A129637 Number of n-step paths that can go {west, southeast, southwest, northwest} on a 240-degree wedge on the equilateral triangular lattice.

1, 3, 11, 41, 157, 607, 2367, 9277, 36505, 144059, 569779, 2257521, 8957109, 35579351, 141460391, 562871557, 2241129905, 8928207987, 35584894299, 141886838329, 565938926669

Offset: 0

Author

Rebecca Xiaoxi Nie (rebecca.nie(AT)utoronto.ca), May 31 2007

Keywords

• nonn

Comments

If we use the "hour hand" positions at 1, 3, 5, 7, 9, and 11 o'clock on a 12-hour clock to specify directions in the triangular lattice, the allowable steps are in directions 5, 7, 9, and 11 and the path is restricted to stay on or above the 1-7 line. In the Mathematica recurrence below, a(n,k) denotes the number of paths of length n ending k units from the 1-7 line, counted by the last step. - David Callan, Jul 22 2008

Examples

			a(1) = 3 because only 3 out of the 4 steps are permissible from the origin;
a(2) = 11 because the northwest and west steps are followed by 4 permissible steps each, but the southwest step is only followed by 3 permissible steps.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..200
• A. Bostan, Computer Algebra for Lattice Path Combinatorics, Séminaire de Combinatoire Ph. Flajolet, March 28 2013.
• Alin Bostan, Andrew Elvey Price, Anthony John Guttmann, Jean-Marie Maillard, Stieltjes moment sequences for pattern-avoiding permutations, arXiv:2001.00393 [math.CO], 2020.

Crossrefs

Cf. A129400.

Programs

Mathematica

a[0,0]=1; a[n_,k_]/;k<0 || k>n := 0; a[n_,k_]/;0<=k<=n := a[n,k] = 2a[n-1,k-1] + a[n-1,k] + a[n-1,k+1]; a[n_]:=Sum[a[n,k],{k,0,n}]; Table[a[n],{n,0,10}] (* David Callan, Jul 22 2008 *)

Maxima

a(n):=sum(binomial(n+1,k)*sum(2^(i+k-1)*binomial(k,i)*(-1)^(n-i)*binomial(n-i-1,k-i-1),i,0,n),k,0,n+1)/(n+1); /* Vladimir Kruchinin, Feb 28 2016 */

PARI

a(n) = {sum(k=0, n+1, binomial(n+1, k) * sum(i=0,n, 2^(i+k-1)*binomial(k,i)*(-1)^(n-i)*binomial(n-i-1,k-i-1)))/(n+1)} \\ Andrew Howroyd, Dec 22 2017

Formula

Recurrence: (-28-28*n)*a(n) + (-13-n)*a(1+n) + (21+6*n)*a(n+2) + (-4-n)*a(n+3), a(0) = 1, a(1) = 3, a(2) = 11.

G.f.: (1/4*i)*sqrt(-1+2*t+7*t^2)/((-1+4*t)*t)-(1/4)*(-1+5*t)/(t*(-1+4*t)), where i is the imaginary unit.

Differential equation: -(1+28*t^3-6*t+t^2)*t*((d/dt)f(t)) + (9*t-12*t^2-1-28*t^3)*f(t) + 1 - 3*t, f(0) = 1.

a(n) = Sum_{k=0..n+1}(binomial(n+1,k)*Sum_{i=0..n}(2^(i+k-1)*binomial(k,i)*(-1)^(n-i)*binomial(n-i-1,k-i-1)))/(n+1). - Vladimir Kruchinin, Feb 28 2016

a(n) ~ 2^(2*n-1). - Vaclav Kotesovec, Feb 28 2016