This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A259663 #84 Jul 01 2023 15:23:04
%S A259663 1,13,11,5,19,7,53,3,55,47,21,35,87,79,31,213,99,23,143,223,191,85,
%T A259663 483,407,15,95,319,127,853,739,663,271,351,63,895,767,341,1251,1175,
%U A259663 1807,863,1599,1407,1279,511
%N A259663 Square array T(n,k) read by antidiagonals upwards: "Dropping Times" in reduced Collatz sequences. (See "Comments" for definitions and explanation.)
%C A259663 Array starts T(2,1).
%C A259663 Coefficients are least residues in congruence classes T(n,k) mod 2^(n+k). Let T"(n,k) be all members of that class.
%C A259663 Let reduced Collatz sequences (R) be Collatz sequences (C) showing only odd terms; and let S be the initial term in R and C, denoted as R(S) and C(S), respectively.
%C A259663 Define "Dropping Time" (D(S)) as the first term in R(S) < its preimage (P(S)). For example, S=159: R(159) starts [159, 239, 359, 539, 809, 607, ...]; therefore, D(159) = 607 and P(159) = 809.
%C A259663 When S is in T"(n,k): k is the number of terms in R(S) from S to D(S), and n is the number of halving steps in C(S) from P(S) to D(S). So for S=159: since 159 == 31 mod 128 and T(2,5) = 31, there are 5 steps from S=159 to D(159) = 607 (239, 359, 539, 809, 607) and 2 halving steps from P(159) = 809 to D(159), i.e., 809*3+1 = 2428; 2428/2 = 1214, 1214/2 = 607.
%C A259663 Generally, when any term in R is in T"(n,k) k>=2, trajectories follow T"(n,k-i) {i=1..k-1}. At T"(n,1), the next term's congruence class is not clearly predictable. So for example S=159: 159 == 31 mod 128 (in T"(2,5)), 239 == 47 mod 64 (in T"(2,4)), 359 == 7 mod 32 (in T"(2,3)), 539 == 11 mod 16 (in T"(2,2)) and 809 == 1 mod 8 (in T"(2,1)). Since 607 == 95 mod 512 (in T"(4,5)), we know the subsequent term will be in T"(4,4), followed by terms in T"(4,3), T"(4,2) and T"(4,1).
%C A259663 Let T"(k) be members of all T"(n,k) holding k constant. Then T"(k) == (2^k-1) mod 2^(k+1). However, since S is the only possible term in R(S) that could be congruent to 0 mod 3, it makes sense to consider only terms congruent to {1,2} mod 3 when evaluating T"(k). Therefore, after the initial term in R(S), all subsequent terms in T"(k) are congruent to either:
%C A259663 i. {(2^k - 1), (6*4^((k-1)/2) - 1)} mod 3*2^(k+1) when k is odd; or
%C A259663 ii. {(3*4^(k/2) - 1), (5*4^(k/2) - 1)} mod 3*2^(k+1) when k is even.
%C A259663 The array yields a wide variety of interesting patterns and sub-patterns associated with the residues and quotients of the congruence classes. Perhaps analysis of these patterns could shed light on the nature of Collatz sequences, including the Collatz conjecture (i.e., all Collatz sequences terminate at 1).
%C A259663 From _Bob Selcoe_, Sep 30 2019: (Start)
%C A259663 From equations i and ii above, terms in T"(k) can be described as follows:
%C A259663 ia. for odd k: {T(2,k), T(m+3,k)} mod 3*2^(k+1) when k == 2^m - 1 mod 2^(m+1), m >= 1; or
%C A259663 iia. for even k: {T(2,k), T(3,k)} mod 3*2^(k+1).
%C A259663 (End)
%F A259663 From _Bob Selcoe_, Jul 15 2017: (Start)
%F A259663 The array is constructed by the following:
%F A259663 T(2,k) = 2^k-1 when k is odd, T(2,k) = 3*2^k-1 when k is even; i.e., A083420((k-1)/2) and A198693(k/2) interleaved.
%F A259663 T(3,k) = 7*2^k-1 when k is odd; T(3,k) = 5*2^k-1 when k is even; i.e., A206372((k-1)/2) and A156760(k/2) interleaved.
%F A259663 For n >= 4: T(n,j) = 2^j-1, j == 2^(n-3) (mod 2^(n-2)); T(n,j-i) = least residue of 2^(j-i)*3^i - 1 mod 2^(n+j-i), 1 <= i < j. (See Example.)
%F A259663 (End)
%e A259663 Array starts T(2,1):
%e A259663 n\k 1 2 3 4 5 6 7 8 9 ...
%e A259663 2: 1 11 7 47 31 191 127 767 511
%e A259663 3: 13 19 55 79 223 319 895 1279 3583
%e A259663 4: 5 3 87 143 95 63 1407 2303 1535
%e A259663 5: 53 35 23 15 351 1599 2431 4351 13823
%e A259663 6: 21 99 407 271 863 575 383 255 22015
%e A259663 7: 213 483 663 1807 3935 2623 12671 8447 5631
%e A259663 8: 85 739 1175 783 5983 14911 20863 57599 38399
%e A259663 9: 853 1251 2199 6927 10079 6719 4479 90367 235007
%e A259663 10: 341 227 151 11023 18271 55871 37247 24831 366079
%e A259663 For n >= 4: e.g., n=4, so j == 2 (mod 4). Select j=6, i=2 to find T(4,4). T(4,6) = 2^6 - 1 = 63. 2^(6-2)*3^2 - 1 mod 2^(4+6-2) = 143 mod 256 = T(4,4) = 143. Now instead select j=10, i=6 to find T(4,4). T(4,10) = 2^10 - 1 = 1023. 2^(10-6)*3^6 - 1 mod 2^(4+10-6) = 11663 mod 256 = 143. - _Bob Selcoe_, Jul 15 2017
%o A259663 (PARI) T(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)));
%o A259663 tabl(nn) = matrix(nn, nn, n, k, T(n+1,k)); \\ _Michel Marcus_, Jul 10 2018
%Y A259663 Cf. A083420, A156760, A198693, A206372.
%K A259663 nonn,tabl
%O A259663 2,2
%A A259663 _Bob Selcoe_, Jul 02 2015