A107265 -id:A107265 - cp's OEIS frontend

A107267 A square array of Motzkin related transforms, read by antidiagonals.

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 6, 3, 1, 0, 9, 20, 12, 4, 1, 0, 21, 72, 54, 20, 5, 1, 0, 51, 272, 261, 112, 30, 6, 1, 0, 127, 1064, 1323, 672, 200, 42, 7, 1, 0, 323, 4272, 6939, 4224, 1425, 324, 56, 8, 1, 0, 835, 17504, 37341, 27456, 10625, 2664, 490, 72, 9, 1

Offset: 0

Paul Barry, May 15 2005

Rows are transforms of k^n, k>=0, under the matrix A107131. As a number triangle, with T(n,k)=if(k<=n,sum{j=0..n-k, (1/(j+1))C(j+1,n-k-j+1)C(n-k,j)k^j},0), row sums are A107268 and diagonal sums are A107269. Rows are series reversions of x/(1+kx+kx^2), k>=0. Conjecture: rows count weighted Motzkin paths.

Row k counts colored Motzkin paths, where H(1,0) and U(1,1) each have k colors and D(1,-1) one color. - Paul Barry, May 16 2005

			Array begins
  1, 0,  0,   0,    0,     0,      0, ...
  1, 1,  2,   4,    9,    21,     51, ...
  1, 2,  6,  20,   72,   272,   1064, ...
  1, 3, 12,  54,  261,  1323,   6939, ...
  1, 4, 20, 112,  672,  4224,  27456, ...
  1, 5, 30, 200, 1425, 10625,  81875, ...
  1, 6, 42, 324, 2664, 22896, 203256, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Rows n=0..6 give A000007, A001006, A071356, A107264, A003645, A107265, A107266.
Main diagonal gives A292716.
Cf. A000108.

Number array T(n,k) = Sum_{j=0..k} n^j * binomial(k,j) * binomial(j+1,k-j+1)/(j+1).
G.f. of row k: 1/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - k*x - k*x^2/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Sep 21 2017
From Seiichi Manyama, May 05 2019: (Start)
T(n,k) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * binomial(2*j,j)/(j+1) = Sum_{j=0..floor(k/2)} n^(k-j) * binomial(k,2*j) * A000108(j).
(k+2) * T(n,k) = n * (2*k+1) * T(n,k-1) - n * (n-4) * (k-1) * T(n,k-2). (End)

A200031 G.f. satisfies: A(x) = 1 + x + 3xA(x) + x*A(x)^2.

Original entry on oeis.org

1, 5, 25, 150, 1000, 7125, 53125, 409375, 3234375, 26059375, 213296875, 1768625000, 14825156250, 125419296875, 1069473046875, 9182583593750, 79319843750000, 688837802734375, 6010580419921875, 52670308222656250, 463321803125000000, 4089876834521484375, 36217014743896484375

Offset: 0

Paul D. Hanna, Nov 12 2011

nonn

Counts colored Motzkin paths starting with a level step H(1,0), where H(1,0) and U(1,1) each have 5 colors and D(1,-1) has one color. - Alexander Burstein, May 27 2021

			G.f.: A(x) = 1 + 5*x + 25*x^2 + 150*x^3 + 1000*x^4 + 7125*x^5 + 53125*x^6 + ...
The g.f. satisfies A(x) = 1 + x*(1 + 3*A(x) + A(x)^2) where:
A(x)^2 = 1 + 10*x + 75*x^2 + 550*x^3 + 4125*x^4 + 31750*x^5 + 250000*x^6 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.

Cf. A107265.

CoefficientList[Series[(1-3*x-Sqrt[1-10*x+5*x^2])/(2*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

{a(n)=polcoeff((1-3*x - sqrt(1-10*x+5*x^2 +x^2*O(x^n)))/(2*x),n)}

G.f.: A(x) = (1-3*x - sqrt(1 - 10*x + 5*x^2))/(2*x).
Recurrence: (n+1)*a(n) = 5*(2*n-1)*a(n-1) - 5*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ sqrt(10+5*sqrt(5))*(5+2*sqrt(5))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012. Equivalently, a(n) ~ 5^((n+1)/2) * phi^(3*n + 3/2) / (2*sqrt(Pi)*n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
a(n) = Sum_{k=0..floor((n-1)/2)}{binomial(n-1,2k)*A000108(k)*5^{n-k}} for n>=1. - Alexander Burstein, May 24 2021
G.f.: Let B(x) = 2 + A(x) and let G(x) be the g.f. for A344623. Then G(x) = 1 + x*G(x)*B(x^2*G(x)). - Alexander Burstein, May 25 2021
a(n) = 5*A107265(n-1) for n >= 1. - Alexander Burstein, May 27 2021

cp's OEIS Frontend

A107267 A square array of Motzkin related transforms, read by antidiagonals.

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A200031 G.f. satisfies: A(x) = 1 + x + 3xA(x) + x*A(x)^2.

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cp's OEIS Frontend

A107267 A square array of Motzkin related transforms, read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A200031 G.f. satisfies: A(x) = 1 + x + 3*x*A(x) + x*A(x)^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A200031 G.f. satisfies: A(x) = 1 + x + 3xA(x) + x*A(x)^2.