A128099 Triangle read by rows: T(n,k) is the number of ways to tile a 3 X n rectangle with k pieces of 2 X 2 tiles and 3n-4k pieces of 1 X 1 tiles (0 <= k <= floor(n/2)).
1, 1, 1, 2, 1, 4, 1, 6, 4, 1, 8, 12, 1, 10, 24, 8, 1, 12, 40, 32, 1, 14, 60, 80, 16, 1, 16, 84, 160, 80, 1, 18, 112, 280, 240, 32, 1, 20, 144, 448, 560, 192, 1, 22, 180, 672, 1120, 672, 64, 1, 24, 220, 960, 2016, 1792, 448, 1, 26, 264, 1320, 3360, 4032, 1792, 128, 1, 28
Offset: 0
Examples
Triangle starts: 1; 1; 1, 2; 1, 4; 1, 6, 4; 1, 8, 12; 1, 10, 24, 8; 1, 12, 40, 32;
References
- Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 80-83, 357-358
Links
- G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened
- Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Intrinsic Properties of a Non-Symmetric Number Triangle, J. Int. Seq., Vol. 26 (2023), Article 23.4.8.
- Richard Fors, Independence Complexes of Certain Families of Graphs, Master's thesis in Mathematics at KTH, presented Aug 19 2011.
- R. J. Mathar, Tiling n x m rectangles with 1 x 1 and s x s squares arXiv:1609.03964 [math.CO] (2016).
- Zagros Lalo, First layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n
- Zagros Lalo, First layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n
- Eric Weisstein's World of Mathematics, Jacobsthal Polynomial
Crossrefs
Programs
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Maple
T := proc(n,k) if k<=n/2 then 2^k*binomial(n-k,k) else 0 fi end: for n from 0 to 16 do seq(T(n,k),k=0..floor(n/2)) od; # yields sequence in triangular form T := proc(n, k) option remember: if k<0 or k > floor(n/2) then return(0) fi: if k = 0 then return(1) fi: 2*procname(n-2, k-1) + procname(n-1, k): end: seq(seq(T(n, k), k=0..floor(n/2)), n=0..13); # Johannes W. Meijer, Aug 28 2013
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Mathematica
Table[2^k*Binomial[n - k, k] , {n,0,25}, {k,0,Floor[n/2]}] // Flatten (* G. C. Greubel, Dec 28 2016 *) t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, t[n - 1, k] + 2 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}] // Flatten (* Zagros Lalo, Jul 31 2018 *)
Formula
T(n, k) = 2^k*binomial(n-k,k) = 2^k*A011973(n,k).
G.f.: 1/(1-z-2*t*z^2).
Sum_{k=0..floor(n/2)} k*T(n,k) = A095977(n-1).
From Johannes W. Meijer, Aug 28 2013: (Start)
T(n, k) = 2*T(n-2, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, k) = 0 for k < 0 and k > floor(n/2).
T(n, k) = A013609(n-k, k), n >= 0 and 0 <= k <= floor(n/2). (End)
Comments