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A109187
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Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k (1,0)-steps.
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8
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1, 0, 1, 2, 0, 1, 0, 6, 0, 1, 6, 0, 12, 0, 1, 0, 30, 0, 20, 0, 1, 20, 0, 90, 0, 30, 0, 1, 0, 140, 0, 210, 0, 42, 0, 1, 70, 0, 560, 0, 420, 0, 56, 0, 1, 0, 630, 0, 1680, 0, 756, 0, 72, 0, 1, 252, 0, 3150, 0, 4200, 0, 1260, 0, 90, 0, 1, 0, 2772, 0, 11550, 0, 9240, 0, 1980, 0, 110, 0, 1
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OFFSET
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0,4
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COMMENTS
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A Grand Motzkin path is a path in the half-plane x >= 0, starting at (0,0), ending at (n,0) and consisting of steps u=(1,1), d=(1,-1) and h=(1,0).
Consider an infinite 1-dimensional integer lattice with an oriented self-loop at each vertex. Then T(n,k) equals the number of walks of length n from a vertex to itself having k loops. There is a bijection between such walks and Grand Motzkin paths which takes a right step and a left step on the lattice to an up step U and a down step D of a Grand Motzkin path respectively, and takes traversing a loop on the lattice to the horizontal step H. See A282252 for the corresponding triangle of walks on a 2-dimensional lattice with self-loops. (End)
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LINKS
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FORMULA
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G.f.: 1/sqrt((1-tz)^2-4z^2).
Row sums yield the central trinomial coefficients (A002426).
T(2n+1, 0) = 0.
T(2n, 0) = binomial(2n,n) (A000984).
Sum_{k=0..n} k*T(n,k) = A109188(n).
Except for the order, same rows as those of A105868.
Column k has e.g.f. (x^k/k!)*Bessel_I(0,2x). - Paul Barry, Mar 11 2006
T(n,k) = binomial((n+k)/2,k)*binomial(n,(n+k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Sep 18 2007
Coefficient array of the polynomials P(n,x) = x^n*hypergeom([1/2-n/2,-n/2], [1], 4/x^2). - Paul Barry, Oct 04 2008
G.f.: 1/(1-xy-2x^2/(1-xy-x^2/(1-xy-x^2/(1-xy-x^2/(1-.... (continued fraction). - Paul Barry, Jan 28 2009
Exponential Riordan array [Bessel_I(0,2x), x].
Coefficient array of the polynomials P(n,x) = Sum_{k=0..floor(n/2)} C(n,2k)*C(2k, k)*x^(n - 2k).
Diagonal sums are the aerated central Delannoy numbers (A001850 with interpolated zeros). (End)
T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))*(1 + (-1)^(n-k))/2.
T(n,k) = (n/k) * T(n-1,k-1).
T(n,k) = the coefficient of H^k in the expansion of (H + U + 1/U)^n.
n-th row polynomial R(n,t) = Sum_{k = 0..floor(n/2)} binomial(n,2*k) * binomial(2*k,k) * t^(n-2*k) = coefficient of x^n in the expansion of (1 + t*x + x^2)^n.
R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(t - 2)^(n-k).
d/dt(R(n,t)) = n*R(n-1,t).
R(n,t) = (1/Pi) * Integral_{x = 0..Pi} (t + 2*cos(x))^n dx.
Moment representation on a finite interval: R(n,t) = 1/Pi * Integral_{x = t-2 .. t+2} x^n/sqrt((t + 2 - x)*(x - t + 2)) dx.
Recurrence: n*R(n,t) = t*(2*n - 1)*R(n-1,t) - (t^2 - 4)*(n - 1)*R(n-2,t) with R(0,t) = 1 and R(1,t) = t.
The zeros of the row polynomials appear to lie on the imaginary axis in the complex plane. Also, the zeros of R(n,t) and R(n+1,t) appear to interlace on the imaginary axis.
The polynomials R(n,1 + t) are the row polynomials of A171128. (End)
These are the coefficients of the polynomials G(n, -n , -x/2) where G(n, a, x) denotes the n-th Gegenbauer polynomial.
These polynomials can also be expressed as C(n, x) = binomial(2*n,n)*hypergeom([-n, -n], [-n+1/2], 1/2-x/4). (End)
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EXAMPLE
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T(3,1)=6 because we have hud,hdu,udh,duh,uhd,dhu, where u=(1,1),d=(1,-1), h=(1,0).
Triangle begins:
n\k [0] [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
[0] 1;
[1] 0, 1;
[2] 2, 0, 1;
[3] 0, 6, 0, 1;
[4] 6, 0, 12, 0, 1;
[5] 0, 30, 0, 20, 0, 1;
[6] 20, 0, 90, 0, 30, 0, 1;
[7] 0, 140, 0, 210, 0, 42, 0, 1;
[8] 70, 0, 560, 0, 420, 0, 56, 0, 1;
[9] 0, 630, 0, 1680, 0, 756, 0, 72, 0, 1;
[10] 252, 0, 3150, 0, 4200, 0, 1260, 0, 90, 0, 1;
[11] ...
The infinitesimal generator begins
0
0 0
2 0 0
0 6 0 0
-6 0 12 0 0
0 -30 0 20 0 0
80 0 -90 0 30 0 0
0 560 0 -210 0 42 0 0
-2310 0 2240 0 -420 0 56 0 0
....
and equals the generalized exponential Riordan array [log(Bessel_I(0,2x)),x], and so has integer entries. (End)
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MAPLE
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G:=1/sqrt((1-t*z)^2-4*z^2):Gser:=simplify(series(G, z=0, 15)): P[0]:=1: for n from 1 to 13 do P[n]:=coeff(Gser, z^n) od: for n from 0 to 13 do seq(coeff(t*P[n], t^k), k=1..n+1) od;
with(PolynomialTools): CL := p -> CoefficientList(simplify(p), x):
C := (n, x) -> binomial(2*n, n)*hypergeom([-n, -n], [-n+1/2], 1/2-x/4):
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MATHEMATICA
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p[0] := 1; p[n_] := GegenbauerC[n, -n , -x/2];
Flatten[Table[CoefficientList[p[n], x], {n, 0, 11}]] (* Peter Luschny, Jan 23 2018 *)
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PROG
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(PARI)
T(n, k) = if ((n-k)%2, 0, binomial(n, k)*binomial(n-k, (n-k)/2));
concat(vector(12, n, vector(n, k, T(n-1, k-1)))) \\ Gheorghe Coserea, Sep 06 2018
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CROSSREFS
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Diagonal of rational function R(x, y, t) = 1/(1 - (x^2 + t*x*y + y^2)) with respect to x,y, i.e., T(n,k) = [(xy)^n*t^k] R(x,y,t). For t=0..7 we have the diagonals: A126869(t=0, column 0), A002426(t=1, row sums), A000984(t=2), A026375(t=3), A081671(t=4), A098409(t=5), A098410(t=6), A104454(t=7).
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KEYWORD
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AUTHOR
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STATUS
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approved
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