A097966 Rectangular array read by rows (n > 0, 1 <= k <= 3): T(n,k) = floor(b(n,k)/6^(2*(A002264(n) + 1)/3)), where b(n,k) = b(n-3,k) + 13*b (n-6,k) + 36*b(n-9,k), with initial values given in comments.
1, 2, 3, 1, 2, 4, 2, 3, 4, 1, 3, 4, 2, 4, 6, 2, 5, 6, 3, 5, 7, 3, 7, 9, 4, 7, 10, 4, 8, 11, 6, 10, 14, 6, 11, 15, 7, 12, 17, 9, 16, 22, 10, 17, 23, 11, 19, 26, 14, 24, 33, 15, 26, 36, 17, 29, 40, 21, 37, 51, 23, 40, 55, 26, 45, 62, 33, 57, 77, 35, 61, 83
Offset: 1
Examples
Triangle begins: 1, 2, 3; 1, 2, 4; 2, 3, 4; 1, 3, 4; 2, 4, 6; 2, 5, 6; 3, 5, 7; 3, 7, 9; 4, 7, 10; 4, 8, 11; 6, 10, 14; 6, 11, 15; ... - _Franck Maminirina Ramaharo_, Nov 08 2018
Crossrefs
Cf. A097964.
Programs
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Mathematica
M = N[(16/9)^(1/3)*({{0, 1, 0}, {1, 1, 0}, {0, 0, 0}}*(1/4) + {{0, 1, 0}, {0, 0, 1}, {1, 1, 0}}*(3/4))]; A[n_] := M.A[n - 1]; A[0] := {{0, 1, 1}, {1, 1, 2}, {1, 2, 2}}; Table[Floor[M.A[n]], {n, 1, 12}]//Flatten
Formula
From Franck Maminirina Ramaharo, Nov 08 2018: (Start)
Let M and A denote the following 3 X 3 matrices:
0, 4, 0
M = 1, 1, 3
3, 3, 0
and
0, 1, 1
A = 1, 1, 2
1, 2, 2.
Then applying floor() to the entries in (h*M)^(n + 1)*A, where h = 1/(6^(2/3)), yields row 3*n - 2 to 3*n. (End)
Extensions
Edited, new name, and offset corrected by Franck Maminirina Ramaharo, Nov 08 2018
Comments