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
Keywords
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
Programs
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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 */
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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
Comments