OFFSET
0,3
COMMENTS
a(n) = number of walks in the integer lattice Z^10 starting and ending at the origin, using only the steps of the form (s_1, ..., s_10) with s_1^2 + ... + s_10^2 = 2, i.e., each possible step has precisely two nonzero entries which can be +1 or -1.
LINKS
Christoph Koutschan, Table of n, a(n) for n = 0..449
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.
FORMULA
a(n) conjecturally satisfies a linear recurrence equation of order 30 with polynomial coefficients of degree 274 (see link above).
The probability generating function P(z) = Sum_{n>=0} a(n)*(z/180)^n is given by the 10-fold integral (1/Pi)^10 Int_{0..Pi} ... Int_{0..Pi} 1/(1-z*lambda_10) dk_1 ... dk_10, where the structure function is defined as lambda_10 = (1/binomial(10,2)) Sum_{i=1..10} Sum_{j=(i+1)..10} cos(k_i)*cos(k_j). The function P(z) conjecturally satisfies a linear ODE of order 22 with polynomial coefficients of degree 300 (see link above).
EXAMPLE
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(10,2).
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 10 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 < 10, Floor[(nmax - n)/2], 0]}], {d1, 3, 10}]; First /@ T
CROSSREFS
KEYWORD
nonn,walk
AUTHOR
Christoph Koutschan, Apr 12 2016
STATUS
approved