cp's OEIS Frontend

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)

Squares of the cotangents of the arcsins of 1/(3*2^n).