OFFSET
0,2
COMMENTS
LINKS
Chai Wah Wu, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,4,-4,0,-6,6,0,4,-4,0,-1,1).
FORMULA
From Chai Wah Wu, Nov 28 2016: (Start)
a(n) = a(n-1) + 4*a(n-3) - 4*a(n-4) - 6*a(n-6) + 6*a(n-7) + 4*a(n-9) - 4*a(n-10) - a(n-12) + a(n-13) for n > 12.
G.f.: -x*(4*x^9 + 20*x^8 + 59*x^7 + 109*x^6 + 96*x^5 + 136*x^4 + 100*x^3 + 28*x^2 + 21*x + 3)/((x - 1)^5*(x^2 + x + 1)^4).
If r = floor(n/3)+1, s = floor((n-1)/3)+1 and t = floor((n-2)/3)+1, then:
a(n) = r^2*s^2 + 2*r^2*s*t + r^2*t^2 + 2*r*s^3 + 6*r*s^2*t + 6*r*s*t^2 + 2*r*t^3 + 2*s^3*t + 2*s*t^3.
If n == 0 mod 3, then a(n) = 4*n^2*(2*n^2 + 6*n + 3)/27.
If n == 1 mod 3, then a(n) = (8*n^4 + 28*n^3 + 33*n^2 + 16*n - 4)/27.
If n == 2 mod 3, then a(n) = 8*(n + 1)^4/27. (End)
MATHEMATICA
t[n_] := t[n] = Flatten[Table[w*z - x*y, {w, a, b}, {x, a, b}, {y, a, b}, {z, a, b}]]
c[n_, k_] := c[n, k] = Count[t[n], k]
u[n_] := u[n] = Sum[c[n, 3 k], {k, -2*n^2, 2*n^2}]
v[n_] := v[n] = Sum[c[n, 3 k + 1], {k, -2*n^2, 2*n^2}]
w[n_] := w[n] = Sum[c[n, 3 k + 2], {k, -2*n^2, 2*n^2}]
Table[u[n], {n, 0, z1}] (* A211033 *)
Table[v[n], {n, 0, z1}] (* A211034 *)
Table[w[n], {n, 0, z1}] (* A211034 *)
LinearRecurrence[{1, 0, 4, -4, 0, -6, 6, 0, 4, -4, 0, -1, 1}, {0, 3, 24, 52, 164, 384, 592, 1131, 1944, 2628, 4128, 6144, 7744}, 60] (* Vincenzo Librandi, Nov 29 2016 *)
PROG
(Python)
from __future__ import division
def A211034(n):
x, y, z = n//3 + 1, (n-1)//3 + 1, (n-2)//3 + 1
return x**2*y**2 + 2*x**2*y*z + x**2*z**2 + 2*x*y**3 + 6*x*y**2*z + 6*x*y*z**2 + 2*x*z**3 + 2*y**3*z + 2*y*z**3 # Chai Wah Wu, Nov 28 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 30 2012
STATUS
approved