A211033 Number of 2 X 2 matrices having all elements in {0,1,...,n} and determinant = 0 (mod 3).
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
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).
Programs
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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 *) -
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
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)
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