A330607 -id:A330607 - cp's OEIS frontend

A318609 a(1) = 2, a(2) = 4, a(3) = 6; for n > 3, a(n) = 3a(n-1) - 3a(n-2) + 9*a(n-3).

2, 4, 6, 24, 90, 252, 702, 2160, 6642, 19764, 58806, 176904, 532170, 1595052, 4780782, 14346720, 43053282, 129146724, 387400806, 1162241784, 3486843450, 10460412252, 31380882462, 94143001680, 282430067922, 847289140884, 2541864234006, 7625595890664, 22876797237930

Offset: 1

Jianing Song, Sep 02 2018

a(n) is the number of solutions to Sum_{i=1..n} x_i^2 == 1 (mod 3).

			a(5) = 90 since M^5 * [1, 0, 0]^T = [81, 90, 72]^T.

Jianing Song, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (3,-3,9).

A101990 gives the number of solutions to Sum_{i=1..n} x_i^2 == 0 (mod 3);
A318610 gives the number of solutions to Sum_{i=1..n} x_i^2 == 2 (mod 3).
Cf. A228921, A228920, A229138, A330607, A330619, A330635.

I:=[2,4,6]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+9*Self(n-3): n in [1..30]]; // Jianing Song, Sep 05 2018

LinearRecurrence[{3, -3, 9}, {2, 4, 6}, 30] (* Jianing Song, Sep 05 2018 *)

Vec(2*x*(1-x)/((1-3*x)*(1+3*x^2)) + O(x^40))

a(n) = ([1, 0, 2 ; 2, 1, 0 ; 0, 2, 1]^n*mattranspose([1, 0, 0]))[2]; \\ Michel Marcus, Dec 20 2019

a(n) = middle term in M^n * [1, 0, 0]^T, where M = the 3 X 3 matrix [1, 0, 2 / 2, 1, 0 / 0, 2, 1] and T denotes transpose. [Edited by Petros Hadjicostas, Dec 19 2019]
O.g.f.: 2*x*(1 - x)/((1 - 3*x)*(1 + 3*x^2)).
E.g.f.: 1/3*(exp(3*x) + 2*cos(sqrt(3)*x - 2*Pi/3)).
a(n) = 3^(n/2 - 1)*((-i)^n*(-1 + sqrt(3)*i)/2 + i^n*(-1 - sqrt(3)*i)/2 + 3^(n/2)), where i is the imaginary unit.
a(n) = 3^(n/2 - 1)*(2*cos(n*Pi/2 - 2*Pi/3) + 3^(n/2)).
a(n) = 3^(n-1) + (-3)^(n/2-1) for even n and 3^(n-1) + (-3)^((n-1)/2) for odd n.
a(n) = a(n-1) + 2*A101990(n-1).
a(n) = A318610(n) for even n and 2*3^(n-1) - A318610(n) for odd n.
a(n) + A101990(n) + A318610(n) = 3^n.

A071304 a(n) = (1/2) * (number of n X n 0..4 matrices M with MM' mod 5 = I, where M' is the transpose of M and I is the n X n identity matrix).

Original entry on oeis.org

1, 4, 120, 14400, 9360000, 29016000000, 457002000000000, 35646156000000000000, 13946558535000000000000000, 27230655539587500000000000000000, 266009466302345390625000000000000000000, 12987912192212013697265625000000000000000000000

Offset: 1

R. H. Hardin, Jun 11 2002

nonn

Also, number of n X n orthogonal matrices over GF(5) with determinant 1. - Max Alekseyev, Nov 06 2022

			From _Petros Hadjicostas_, Dec 17 2019: (Start)
For n = 2, the 2*a(2) = 8 n X n matrices M with elements in {0,1,2,3,4} that satisfy MM' mod 5 = I are the following:
(a) those with 1 = det(M) mod 5:
[[1,0],[0,1]]; [[0,4],[1,0]]; [[0,1],[4,0]]; [[4,0],[0,4]].
These form the abelian group SO(2, Z_5). See the comments for sequence A060968.
(b) those with 4 = det(M) mod 5:
[[0,1],[1,0]]; [[0,4],[4,0]]; [[1,0],[0,4]]; [[4,0],[0,1]].
Note that, for n = 3, we have 2*a(3) = 2*120 = 240 = A264083(5). (End)

Jessie MacWilliams, Orthogonal Matrices Over Finite Fields, The American Mathematical Monthly 76:2 (1969), 152-164.
Jianing Song, Structure of the group SO(2,Z_n).

