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%I #14 Nov 09 2018 07:31:00
%S 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,
%T 14,6,11,15,7,12,17,9,16,22,10,17,23,11,19,26,14,24,33,15,26,36,17,29,
%U 40,21,37,51,23,40,55,26,45,62,33,57,77,35,61,83
%N 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.
%C From _Franck Maminirina Ramaharo_, Nov 08 2018: (Start)
%C The initial values for b(n,k), 1 <= n <= 9, 1 <= k <= 3, are
%C n\k | 1 2 3
%C ----+---------------
%C 1 | 16 32 36
%C 2 | 17 30 44
%C 3 | 24 36 51
%C 4 | 68 120 176
%C 5 | 105 170 233
%C 6 | 99 186 240
%C 7 | 420 680 932
%C 8 | 470 848 1129
%C 9 | 519 870 1227. (End)
%F From _Franck Maminirina Ramaharo_, Nov 08 2018: (Start)
%F Let M and A denote the following 3 X 3 matrices:
%F 0, 4, 0
%F M = 1, 1, 3
%F 3, 3, 0
%F and
%F 0, 1, 1
%F A = 1, 1, 2
%F 1, 2, 2.
%F 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)
%e Triangle begins:
%e 1, 2, 3;
%e 1, 2, 4;
%e 2, 3, 4;
%e 1, 3, 4;
%e 2, 4, 6;
%e 2, 5, 6;
%e 3, 5, 7;
%e 3, 7, 9;
%e 4, 7, 10;
%e 4, 8, 11;
%e 6, 10, 14;
%e 6, 11, 15;
%e ... - _Franck Maminirina Ramaharo_, Nov 08 2018
%t 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))];
%t A[n_] := M.A[n - 1]; A[0] := {{0, 1, 1}, {1, 1, 2}, {1, 2, 2}};
%t Table[Floor[M.A[n]], {n, 1, 12}]//Flatten
%Y Cf. A097964.
%K nonn,tabf,less
%O 1,2
%A _Roger L. Bagula_, Sep 06 2004
%E Edited, new name, and offset corrected by _Franck Maminirina Ramaharo_, Nov 08 2018