A094718 Array T read by antidiagonals: T(n,k) is the number of involutions avoiding 132 and 12...k.
0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 2, 2, 1, 0, 1, 2, 3, 4, 1, 0, 1, 2, 3, 5, 4, 1, 0, 1, 2, 3, 6, 8, 8, 1, 0, 1, 2, 3, 6, 9, 13, 8, 1, 0, 1, 2, 3, 6, 10, 18, 21, 16, 1, 0, 1, 2, 3, 6, 10, 19, 27, 34, 16, 1, 0, 1, 2, 3, 6, 10, 20, 33, 54, 55, 32, 1, 0, 1, 2, 3, 6, 10, 20, 34, 61, 81, 89, 32, 1
Offset: 1
Examples
Array begins 0 0 0 0 0 0 0 0 0 0 ... 1 1 1 1 1 1 1 1 1 1 ... 1 2 2 4 4 8 8 16 16 32 ... 1 2 3 5 8 13 21 34 55 89 ... 1 2 3 6 9 18 27 54 81 162 ... 1 2 3 6 10 19 33 61 108 197 ... 1 2 3 6 10 20 34 68 116 232 ... 1 2 3 6 10 20 35 69 124 241 ... 1 2 3 6 10 20 35 70 125 250 ... 1 2 3 6 10 20 35 70 126 251 ... ...
Links
- Shaun V. Ault and Charles Kicey, Counting paths in corridors using circular Pascal arrays, Discrete Mathematics, Volume 332, October 2014, Pages 45-54.
- Nachum Dershowitz, Between Broadway and the Hudson, arXiv:2006.06516 [math.CO], 2020.
- Robert Dutton Fray and David Paul Roselle, Weighted lattice paths, Pacific Journal of Mathematics, 37(1) (1971), 85-96.
- O. Guibert and T. Mansour, Restricted 132-involutions, Séminaire Lotharingien de Combinatoire, B48a (2002), 23 pp.
- T. Mansour, Restricted even permutations and Chebyshev polynomials, arXiv:math/0302014 [math.CO], 2003.
Crossrefs
Programs
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Maple
X := (j, n) -> (-1)^(j/(n+1)) - (-1)^((n-j+1)/(n+1)): R := n -> select(k -> type(k, odd), [$(1..n)]): T := (n, k) -> add((2 + X(j,n))*X(j,n)^k, j in R(n))/(n+1): seq(lprint([n], seq(simplify(T(n, k)), k=0..10)), n=0..9); # Peter Luschny, Sep 17 2020
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Mathematica
U = ChebyshevU; m = maxExponent = 14; row[1] = Array[0&, m]; row[k_] := 1/(x U[k, 1/(2x)])*Sum[U[j, 1/(2x)], {j, 0, k-1}] + O[x]^m // CoefficientList[#, x]& // Rest; T = Table[row[n], {n, 1, m}]; Table[T[[n-k+1, k]], {n, 1, m-1}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 17 2018 *)
Formula
G.f. for k-th row: 1/(x*U(k, 1/(2*x))) * Sum_{j=0..k-1} U(j, 1/(2*x)), with U(k, x) the Chebyshev polynomials of second kind. [Clarified by Jean-François Alcover, Nov 17 2018]
T(n, k) = (1/(n+1))*Sum_{j=1..n, j odd} (2 + [j, n]) * [j, n]^k where [j, n] := (-1)^(j/(n+1)) - (-1)^((n-j+1)/(n+1)). - Peter Luschny, Sep 17 2020
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