OFFSET
2,2
COMMENTS
Array starts T(2,1).
Coefficients are least residues in congruence classes T(n,k) mod 2^(n+k). Let T"(n,k) be all members of that class.
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.
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.
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.
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).
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:
i. {(2^k - 1), (6*4^((k-1)/2) - 1)} mod 3*2^(k+1) when k is odd; or
ii. {(3*4^(k/2) - 1), (5*4^(k/2) - 1)} mod 3*2^(k+1) when k is even.
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).
From Bob Selcoe, Sep 30 2019: (Start)
From equations i and ii above, terms in T"(k) can be described as follows:
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
iia. for even k: {T(2,k), T(3,k)} mod 3*2^(k+1).
(End)
FORMULA
From Bob Selcoe, Jul 15 2017: (Start)
The array is constructed by the following:
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.
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.
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.)
(End)
EXAMPLE
Array starts T(2,1):
n\k 1 2 3 4 5 6 7 8 9 ...
2: 1 11 7 47 31 191 127 767 511
3: 13 19 55 79 223 319 895 1279 3583
4: 5 3 87 143 95 63 1407 2303 1535
5: 53 35 23 15 351 1599 2431 4351 13823
6: 21 99 407 271 863 575 383 255 22015
7: 213 483 663 1807 3935 2623 12671 8447 5631
8: 85 739 1175 783 5983 14911 20863 57599 38399
9: 853 1251 2199 6927 10079 6719 4479 90367 235007
10: 341 227 151 11023 18271 55871 37247 24831 366079
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
PROG
(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)));
tabl(nn) = matrix(nn, nn, n, k, T(n+1, k)); \\ Michel Marcus, Jul 10 2018
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Bob Selcoe, Jul 02 2015
STATUS
approved