%I #64 Aug 21 2020 13:12:01
%S 1,1,2,1,3,2,5,5,1,8,10,3,13,20,9,1,21,38,22,4,34,71,51,14,1,55,130,
%T 111,40,5,89,235,233,105,20,1,144,420,474,256,65,6,233,744,942,594,
%U 190,27,1,377,1308,1836,1324,511,98,7,610,2285,3522,2860,1295,315,35,1,987,3970
%N Triangle read by rows: T(n,k) is the number of ways to tile a 2 X n rectangle with k pieces of 2 X 2 tiles and n-2k pieces of 1 X 2 tiles (0 <= k <= floor(n/2)).
%C Row sums are the Jacobsthal numbers (A001045). Column 0 yields the Fibonacci numbers (A000045); the other columns yield convolved Fibonacci numbers (A001629, A001628, A001872, A001873, etc.). Sum_{k=0..floor(n/2)} k*T(n,k) = A073371(n-2).
%C Triangle T(n,k), with zeros omitted, given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Jan 24 2012
%C Riordan array (1/(1-x-x^2), x^2/(1-x-x^2)), with zeros omitted. - _Philippe Deléham_, Feb 06 2012
%C Diagonal sums are A000073(n+2) (tribonacci numbers). - _Philippe Deléham_, Feb 16 2014
%C Number of induced subgraphs of the Fibonacci cube Gamma(n-1) that are isomorphic to the hypercube Q_k. Example: row n=4 is 5, 5, 1; indeed, the Fibonacci cube Gamma(3) is a square with an additional pendant edge attached to one of its vertices; it has 5 vertices (i.e., Q_0's), 5 edges (i.e., Q_1's) and 1 square (i.e., Q_2). - _Emeric Deutsch_, Aug 12 2014
%C Row n gives the coefficients of the polynomial p(n,x) defined as the numerator of the rational function given by f(n,x) = 1 + (x + 1)/f(n-1,x), where f(x,0) = 1. Conjecture: for n > 2, p(n,x) is irreducible if and only if n is a (prime - 2). - _Clark Kimberling_, Oct 22 2014
%H C.-P. Chou and H. A. Witek, <a href="http://match.pmf.kg.ac.rs/electronic_versions/Match68/n1/match68n1_3-30.pdf">An Algorithm and FORTRAN Program for Automatic Computation of the Zhang-Zhang Polynomial of Benzenoids</a>, MATCH: Commun. Math. Comput. Chem, 68 (2012) 3-30. See Eq. (9). - From _N. J. A. Sloane_, Dec 23 2012
%H S. Klavzar, M. Mollard, <a href="http://www-fourier.ujf-grenoble.fr/~mollard/accepte/CubePolyRevised.pdf">Cube polynomial of Fibonacci and Lucas cubes</a>, preprint.
%H S. Klavzar, M. Mollard, <a href="http://dx.doi.org/10.1007/s10440-011-9652-4">Cube polynomial of Fibonacci and Lucas cubes</a>, Acta Appl. Math. 117, 2012, 93-105. - _Emeric Deutsch_, Aug 12 2014
%F G.f.: 1/(1-z-(1+t)z^2).
%F Sum_{k=0..n} T(n,k)*x^k = A053404(n), A015447(n), A015446(n), A015445(n), A015443(n), A015442(n), A015441(n), A015440(n), A006131(n), A006130(n), A001045(n+1), A000045(n+1), A000012(n), A010892(n), A107920(n+1), A106852(n), A106853(n), A106854(n), A145934(n), A145976(n), A145978(n), A146078(n), A146080(n), A146083(n), A146084(n) for x = 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6, -7, -8, -9, -10, -11, -12, and -13, respectively. - _Philippe Deléham_, Jan 24 2012
%F T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - _Philippe Deléham_, Jan 24 2012
%F G.f.: T(0)/2, where T(k) = 1 + 1/(1 - (2*k+1+ x*(1+y))*x/((2*k+2+ x*(1+y))*x + 1/T(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Nov 06 2013
%F T(n,k) = Sum_{i=k..floor(n/2)} binomial(n-i,i)*binomial(i,k). See Corollary 3.3 in the Klavzar et al. link. - _Emeric Deutsch_, Aug 12 2014
%e Triangle starts:
%e 1;
%e 1;
%e 2, 1;
%e 3, 2;
%e 5, 5, 1;
%e 8, 10, 3;
%e 13, 20, 9, 1;
%e 21, 38, 22, 4;
%e From _Philippe Deléham_, Jan 24 2012: (Start)
%e Triangle (1, 1, -1, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, ...) begins:
%e 1;
%e 1, 0;
%e 2, 1, 0;
%e 3, 2, 0, 0;
%e 5, 5, 1, 0, 0;
%e 8, 10, 3, 0, 0, 0;
%e 13, 20, 9, 1, 0, 0, 0;
%e 21, 38, 22, 4, 0, 0, 0, 0; (End)
%e From _Clark Kimberling_, Oct 22 2014: (Start)
%e Here are the first 4 polynomials p(n,x) as in Comment and generated by Mathematica program:
%e 1
%e 2 + x
%e 3 + 2x
%e 5 + 5x + x^2. (End)
%p G:=1/(1-z-(1+t)*z^2): Gser:=simplify(series(G,z=0,19)): for n from 0 to 16 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 16 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form
%t p[x_, n_] := 1 + (x + 1)/p[x, n - 1]; p[x_, 1] = 1;
%t Numerator[Table[Factor[p[x, n]], {n, 1, 20}]] (* _Clark Kimberling_, Oct 22 2014 *)
%Y Cf. A001045, A000045, A001629, A001628, A001872, A001873, A073371.
%Y Cf. A109466, A119473.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Feb 18 2007