A337905 The number of walks of n steps on the hexagonal lattice that start at the origin and end at the adjacent vertex (1,0).

1, 2, 15, 60, 340, 1680, 9135, 48440, 264726, 1446060, 7996296, 44396352, 248133600, 1392623232, 7850732175, 44413669872, 252098234674, 1435074678180, 8190821465970, 46860693370920, 268676908816680, 1543504863288960, 8883248453674920, 51210412534906560

Offset: 1

Author

R. J. Mathar, Sep 29 2020

Keywords

• nonn
• walk

Examples

			There are a(2)=2 paths with 2 steps: UD or DU, where R=(1,0), L=(-1,0), U=(1/2,sqrt(3)/2), u=(-1/2,sqrt(3)/2), D=(1/2,-sqrt(3)/2), d=(-1/2,-sqrt(3)/2).
There are a(3)=15 paths with 3 steps: 6 paths permutations of RuD, 6 permutations of RUd, and 3 permutations of RRL.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1000
• Yen Lee Loh, A general method for calculating lattice Green functions on the branch cut, arXiv:1706.03083 [math-ph], 2017, see W_{xyzn}^{hon}.

Crossrefs

Cf. A002898 (returns to origin), A337906, A337907.

Programs

Maple

HexLat := proc(n,finx,finy)
    local a,L,R;
    a := 0 ;
    for L from 0 to n do
    for R from modp(n+finy-L,2) to n-L by 2 do
        a := a+ binomial(n,L) *binomial(n-L,R) *binomial(n-L-R,n/2+L/2-3*R/2+finx) *binomial(n-L-R,(n-L-R-finy)/2) ;
    end do:
    end do:
    a ;
end proc:
seq(HexLat(n,0,0),n=0..15) ; # A002898
seq(HexLat(n,1,0),n=0..15) ; # A337905
seq(HexLat(n,2,0),n=0..15) ; # A337906
seq(HexLat(n,1/2,1),n=0..15) ; # A337905

Mathematica

HexLat[n_, finx_, finy_] := Module[{a = 0, L, R}, For[L = 0, L <= n, L++, For[R = Mod[n + finy - L, 2], R <= n - L, R += 2, a = a + Binomial[n, L]*Binomial[n - L, R]*Binomial[n - L - R, n/2 + L/2 - 3*R/2 + finx]*Binomial[n - L - R, (n - L - R - finy)/2]]]; a];
Table[HexLat[n, 1, 0], {n, 1, 24}] (* Jean-François Alcover, Jun 25 2023, after R. J. Mathar *)

PARI

seq(n)={my(g=sum(m=1, n+1, (3*m)!/m!^3*x^(2*m)*(1+2*x)^m, O(x^(n+2)))); Vec(g/6)} \\ Andrew Howroyd, Aug 09 2025

Formula

D-finite with recurrence (n+1)^2*a(n) -n*(n+1)*a(n-1) -24n^2*a(n-2) -36*n*(n-1)*a(n-3)=0.

a(n+1) = A002898(n)+2*a(n)+2*A337907(n)+A337906(n).

a(n) ~ 2^(n-1) * 3^(n + 1/2) / (Pi*n). - Vaclav Kotesovec, Apr 30 2024

a(n) = A002898(n+1) / 6. - Andrew Howroyd, Aug 09 2025