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A211033
Number of 2 X 2 matrices having all elements in {0,1,...,n} and determinant = 0 (mod 3).
4
1, 10, 33, 152, 297, 528, 1217, 1834, 2673, 4744, 6385, 8448, 13073, 16506, 20625, 29336, 35545, 42768, 57457, 67642, 79233, 102152, 117729, 135168, 168929, 191530, 216513, 264088, 295561, 330000, 394721, 437130, 483153, 568712, 624337, 684288, 794737, 866074
OFFSET
0,2
COMMENTS
A211033(n) + 2*A211034(n)=n^4 for n>0. For a guide to related sequences, see A210000.
LINKS
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^11 - 7*x^10 - 25*x^9 - 53*x^8 - 91*x^7 - 219*x^6 - 139*x^5 - 109*x^4 - 115*x^3 - 23*x^2 - 9*x - 1)/((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^4 + 4*r^3*s + 4*r^3*t + 4*r^2*s^2 + 8*r^2*s*t + 4*r^2*t^2 + s^4 + 6*s^2*t^2 + t^4.
If n == 0 mod 3, then a(n) = (11*n^4 + 60*n^3 + 138*n^2 + 108*n)/27 + 1.
If n == 1 mod 3, then a(n) = (11*n^4 + 52*n^3 + 96*n^2 + 76*n + 35)/27.
If n == 2 mod 3, then a(n) = 11*(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 *)
PROG
(Python)
from __future__ import division
def A211033(n):
x, y, z = n//3 + 1, (n-1)//3 + 1, (n-2)//3 + 1
return x**4 + 4*x**3*y + 4*x**3*z + 4*x**2*y**2 + 8*x**2*y*z + 4*x**2*z**2 + y**4 + 6*y**2*z**2 + z**4 # Chai Wah Wu, Nov 28 2016
CROSSREFS
Sequence in context: A094170 A373129 A004638 * A020479 A219818 A264251
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 30 2012
STATUS
approved