A175934 Number of lattice paths from (0,0) to (n,n) using steps S={(1,0),(0,1),(r,r)|0

1, 2, 7, 27, 116, 532, 2554, 12675, 64507, 334836, 1765833, 9434779, 50962640, 277839361, 1526834471, 8448751385, 47035469902, 263260232668, 1480527858436, 8361881707770, 47409359120684, 269736194796537, 1539547726712437, 8812663513352489, 50579825742416942, 291012706492224315, 1678146650028389023

Offset: 0

Author

Eric Werley, Dec 06 2010

Keywords

• nonn

Comments

Number of lattice paths of weight n that start in (0,0), end on the horizontal axis, never go below this axis, whose steps are of the following four kinds: h=(1,0) of weight 1, H=(1,0) of weight 2, u=(1,1) of weight 1, and d=(1,-1) of weight 0; the weight of a path is the sum of the weights of its steps. Example: a(2)=7 because we have hh, H, hud, udh, udud, uhd, and uudd. - Emeric Deutsch, Sep 09 2014

Examples

			a(2)=7 because we can reach (2,2) in the following ways:
(1,0),(1,0),(0,1),(0,1);
(1,0),(0,1),(1,0),(0,1);
(1,0),(0,1),(1,1);
(1,0),(1,1),(0,1);
(1,1),(1,0),(0,1);
(1,1),(1,1);
(2,2).
G.f. = 1 + 2*x + 7*x^2 + 27*x^3 + 116*x^4 + 532*x^5 + 2554*x^6 + ...
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• J. P. S. Kung and A. de Mier, Catalan lattice paths with rook, bishop and spider steps, Journal of Combinatorial Theory, Series A 120 (2013) 379-389. - From _N. J. A. Sloane_, Dec 27 2012

Crossrefs

Cf. A175935, A175936, A175937, A175939.

Programs

Maxima

a(n):=if n<2 then n+1 else sum((binomial(2*k,k)*sum(binomial(k+1,j)*sum(3^(j-i)*binomial(j,i)*binomial(k-j+1,n-3*k+2*j-i-2)*(2)^(-n+2*(k-j)+1)*(-1)^(k-j+1),i,0,n-k),j,0,k+1))/(k+1),k,0,n); /* Vladimir Kruchinin, Mar 01 2016 */

Sage

n=20
M=[]
for posx in range(0,n+1):
        for posy in range(0,n+1):
            if posx==0:
                if posy==0:
                    M.append([1])
                else:
                    M.append([0])
            else:
                if posy==0:
                    M[0].append(0)
                    M[0][posx]=M[0][posx]+M[0][posx-1]
                else:
                    if posy>posx:
                        M[posy].append(0)
                    else:
                        M[posy].append(0)
                        M[posy][posx]=M[posy][posx-1]+M[posy][posx]
                        M[posy][posx]=M[posy-1][posx]+M[posy][posx]
                        for r in range(1,min(posx,posy,2)+1):
                            M[posy][posx]=M[posy-r][posx-r]+M[posy][posx]
L=[]
for k in range(0,n+1):
    L.append(M[k][k])
print(L)

Formula

G.f.: (1 - x - x^2 - sqrt(x^4 + 2*x^3 - x^2 - 6*x + 1))/(2*x). - Mark van Hoeij, Apr 16 2013

G.f.: 1/(1 - x*(1 + x + 1/(1 - x*(1 + x + 1/...)))). - Michael Somos, Mar 30 2014

G.f. g = g(x) satisfies g = 1 + x*g + x^2*g + x*g^2. - Emeric Deutsch, Sep 09 2014

a(n) = Sum_{k=0..n} ((C(k)*Sum_{j=0..k+1} (binomial(k+1,j)*Sum_{i=0..n-k} (3^(j-i)*binomial(j,i)*binomial(k-j+1,n-3*k+2*j-i-2)*2^(2*(k-j)-n+1)*(-1)^(k-j+1))))), a(0)=1, a(1)=2, where C(k) = A000108(k). - Vladimir Kruchinin, Mar 01 2016

a(0) = 1, a(1) = 2; a(n) = 2 * a(n-1) + 3 * a(n-2) + Sum_{k=2..n-1} a(k) * a(n-k-1). - Ilya Gutkovskiy, Jul 20 2021

G.f.: 1/G(0), where G(k) = 1 - (2*x+x^2)/(1-x/G(k+1)) (continued fraction). - Nikolaos Pantelidis, Jan 08 2023