A271671 Number of n-step excursions on the 8-dimensional f.c.c. lattice.

1, 0, 112, 2688, 126000, 6316800, 364887040, 23038364160, 1562288430640, 112014905049600, 8399872737107712, 653454438359331840, 52412319029000899584, 4313870772211888183296, 362994066330649023029760

Offset: 0

Author

Christoph Koutschan, Apr 12 2016

Keywords

• nonn
• walk

Comments

a(n) = number of walks in the integer lattice Z^8 starting and ending at the origin, using only the steps of the form (s_1, ..., s_8) with s_1^2 + ... + s_8^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.

Examples

			There is one walk with no steps.
No walk with a single steps returns to the origin.
The number of returning walks with two steps is exactly the number of allowed steps (called the coordination number of the lattice): a(2) = 4*binomial(8,2).
		

Links

• Christoph Koutschan, Table of n, a(n) for n = 0..493
• S. Hassani, C. Koutschan, J-M. Maillard, N. Zenine, Lattice Green Functions: the d-dimensional face-centred cubic lattice, d = 8, 9, 10, 11, 12, arXiv:1601.05657 [math-ph], 2016.
• S. Hassani, C. Koutschan, J-M. Maillard, N. Zenine, Lattice Green functions: the d-dimensional face-centred cubic lattice, d = 8, 9, 10, 11, 12, Journal of Physics A: Mathematical and Theoretical 49(16) (2016), 164003.
• C. Koutschan, Computations for higher-dimensional fcc lattices.
• C. Koutschan, Differential operator annihilating the generating function.
• C. Koutschan, Recurrence equation.

Crossrefs

Cf. A002899 (d = 3, i.e., excursions on the 3-dimensional f.c.c. lattice), A271432 (d = 4), A271650 (d = 5), A271651 (d = 6), A271670 (d = 7), this sequence (d = 8), A271672 (d = 9), A271673 (d = 10), A271674 (d = 11).

Programs

Maple

nmax := 50: tt := [seq([seq(add(binomial(2*p,p)*binomial(2*j,2*p-n)*binomial(2*n+2*j-2*p,n+j-p), p = floor((n+1)/2)..floor((n+2*j)/2)), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: for d1 from 3 to 8 do tt := [seq([seq(add(binomial(n,p)*add(binomial(2*j,2*q-p)*binomial(2*j+2*p-2*q,j+p-q)*tt[n-p+1,q+1], q = floor((p+1)/2)..floor((p+2*j)/2)), p = 0..n), j = 0..floor((nmax-n)/2))], n = 0..nmax)]: od: [seq(tt[n+1,1], n = 0..nmax)];

Mathematica

nmax = 50; T = Table[Sum[Binomial[2 p, p]*Binomial[2 j, 2 p - n]*Binomial[2 n + 2 j - 2 p, n + j - p], {p, Floor[(n + 1)/2], Floor[(n + 2 j)/2]}], {n, 0, nmax}, {j, 0, Floor[(nmax - n)/2]}]; Do[T = Table[Sum[Binomial[n, p]*Sum[Binomial[2 j, 2 q - p]*Binomial[2 j + 2 p - 2 q, j + p - q]*T[[n - p + 1, q + 1]], {q, Floor[(p + 1)/2], Floor[(p + 2 j)/2]}], {p, 0, n}], {n, 0, nmax}, {j, 0, If[d1 < 8, Floor[(nmax - n)/2], 0]}], {d1, 3, 8}]; First /@ T

Formula

a(n) conjecturally satisfies a linear recurrence equation of order 20 with polynomial coefficients of degree 109 (see link above).

The probability generating function P(z) = Sum_{n>=0} a(n)*(z/112)^n is given by the 8-fold integral (1/Pi)^8 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_8) dk_1 ... dk_8, where the structure function is defined as lambda_8 = (1/binomial(8,2)) Sum_{i=1..8} Sum_{j=(i+1)..8} cos(k_i)*cos(k_j). The function P(z) conjecturally satisfies a linear ODE of order 14 with polynomial coefficients of degree 126 (see link above).