OFFSET
0,4
COMMENTS
Also the triangle of the coefficients of the squares of the Fibonacci polynomials. Row n has 1+2*floor(n/2) terms. Sum of terms in row n = (Fibonacci(n+1))^2 (A007598).
From Michael A. Allen, Jun 24 2020: (Start)
T(n,k) is the number of tilings of an n-board (a board with dimensions n X 1) using k (1/2, 1/2)-fence tiles and 2*(n-k) half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal). A (1/2, 1/2)-fence is a tile composed of two 1/2 X 1 pieces separated by a gap of width 1/2.
T(n,k) is the (n, (n-k))-th entry of the (1/(1-x^2), x/(1-x)^2) Riordan array.
(-1)^(n+k)*T(n,k) is the (n, (n-k))-th entry of the (1/(1-x^2), x/(1+x)^2) Riordan array (A158454). (End)
REFERENCES
Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
LINKS
G. C. Greubel, Rows n = 0..100 of the irregula triangle, flattened
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, arXiv:2009.04649 [math.CO], 2020.
Kenneth Edwards and Michael A. Allen, New combinatorial interpretations of the Fibonacci numbers squared, golden rectangle numbers, and Jacobsthal numbers using two types of tile, JIS 24 (2021) Article 21.3.8.
FORMULA
G.f.: G = (1-t*z)/((1+t*z)*(1-z-2*t*z+t^2*z^2)). G = 1/(1-g), where g = z+t^2*z^2+2*t*z^2/(1-t*z) is the g.f. of the indecomposable tilings, i.e., of those that cannot be split vertically into smaller tilings. The row generating polynomials are P(n) = (Fibonacci(n))^2. They satisfy the recurrence relation P(n) = (1+t)*(P(n-1) + t*P(n-2)) - t^3*P(n-3).
T(n,k) = T(n-2,k-2) + binomial(2*n-k-1, 2*n-2*k-1). - Michael A. Allen, Jun 24 2020
EXAMPLE
T(3,1)=4 because the 1 X 2 tile can be placed in any of the four corners of the 2 X 3 grid.
The irregular triangle begins as:
1;
1;
1, 2, 1;
1, 4, 4;
1, 6, 11, 6, 1;
1, 8, 22, 24, 9;
1, 10, 37, 62, 46, 12, 1;
1, 12, 56, 128, 148, 80, 16;
1, 14, 79, 230, 367, 314, 130, 20, 1;
1, 16, 106, 376, 771, 920, 610, 200, 25;
1, 18, 137, 574, 1444, 2232, 2083, 1106, 295, 30, 1;
1, 20, 172, 832, 2486, 4744, 5776, 4352, 1897, 420, 36;
MAPLE
G:=(1-t*z)/(1+t*z)/(1-z-2*t*z+t^2*z^2): Gser:=simplify(series(G, z=0, 14)): for n from 0 to 11 do P[n]:=sort(coeff(Gser, z, n)) od: for n from 0 to 11 do seq(coeff(P[n], t, k), k=0..2*floor(n/2)) od; # yields sequence in triangular form
MATHEMATICA
Block[{T}, T[0, 0]= T[1, 0]= 1; T[n_, k_]:= Which[k==0, 1, k==1, 2(n-1), True, T[n -2, k-2] + Binomial[2n-k-1, 2n-2k-1]]; Table[T[n, k], {n, 0, 14}, {k, 0, 2 Floor[n/2]}]] // Flatten (* Michael De Vlieger, Jun 24 2020 *)
PROG
(Magma)
function A123521(n, k)
if k eq 0 then return 1;
elif k eq 1 then return 2*(n-1);
else return A123521(n-2, k-2) + Binomial(2*n-k-1, 2*n-2*k-1);
end if; return A123521;
end function;
[A123521(n, k): k in [0..2*Floor(n/2)], n in [0..14]]; // G. C. Greubel, Sep 01 2022
(SageMath)
@CachedFunction
def T(n, k): # T = A123521
if (k==0): return 1
elif (k==1): return 2*(n-1)
else: return T(n-2, k-2) + binomial(2*n-k-1, 2*n-2*k-1)
flatten([[T(n, k) for k in (0..2*(n//2))] for n in (0..12)]) # G. C. Greubel, Sep 01 2022
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Oct 16 2006
STATUS
approved