Cf. A060968, A071302, A071303, A071305, A071306, A071307, A071308, A071309, A071310, A071900, A087784, A208895, A264083, A318609, A330607.

{ a071304(n) = my(t=n\2); prod(i=0, t-1, 5^(2*t)-5^(2*i)) * if(n%2, 5^t, 1/(5^t+1)); } \\ Max Alekseyev, Nov 06 2022

a(2k+1) = 5^k * Product_{i=0..k-1} (5^(2k) - 5^(2i)); a(2k) = (5^k - 1) * Product_{i=1..k-1} (5^(2k) - 5^(2i)) (see MacWilliams, 1969). - Max Alekseyev, Nov 06 2022
From Petros Hadjicostas, Dec 20 2019: (Start)
Let b(n) be the number of solutions to the equation Sum_{i = 1..n} x_i^2 = 1 (mod 5) with x_i in 0..4. We have that b(n) = 5*b(n-1) + 5*b(n-2) - 25*b(n-3) for n >= 3 with b(0) = 0, b(1) = 2, and b(2) = 4.
We have b(n) = A330607(n, k=1) for n >= 0.
We conjecture that a(n+1) = a(n)*b(n+1) for n >= 1. (End)

Terms a(7) onward from Max Alekseyev, Nov 06 2022

A318610 a(1) = 0, a(2) = 4, a(3) = 12; for n > 3, a(n) = 3a(n-1) - 3a(n-2) + 9*a(n-3).

Original entry on oeis.org

0, 4, 12, 24, 72, 252, 756, 2160, 6480, 19764, 59292, 176904, 530712, 1595052, 4785156, 14346720, 43040160, 129146724, 387440172, 1162241784, 3486725352, 10460412252, 31381236756, 94143001680, 282429005040, 847289140884, 2541867422652, 7625595890664, 22876787671992

Offset: 1

Jianing Song, Sep 02 2018

a(n) is the number of solutions to Sum_{i=1..n} x_i^2 == 2 (mod 3).

			a(5) = 72 since M^5 * [1, 0, 0]^T = [81, 90, 72]^T.

Jianing Song, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (3,-3,9).

A101990 gives the number of solutions to Sum_{i=1..n} x_i^2 == 0 (mod 3);
A318609 gives the number of solutions to Sum_{i=1..n} x_i^2 == 1 (mod 3).
Cf. A228921, A228920, A229138, A330607, A330619, A330635.

I:=[0,4,12]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+9*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 04 2018

LinearRecurrence[{3, -3, 9}, {0, 4, 12}, 30] (* Vincenzo Librandi, Sep 04 2018 *)

concat([0], Vec(4*x^2/((1-3*x)*(1+3*x^2)) + O(x^40)))

a(n) = ([1, 0, 2 ; 2, 1, 0 ; 0, 2, 1]^n*mattranspose([1, 0, 0]))[3]; \\ Michel Marcus, Dec 20 2019

a(n) = last term in M^n * [1, 0, 0]^T, where M = the 3 X 3 matrix [1, 0, 2 /  2, 1, 0 /  0, 2, 1] and T denotes transpose. [Edited by Petros Hadjicostas, Dec 19 2019]
O.g.f.: 4*x^2/((1 - 3x)*(1 + 3*x^2)).
E.g.f.: 1/3*(exp(3*x) + 2*cos(sqrt(3)*x + 2*Pi/3)).
a(n) = 3^(n/2 - 1)*((-i)^n*(-1 - sqrt(3)*i)/2 + i^n*(-1 + sqrt(3)*i)/2 + 3^(n/2)), where i is the imaginary unit.
a(n) = 3^(n/2 - 1)*(2*cos(n*Pi/2 + 2*Pi/3) + 3^(n/2)).
a(n) = 3^(n-1) + (-3)^(n/2-1) for even n and 3^(n-1) - (-3)^((n-1)/2) for odd n.
a(n) = a(n-1) + 2*A318609(n-1).
a(n) = A318609(n) for even n and 2*3^(n-1) - A318609(n) for odd n.
a(n) + A101990(n) + A318609(n) = 3^n.

A229136 Number of solutions to Sum_{i=1..n} x_i^2 == 1 (mod 4) with x_i in 0..3.

