A055252 Triangle of partial row sums (prs) of triangle A055249.
1, 4, 1, 13, 5, 1, 38, 18, 6, 1, 104, 56, 24, 7, 1, 272, 160, 80, 31, 8, 1, 688, 432, 240, 111, 39, 9, 1, 1696, 1120, 672, 351, 150, 48, 10, 1, 4096, 2816, 1792, 1023, 501, 198, 58, 11, 1, 9728, 6912, 4608, 2815, 1524, 699, 256, 69, 12, 1, 22784, 16640, 11520
Offset: 0
Examples
[0] 1 [1] 4, 1 [2] 13, 5, 1 [3] 38, 18, 6, 1 [4] 104, 56, 24, 7, 1 [5] 272, 160, 80, 31, 8, 1 [6] 688, 432, 240, 111, 39, 9, 1 [7] 1696, 1120, 672, 351, 150, 48, 10, 1 Fourth row polynomial (n = 3): p(3, x) = 38 + 18*x + 6*x^2 + x^3.
Programs
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Maple
T := (n, k) -> binomial(n, k)*hypergeom([3, k - n], [k + 1], -1): for n from 0 to 7 do seq(simplify(T(n, k)), k = 0..n) od; # Peter Luschny, Sep 23 2024
Formula
a(n, m)=sum(A055249(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n
Column m recursion: a(n, m)= sum(a(j, m), j=m..n-1)+ A055249(n, m), n >= m >= 0, a(n, m) := 0 if n
G.f. for column m: (((1-x)^2)/(1-2*x)^3)*(x/(1-x))^m, m >= 0.
T(n, k) = binomial(n, k)*hypergeom([3, k - n], [k + 1], -1). - Peter Luschny, Sep 23 2024
A331969 T(n, k) = [x^(n-k)] 1/(((1 - 2*x)^k)*(1 - x)^(k + 1)). Triangle read by rows, for 0 <= k <= n.
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, 16, 1, 1, 247, 825, 955, 515, 141, 19, 1, 1, 502, 2116, 3178, 2310, 906, 196, 22, 1, 1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1
Offset: 0
Comments
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.
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
Examples
Triangle starts: [0] [1] [1] [1, 1] [2] [1, 4, 1] [3] [1, 11, 7, 1] [4] [1, 26, 30, 10, 1] [5] [1, 57, 102, 58, 13, 1] [6] [1, 120, 303, 256, 95, 16, 1] [7] [1, 247, 825, 955, 515, 141, 19, 1] [8] [1, 502, 2116, 3178, 2310, 906, 196, 22, 1] [9] [1, 1013, 5200, 9740, 9078, 4746, 1456, 260, 25, 1] ... Seen as a square array (the triangle is formed by descending antidiagonals): 1, 1, 1, 1, 1, 1, 1, 1, 1, ... [A000012] 1, 4, 11, 26, 57, 120, 247, 502, 1013, ... [A000295] 1, 7, 30, 102, 303, 825, 2116, 5200, 12381, ... [A045889] 1, 10, 58, 256, 955, 3178, 9740, 28064, 77093, ... [A055583] 1, 13, 95, 515, 2310, 9078, 32354, 106970, 333295, ... 1, 16, 141, 906, 4746, 21504, 87374, 326084, 1136799, ... 1, 19, 196, 1456, 8722, 44758, 204204, 849180, 3275931, ...
Links
- Donghyun Kim and Lauren Williams, Schubert polynomials and the inhomogeneous TASEP on a ring, arXiv:2102.00560 [math.CO], 2021.
Crossrefs
Programs
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Maple
gf := k -> 1/(((1-2*x)^k)*(1-x)^(k+1)): ser := k -> series(gf(k), x, 32): # Prints the triangle: seq(lprint(seq(coeff(ser(k), x, n-k), k=0..n)), n=0..6); # Prints the square array: seq(lprint(seq(coeff(ser(k), x, n), n=0..8)), k=0..6);
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Mathematica
(* The function RiordanSquare is defined in A321620; returns the triangle as a lower triangular matrix. *) M := RiordanSquare[(1 + x - Sqrt[1 - 6 x + x^2])/(4 x), 9]; Abs[#] & /@ Inverse[PadRight[M]]
Comments