%I #28 Aug 27 2021 16:29:10

%S 1,1,1,1,4,1,1,11,7,1,1,26,30,10,1,1,57,102,58,13,1,1,120,303,256,95,

%T 16,1,1,247,825,955,515,141,19,1,1,502,2116,3178,2310,906,196,22,1,1,

%U 1013,5200,9740,9078,4746,1456,260,25,1

%N T(n, k) = [x^(n-k)] 1/(((1 - 2*x)^k)*(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.

%C The triangle is the matrix inverse of the Riordan square (see A321620) generated by (1 + x - sqrt(1 - 6*x + x^2))/(4*x) (see A172094), where we take the absolute value of the terms.

%C T(n,k) is the number of evil-avoiding (2413, 3214, 4132, and 4213 avoiding) permutations of length (n+2) that start with 1 and whose inverse has k descents. - _Donghyun Kim_, Aug 16 2021

%H Donghyun Kim and Lauren Williams, <a href="https://arxiv.org/abs/2102.00560">Schubert polynomials and the inhomogeneous TASEP on a ring</a>, arXiv:2102.00560 [math.CO], 2021.

%e Triangle starts:

%e [0] [1]

%e [1] [1, 1]

%e [2] [1, 4, 1]

%e [3] [1, 11, 7, 1]

%e [4] [1, 26, 30, 10, 1]

%e [5] [1, 57, 102, 58, 13, 1]

%e [6] [1, 120, 303, 256, 95, 16, 1]

%e [7] [1, 247, 825, 955, 515, 141, 19, 1]

%e [8] [1, 502, 2116, 3178, 2310, 906, 196, 22, 1]

%e [9] [1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1]

%e ...

%e Seen as a square array (the triangle is formed by descending antidiagonals):

%e 1, 1, 1, 1, 1, 1, 1, 1, 1, ... [A000012]

%e 1, 4, 11, 26, 57, 120, 247, 502, 1013, ... [A000295]

%e 1, 7, 30, 102, 303, 825, 2116, 5200, 12381, ... [A045889]

%e 1, 10, 58, 256, 955, 3178, 9740, 28064, 77093, ... [A055583]

%e 1, 13, 95, 515, 2310, 9078, 32354, 106970, 333295, ...

%e 1, 16, 141, 906, 4746, 21504, 87374, 326084, 1136799, ...

%e 1, 19, 196, 1456, 8722, 44758, 204204, 849180, 3275931, ...

%p gf := k -> 1/(((1-2*x)^k)*(1-x)^(k+1)): ser := k -> series(gf(k), x, 32):

%p # Prints the triangle:

%p seq(lprint(seq(coeff(ser(k), x, n-k), k=0..n)), n=0..6);

%p # Prints the square array:

%p seq(lprint(seq(coeff(ser(k), x, n), n=0..8)), k=0..6);

%t (* The function RiordanSquare is defined in A321620; returns the triangle as a lower triangular matrix. *)

%t M := RiordanSquare[(1 + x - Sqrt[1 - 6 x + x^2])/(4 x), 9];

%t Abs[#] & /@ Inverse[PadRight[M]]

%Y Row sums A006012, alternating row sums A118434 with different signs, central column A091527.

%Y T(n, 1) = A000295(n+1) for n >= 1, T(n, 2) = A045889(n-2) for n >= 2, T(n, 3) = A055583(n-3) for n >= 3.

%Y Cf. A172094 (inverse up to sign).

%K nonn,tabl

%O 0,5

%A _Peter Luschny_, Feb 03 2020