Example below). Thus, each row n is defined here as a "quotient trajectory" for the reduced sequence with starting term 2n+1.
%e Triangle starts:
%e 0;
%e 0, 0, 0;
%e 0, 0;
%e 0, 0, 2, 0, 0, 0;
%e 1, 0, 0, 2, 0, 0, 0;
%e 0, 2, 0, 0, 0;
%e 0, 0, 0;
%e 0, 0, 0, 0, 0, 0;
%e 2, 0, 0, 0;
%e 0, 1, 0, 2, 0, 0, 0;
%e 0, 0;
%e 0, 0, 0, 0, 0;
%e 3, 0, 1, 0, 2, 0, 0, 0;
%e ...
%e n=13 starts with 27 = T"(2,2)_1 and takes 41 steps: 1, 5, 0, 0, 2, 6, 20, 15, 5, ..., 0, 0, 0.
%e Row n=12 maps to the reduced sequence n=12 in A256598: 25 -> 19 -> 29 -> 11 -> 17 -> 13 -> 5 -> 1, which is T"(2,1)_3 -> T"(3,2)_0 -> T"(3,1)_1 -> T"(2,2)_0 -> T"(2,1)_2 -> T"(3,1)_0 -> T"(4,1)_0 -> T"(2,1)_0.
%o (PARI) Tdt(n, k) = if (n==2, if (k%2, 2^k-1, 3*2^k-1), if (n==3, if (k%2, 7*2^k-1, 5*2^k-1), mj = 2^(n-3) % 2^(n-2); mk = k % 2^(n-2); (2^k*3^(mj-mk) - 1) % 2^(n+k))); \\ A259663
%o qvalue(m) = {my(line = 2, i, md); while (1, i = line; for (j=1, line-1, md = Tdt(i, j); if (m % (2^(i+j)) == md % (2^(i+j)), return((m-md)/2^(i+j))); i--;); line ++;);}
%o row(n) = {my(oddn = 2*n+1, vl = List(oddn), x); while (oddn != 1, x = 3*oddn+1; oddn = x >> valuation(x, 2); listput(vl, oddn)); my(v = Vec(vl)); for (i=1, #v, v[i] = qvalue(v[i]);); v;} \\ A256598
%o tabf(nn) = {for (n=0, nn, my(rown = row(n)); for (k=1, #rown, print1(rown[k], ", ")); print;);} \\ _Michel Marcus_, Oct 04 2019
%Y Cf. A256598, A259663.
%K nonn,tabf
%O 0,9
%A _Bob Selcoe_, Oct 03 2019
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