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%I #80 May 04 2021 19:00:55
%S 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,
%T 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,
%U 252,0,3150,0,4200,0,1260,0,90,0,1,0,2772,0,11550,0,9240,0,1980,0,110,0,1
%N Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k (1,0)-steps.
%C 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).
%C From _Peter Bala_, Feb 11 2017: (Start)
%C 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)
%H Gheorghe Coserea, <a href="/A109187/b109187.txt">Rows n = 0..100, flattened</a>
%H Paul Barry, <a href="https://arxiv.org/abs/2101.10218">On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths</a>, arXiv:2101.10218 [math.CO], 2021.
%F G.f.: 1/sqrt((1-tz)^2-4z^2).
%F Row sums yield the central trinomial coefficients (A002426).
%F T(2n+1, 0) = 0.
%F T(2n, 0) = binomial(2n,n) (A000984).
%F Sum_{k=0..n} k*T(n,k) = A109188(n).
%F Except for the order, same rows as those of A105868.
%F Column k has e.g.f. (x^k/k!)*Bessel_I(0,2x). - _Paul Barry_, Mar 11 2006
%F T(n,k) = binomial((n+k)/2,k)*binomial(n,(n+k)/2)*(1+(-1)^(n-k))/2. - _Paul Barry_, Sep 18 2007
%F 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
%F 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
%F From _Paul Barry_, Apr 21 2010: (Start)
%F Exponential Riordan array [Bessel_I(0,2x), x].
%F Coefficient array of the polynomials P(n,x) = Sum_{k=0..floor(n/2)} C(n,2k)*C(2k, k)*x^(n - 2k).
%F Diagonal sums are the aerated central Delannoy numbers (A001850 with interpolated zeros). (End)
%F From _Peter Bala_, Feb 11 2017: (Start)
%F T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))*(1 + (-1)^(n-k))/2.
%F T(n,k) = (n/k) * T(n-1,k-1).
%F T(n,k) = the coefficient of H^k in the expansion of (H + U + 1/U)^n.
%F 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.
%F R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(t - 2)^(n-k).
%F d/dt(R(n,t)) = n*R(n-1,t).
%F R(n,t) = (1/Pi) * Integral_{x = 0..Pi} (t + 2*cos(x))^n dx.
%F 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.
%F 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.
%F R(n,t) = A002426 (t = 1), A000984 (t = 2), A026375 (t = 3), A081671 (t = 4), A098409 (t = 5), A098410 (t = 6) and A104454(t = 7).
%F 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.
%F The polynomials R(n,1 + t) are the row polynomials of A171128. (End)
%F From _Peter Luschny_, Jan 23 2018: (Start)
%F These are the coefficients of the polynomials G(n, -n , -x/2) where G(n, a, x) denotes the n-th Gegenbauer polynomial.
%F 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)
%e T(3,1)=6 because we have hud,hdu,udh,duh,uhd,dhu, where u=(1,1),d=(1,-1), h=(1,0).
%e Triangle begins:
%e n\k [0] [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
%e [0] 1;
%e [1] 0, 1;
%e [2] 2, 0, 1;
%e [3] 0, 6, 0, 1;
%e [4] 6, 0, 12, 0, 1;
%e [5] 0, 30, 0, 20, 0, 1;
%e [6] 20, 0, 90, 0, 30, 0, 1;
%e [7] 0, 140, 0, 210, 0, 42, 0, 1;
%e [8] 70, 0, 560, 0, 420, 0, 56, 0, 1;
%e [9] 0, 630, 0, 1680, 0, 756, 0, 72, 0, 1;
%e [10] 252, 0, 3150, 0, 4200, 0, 1260, 0, 90, 0, 1;
%e [11] ...
%e From _Peter Bala_, Feb 11 2017: (Start)
%e The infinitesimal generator begins
%e 0
%e 0 0
%e 2 0 0
%e 0 6 0 0
%e -6 0 12 0 0
%e 0 -30 0 20 0 0
%e 80 0 -90 0 30 0 0
%e 0 560 0 -210 0 42 0 0
%e -2310 0 2240 0 -420 0 56 0 0
%e ....
%e and equals the generalized exponential Riordan array [log(Bessel_I(0,2x)),x], and so has integer entries. (End)
%p 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;
%p with(PolynomialTools): CL := p -> CoefficientList(simplify(p), x):
%p C := (n,x) -> binomial(2*n,n)*hypergeom([-n,-n],[-n+1/2],1/2-x/4):
%p seq(print(CL(C(n,x))), n=0..11); # _Peter Luschny_, Jan 23 2018
%t p[0] := 1; p[n_] := GegenbauerC[n, -n , -x/2];
%t Flatten[Table[CoefficientList[p[n], x], {n, 0, 11}]] (* _Peter Luschny_, Jan 23 2018 *)
%o (PARI)
%o T(n,k) = if ((n-k)%2, 0, binomial(n,k)*binomial(n-k, (n-k)/2));
%o concat(vector(12, n, vector(n, k, T(n-1, k-1)))) \\ _Gheorghe Coserea_, Sep 06 2018
%Y 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).
%Y Cf. A089627, A109188, A105868, A171128, A108666, A282252, A298608.
%K nonn,tabl
%O 0,4
%A _Emeric Deutsch_, Jun 21 2005
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