A128100 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)).

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, 111, 40, 5, 89, 235, 233, 105, 20, 1, 144, 420, 474, 256, 65, 6, 233, 744, 942, 594, 190, 27, 1, 377, 1308, 1836, 1324, 511, 98, 7, 610, 2285, 3522, 2860, 1295, 315, 35, 1, 987, 3970

Offset: 0

Author

Emeric Deutsch, Feb 18 2007

Keywords

• nonn
• tabf

Comments

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).

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

Riordan array (1/(1-x-x^2), x^2/(1-x-x^2)), with zeros omitted. - Philippe Deléham, Feb 06 2012

Diagonal sums are A000073(n+2) (tribonacci numbers). - Philippe Deléham, Feb 16 2014

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

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

Examples

			Triangle starts:
   1;
   1;
   2,  1;
   3,  2;
   5,  5,  1;
   8, 10,  3;
  13, 20,  9,  1;
  21, 38, 22,  4;
From _Philippe Deléham_, Jan 24 2012: (Start)
Triangle (1, 1, -1, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, ...) begins:
   1;
   1,  0;
   2,  1,  0;
   3,  2,  0,  0;
   5,  5,  1,  0,  0;
   8, 10,  3,  0,  0,  0;
  13, 20,  9,  1,  0,  0,  0;
  21, 38, 22,  4,  0,  0,  0,  0; (End)
From _Clark Kimberling_, Oct 22 2014: (Start)
Here are the first 4 polynomials p(n,x) as in Comment and generated by Mathematica program:
  1
  2 +  x
  3 + 2x
  5 + 5x + x^2. (End)
		

Links

• C.-P. Chou and H. A. Witek, An Algorithm and FORTRAN Program for Automatic Computation of the Zhang-Zhang Polynomial of Benzenoids, MATCH: Commun. Math. Comput. Chem, 68 (2012) 3-30. See Eq. (9). - From _N. J. A. Sloane_, Dec 23 2012
• S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, preprint.
• S. Klavzar, M. Mollard, Cube polynomial of Fibonacci and Lucas cubes, Acta Appl. Math. 117, 2012, 93-105. - _Emeric Deutsch_, Aug 12 2014

Crossrefs

Cf. A001045, A000045, A001629, A001628, A001872, A001873, A073371.

Cf. A109466, A119473.

Programs

Maple

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

Mathematica

p[x_, n_] := 1 + (x + 1)/p[x, n - 1]; p[x_, 1] = 1;
Numerator[Table[Factor[p[x, n]], {n, 1, 20}]]  (* Clark Kimberling, Oct 22 2014 *)

Formula

G.f.: 1/(1-z-(1+t)z^2).

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

T(n,k) = T(n-1,k) + T(n-2,k) + T(n-2,k-1). - Philippe Deléham, Jan 24 2012

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

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