A179943 Triangle read by rows, antidiagonals of an array (r,k), r=(0,1,2,...), generated from 2 X 2 matrices of the form [1,r; 1,(r+1)].

1, 1, 2, 1, 3, 3, 1, 4, 8, 4, 1, 5, 15, 21, 5, 1, 6, 24, 56, 55, 6, 1, 7, 35, 115, 209, 144, 7, 1, 8, 48, 204, 551, 780, 377, 8, 1, 9, 63, 329, 1189, 2640, 2911, 987, 9, 1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10, 1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11

Offset: 0

Author

Gary W. Adamson, Aug 07 2010

Keywords

• nonn
• tabl

Comments

Row sums = A179944: (1, 3, 7, 17, 47, 148, 518,...)

Row 1 = A001906, row 2 = A001353, row 3 = A004254, row 4 = A001109, row 5 = A004187, row 6 = A001090, row 7 = A018913, row 9 = A004189.

Let S_m(x) be the m-th Chebyshev S-polynomial, described by Wolfdieter Lang in his draft [Lang], defined by S_0(x)=1, S_1(x)=x and S_m(x)=x*S_{m-1}(x)-S_{m-2}(x) (m>1). Let A = (A(r,c)) denote the rectangular array (not the triangle). Then A(r,c) = S_c(r+2), r,c=0,1,2,.... - L. Edson Jeffery, Aug 14 2011

As to the array, (n+1)-th row is the INVERT transform of n-th row. - Gary W. Adamson, Jun 30 2013

If the array sequences are labeled (2,3,4,...) for the n-th sequence, convergence tends to (n + sqrt(n^2 - 4))/2. - Gary W. Adamson, Aug 20 2013

Examples

			First few rows of the array:
  1, 2,  3,   4,    5,    6,     7,...
  1, 3,  8,  21,   55,  144,   377,...
  1, 4, 15,  56,  209,  780,  2911,...
  1, 5, 24, 115,  551, 2640, 12649,...
  1, 6, 35, 204, 1189, 6930, 40391,...
Taking antidiagonals, we obtain triangle A179943:
  1;
  1, 2;
  1, 3, 3;
  1, 4, 8, 4;
  1, 5, 15, 21, 5;
  1, 6, 24, 56, 55, 6;
  1, 7, 35, 115, 209, 144, 7;
  1, 8, 48, 204, 551, 780, 377, 8;
  1, 9, 63, 329, 1189, 2640, 2911, 987, 9;
  1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10;
  1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11;
  1, 12, 120, 980, 6319, 30744, 105937, 235416, 290376, 151316, 17711, 12;
  ...
Examples: Row 1 of the array: (1, 3, 8, 21, 55, 144,...); 144 = term (1,5) of the array = term (2,1) of M^6; where M = the 2 X 2 matrix [1,1; 1,2] and M^6 = [89,144; 144,233].
Term (1,5) of the array = 144 = (r+2)*(term (1,4)) - (term (1,3)) = 3*55 - 21.
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened
• Robert G. Donnelly, Molly W. Dunkum, Sasha V. Malone, and Alexandra Nance, Symmetric Fibonaccian distributive lattices and representations of the special linear Lie algebras, arXiv:2012.14991 [math.CO], 2020.
• Wolfdieter Lang, Chebyshev S-polynomials: ten applications.

Crossrefs

Cf. A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

Programs

Maple

invtr:= proc(b) local a;
          a:= proc(n) option remember; local i;
          `if`(n<1, 1, add(a(n-i) *b(i-1), i=1..n+1)) end
        end:
A:= proc(n) A(n):= `if`(n=0, k->k+1, invtr(A(n-1))) end:
seq(seq(A(d-k)(k), k=0..d), d=0..10);  # Alois P. Heinz, Jul 17 2013
# using observation by Gary W. Adamson

Mathematica

a[, 0] = 0; a[, 1] = 1; a[r_, k_] := a[r, k] = (r+1)*a[r, k-1] - a[r, k-2]; Table[a[r-k+2, k], {r, 0, 10}, {k, 1, r+1}] // Flatten (* Jean-François Alcover, Feb 23 2015 *)

Formula

Antidiagonals of an array, (r,k), a(k) = (r+2)*a(k-1) - a*(k-2), r=0,1,2,... where (r,k) = term (2,1) in the 2 X 2 matrix [1,r; 1,r+1]^(k+1).

G.f. for row r of array: 1/(1 - (r+2)*x + x^2). - L. Edson Jeffery, Oct 26 2012