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%I #10 Dec 11 2019 09:47:45
%S 1,1,2,1,2,5,1,2,5,13,1,2,5,14,34,1,2,5,14,42,89,1,2,5,14,42,131,233,
%T 1,2,5,14,42,132,417,610,1,2,5,14,42,132,429,1341,1597,1,2,5,14,42,
%U 132,429,1429,4334,4181
%N Triangle read by rows: number of Catalan paths of 2n steps of all values less than or equal to m.
%C Convergents of k-th diagonals relate to (2k+3)-polygons; e.g., right border relates to the pentagon (N=5), next border relates to the heptagon (N=7). Convergents of the diagonals are 2 + 2*cos(2*Pi/N) and are roots to Morgan-Voyce polynomials. k2 diagonal = A080937, number of Catalan paths of 2n steps of all values less than or equal to 5. k3 diagonal = A080938, number of Catalan paths of 2n steps of all values less than or equal to 7.
%F Begin with polygonal matrices of the form (exemplified by the Heptagonal matrix M3: [1, 1, 1; 1, 1, 0; 1, 0, 0]). Let matrix P3 = 1 / M3^2; then for n X n matrices P2, P3, P4...perform P^n * [1, 0, 0] letting this vector = k-th diagonal of the triangle.
%e For the right border, odd-indexed Fibonacci numbers (1, 2, 5, 13, 34...), we begin with (M2) = [1, 1; 1, 0], then P2 = [1, -1; -1, 2] = 1/(M2)^2. Performing (P2)^n * [1,0] we extract the left vector (1, 2, 5, 13, ...), making it the right border of the triangle, k1 diagonal.
%e For the next diagonal going to the left, we begin with the Heptagonal matrix M3 = [1, 1, 1; 1, 1, 0; 1, 0, 0], take the inverse square (P3) and then perform the analogous operation getting 1, 2, 5, 14, 42, ...
%e First few rows of the triangle are:
%e 1;
%e 1, 2;
%e 1, 2, 5;
%e 1, 2, 5, 13;
%e 1, 2, 5, 14, 34;
%e 1, 2, 5, 14, 42, 89;
%e 1, 2, 5, 14, 42, 131, 233;
%e 1, 2, 5, 14, 42, 132, 417, 610;
%e ...
%Y Cf. A112880, A001519, A000108, A080937, A080938.
%K nonn,tabl
%O 1,3
%A _Gary W. Adamson_, Sep 16 2006
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