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A123519
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Triangle read by rows: T(n,k) number of tilings of a 2n X 3 grid by dominoes, 2k of which are in a vertical position (0<=k<=n).
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4
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1, 1, 2, 1, 6, 4, 1, 12, 20, 8, 1, 20, 60, 56, 16, 1, 30, 140, 224, 144, 32, 1, 42, 280, 672, 720, 352, 64, 1, 56, 504, 1680, 2640, 2112, 832, 128, 1, 72, 840, 3696, 7920, 9152, 5824, 1920, 256, 1, 90, 1320, 7392, 20592, 32032, 29120, 15360, 4352, 512, 1, 110, 1980, 13728, 48048, 96096, 116480, 87040, 39168, 9728, 1024
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OFFSET
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0,3
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COMMENTS
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Sum of terms in row n = A001835(n+1). Sum(k*T(n,k), k=0..n)=A123520(n) (n>=1).
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LINKS
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FORMULA
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T(n,k) = 2^k * binomial(n+k,2*k).
G.f.: (1-z)/(1 - 2*z + z^2 - 2*t*z).
Sum_{k=0..n} k*T(n,k) = A123520(n) (n>=1).
Row polynomials are b(n,2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k) * x^k are the Morgan-Voyce polynomials of A085478. The triangle is made up of the odd-indexed rows of A211956. - Peter Bala, May 01 2012
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EXAMPLE
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T(1,1)=2 because a 2 X 3 grid can be tiled in 2 ways with dominoes so that exactly 2 dominoes are in vertical position: place a horizontal domino above or below two adjacent vertical dominoes.
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MAPLE
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T:=(n, k)->2^k*binomial(n+k, 2*k): for n from 0 to 10 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
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MATHEMATICA
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Table[2^k*Binomial[n + k, 2*k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 14 2017 *)
CoefficientList[Table[Sqrt[2] Cosh[(2 n + 1) ArcSinh[Sqrt[x/2]]]/Sqrt[2 + x], {n, 0, 10}] // FunctionExpand // Simplify, x] // Flatten (* Eric W. Weisstein, Apr 04 2018 *)
CoefficientList[Table[ChebyshevT[2 n - 1, Sqrt[1 + x/2]]/Sqrt[1 + x/2], {n, 10}], x] (* Eric W. Weisstein, Apr 04 2018 *)
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PROG
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(PARI) for(n=0, 10, for(k=0, n, print1(2^k*binomial(n+k, 2*k), ", "))) \\ G. C. Greubel, Oct 14 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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