OFFSET
1,3
COMMENTS
Let R_s be the reduced Collatz sequence (cf. A259663) starting with s and let R_s(k), k >= 0 be the k-th term in R_s. Then R_(2n-1)(k) = (3^k*(2n-1) + T(n,k))/2^j, where j is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k). T(n,k) is defined here as the "residual summand".
The sequence without duplicates is a permutation of A116641.
FORMULA
T(n,k) = 2^j*R_(2n-1)(k) - 3^k*(2n-1), as defined in Comments.
T(n,1) = 1; for k>1: T(n,k) = 3*T(n,k-1) + 2^i, where i is the total number of halving steps from R_(2n-1)(0) to R_(2n-1)(k-1).
EXAMPLE
Triangle starts:
1;
1, 5;
1;
1, 5, 19, 73, 347;
1, 7, 29, 103, 373, 1631;
1, 5, 23, 133;
1, 11;
1, 5, 19, 65, 451;
1, 7, 53;
1, 5, 31, 125, 503, 2533;
1;
1, 5, 19, 185;
1, 7, 29, 151, 581, 2255, 10861;
...
T(5,4)=103 because R_9(4) = 13; the number of halving steps from R_9(0) to R_9(4) is 6, and 13 = (81*9 + 103)/64.
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Bob Selcoe, Sep 15 2019
STATUS
approved