A210698 Number of 2 X 2 matrices having all terms in {1,...,n} and determinant = 0 (mod 3).
1, 8, 33, 90, 209, 528, 889, 1432, 2673, 3802, 5297, 8448, 11025, 14216, 20625, 25546, 31393, 42768, 51145, 60824, 79233, 92394, 107297, 135168, 154657, 176392, 216513, 244090, 274481, 330000, 367641, 408728, 483153, 533050, 587089
Offset: 1
Keywords
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..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
a = 1; b = n; z1 = 45; 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, 1, z1}] (* A210698 *) Table[v[n], {n, 1, z1}] (* A211071 *) Table[w[n], {n, 1, z1}] (* A211071 *) LinearRecurrence[{1, 0, 4, -4, 0, -6, 6, 0, 4, -4, 0, -1, 1}, {1, 8, 33, 90, 209, 528, 889, 1432, 2673, 3802, 5297, 8448, 11025}, 40] (* Vincenzo Librandi, Dec 01 2016 *) -
Python
from _future_ import division def A210698(n): if n % 3 == 0: return 11*n**4//27 elif n % 3 == 1: return (11*n**4 - 8*n**3 + 6*n**2 + 4*n + 14)//27 else: return (11*n**4 - 16*n**3 + 24*n**2 + 32*n + 8)//27 # Chai Wah Wu, Nov 30 2016
Formula
From Chai Wah Wu, Nov 30 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 > 13.
G.f.: x*(-x^11 - 9*x^10 - 23*x^9 - 115*x^8 - 109*x^7 - 139*x^6 - 219*x^5 - 91*x^4 - 53*x^3 - 25*x^2 - 7*x - 1)/((x - 1)^5*(x^2 + x + 1)^4).
If r = floor(n/3), 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^4/27.
If n == 1 mod 3, then a(n) = (11*n^4 - 8*n^3 + 6*n^2 + 4*n + 14)/27.
If n == 2 mod 3, then a(n) = (11*n^4 - 16*n^3 + 24*n^2 + 32*n + 8)/27. (End)
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