A327283 Irregular triangle T(n,k) read by rows: "residual summands" in reduced Collatz sequences (see Comments for definition and explanation).

1, 1, 5, 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, 1, 5, 23, 85, 287, 925

Offset: 1

Author

Bob Selcoe, Sep 15 2019

Keywords

• nonn
• tabf

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.

Examples

			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

Cf. A116623, A116641, A259663.

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).