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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)].

%I A179943 #64 Feb 10 2025 01:10:42
%S 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,
%T A179943 1,8,48,204,551,780,377,8,1,9,63,329,1189,2640,2911,987,9,1,10,80,496,
%U A179943 2255,6930,12649,10864,2584,10,1,11,99,711,3905,15456,40391,60605,40545,6765,11
%N 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)].
%C A179943 Row sums = A179944: (1, 3, 7, 17, 47, 148, 518,...)
%C A179943 Row 1 = A001906, row 2 = A001353, row 3 = A004254, row 4 = A001109, row 5 = A004187, row 6 = A001090, row 7 = A018913, row 9 = A004189.
%C A179943 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
%C A179943 As to the array, (n+1)-th row is the INVERT transform of n-th row. - _Gary W. Adamson_, Jun 30 2013
%C A179943 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
%H A179943 Alois P. Heinz, <a href="/A179943/b179943.txt">Rows n = 0..140, flattened</a>
%H A179943 Robert G. Donnelly, Molly W. Dunkum, Sasha V. Malone, and Alexandra Nance, <a href="https://arxiv.org/abs/2012.14991">Symmetric Fibonaccian distributive lattices and representations of the special linear Lie algebras</a>, arXiv:2012.14991 [math.CO], 2020.
%H A179943 Wolfdieter Lang, <a href="/A049310/a049310appl.pdf">Chebyshev S-polynomials: ten applications</a>.
%F A179943 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).
%F A179943 G.f. for row r of array: 1/(1 - (r+2)*x + x^2). - _L. Edson Jeffery_, Oct 26 2012
%e A179943 First few rows of the array:
%e A179943   1, 2,  3,   4,    5,    6,     7,...
%e A179943   1, 3,  8,  21,   55,  144,   377,...
%e A179943   1, 4, 15,  56,  209,  780,  2911,...
%e A179943   1, 5, 24, 115,  551, 2640, 12649,...
%e A179943   1, 6, 35, 204, 1189, 6930, 40391,...
%e A179943 Taking antidiagonals, we obtain triangle A179943:
%e A179943   1;
%e A179943   1, 2;
%e A179943   1, 3, 3;
%e A179943   1, 4, 8, 4;
%e A179943   1, 5, 15, 21, 5;
%e A179943   1, 6, 24, 56, 55, 6;
%e A179943   1, 7, 35, 115, 209, 144, 7;
%e A179943   1, 8, 48, 204, 551, 780, 377, 8;
%e A179943   1, 9, 63, 329, 1189, 2640, 2911, 987, 9;
%e A179943   1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10;
%e A179943   1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11;
%e A179943   1, 12, 120, 980, 6319, 30744, 105937, 235416, 290376, 151316, 17711, 12;
%e A179943   ...
%e A179943 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].
%e A179943 Term (1,5) of the array = 144 = (r+2)*(term (1,4)) - (term (1,3)) = 3*55 - 21.
%p A179943 invtr:= proc(b) local a;
%p A179943           a:= proc(n) option remember; local i;
%p A179943           `if`(n<1, 1, add(a(n-i) *b(i-1), i=1..n+1)) end
%p A179943         end:
%p A179943 A:= proc(n) A(n):= `if`(n=0, k->k+1, invtr(A(n-1))) end:
%p A179943 seq(seq(A(d-k)(k), k=0..d), d=0..10);  # _Alois P. Heinz_, Jul 17 2013
%p A179943 # using observation by _Gary W. Adamson_
%t A179943 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 *)
%Y A179943 Cf. A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.
%K A179943 nonn,tabl
%O A179943 0,3
%A A179943 _Gary W. Adamson_, Aug 07 2010