A123521 Triangle read by rows: T(n,k)=number of tilings of a 2 X n grid with k pieces of 1 X 2 tiles (in horizontal position) and 2n-2k pieces of 1 X 1 tiles (0<=k<=n).
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
Offset: 0
Examples
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;
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.
Crossrefs
Programs
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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
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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
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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 *) -
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
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
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