Original entry on oeis.org

2, 8, 24, 64, 192, 768, 3584, 16384, 69632, 278528, 1081344, 4194304, 16515072, 66060288, 266338304, 1073741824, 4311744512, 17246978048, 68853694464, 274877906944, 1098437885952, 4393751543808, 17583596109824, 70368744177664, 281543696187392, 1126174784749568

Offset: 1

José María Grau Ribas, Sep 15 2013

Conjecture: a(n) = A131885(n)*2^(n-1) for n >= 1. [Corrected by Petros Hadjicostas, Dec 20 2019]
From Petros Hadjicostas, Dec 20 2019: (Start)
Since this sequence is column k = 1 of A330619, we have a(n) = 4*a(n-1) - 8*a(n-2) + 2^(2*n-3) for n >= 3. (This follows from the theory developed in A330619.) If we let b(n) = a(n)/2^(n-1) for n >= 1, we get b(n) = 2*b(n-1) - 2*b(n-2) + 2^(n-2) for n >= 3.
It follows that 2*b(n-1) = 4*b(n-2) - 4*b(n-3) + 2^(n-2) for n >= 4. Subtracting the last equation from the previous one, we get (after some algebra) b(n) = 4*b(n-1) - 6*b(n-2) + 4*b(n-3) for n >= 4. We can easily verify that b(1) = 2, b(2) = 4, and b(3) = 6, and this proves that b(n) = A131885(n) for n >= 1. This proves the above conjecture. (End)

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (8,-24,32).

Cf. A101990, A131885, A228921, A228920, A229138, A318609, A318610, A330607.

Column k = 1 of A330619.

a[1] = 2; a[2] = 8; a[3] = 24; a[n_] := a[n-1]*8 + a[n-2]*(-24) + 32*a[n - 3]; Table[a[n], {n, 15}]

Vec(-2*x*(2*x-1)^2/((4*x-1)*(8*x^2-4*x+1)) + O(x^100)) \\ Colin Barker, Nov 10 2014

G.f.: 1/(1 - 4*x) + Q(0)/(2 - 4*x), where Q(k) = 1 + 1/(1 - 2*x*(k+1)/(2*x*(k+2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 27 2013
G.f.: -2*x*(2*x - 1)^2 / ((4*x - 1)*(8*x^2 - 4*x + 1)). - Colin Barker, Nov 10 2014
a(n) = 4*a(n-1) - 8*a(n-2) + 2^(2*n-3) for n >= 3. - Petros Hadjicostas, Dec 20 2019

A229138 Number of solutions to Sum_{i=1...n} x_i^2 == 1 (mod 8) with x_i in 0..7.

Original entry on oeis.org

4, 16, 96, 512, 2560, 24576, 229376, 2097152, 17956864, 142606336, 1107296256, 8589934592, 67612180480, 541165879296, 4363686772736, 35184372088832, 282583078273024, 2260595906707456, 18049582881570816, 144115188075855872, 1151793405676748800

Offset: 1

José María Grau Ribas, Sep 15 2013

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (16,-96,256,-256,4096,-24576,65536).

Cf. A101990, A228920, A228921, A229136, A318609, A318610, A330607, A330619.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( 4*x*(1 -12*x +56*x^2 -128*x^3 +128*x^4 -1024*x^5 +2048*x^6)/((1-8*x)*(1+256*x^4)*(1-8*x+32*x^2)) )); // G. C. Greubel, Dec 21 2019

seq(coeff(series(4*x*(1 -12*x +56*x^2 -128*x^3 +128*x^4 -1024*x^5 +2048*x^6)/( (1-8*x)*(1+256*x^4)*(1-8*x+32*x^2)), x, n+1), x, n), n = 1..30); # G. C. Greubel, Dec 21 2019

a[n_]:= a[n]= 16a[n-1] -96a[n-2] +256a[n-3] -256a[n-4] +4096a[n-5] -24576 a[n-6] +65536 a[n-7]; Do[a[i]={4, 16, 96, 512, 2560, 24576, 229376}[[i]], {i,7}]; Array[a, 33]

Vec(4*x*(1-12*x+56*x^2-128*x^3+128*x^4-1024*x^5+2048*x^6)/((1-8*x)*(1+256*x^4)*(1-8*x+32*x^2)) + O(x^30)) \\ Colin Barker, Nov 10 2014

def A229138_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 4*x*(1 -12*x +56*x^2 -128*x^3 +128*x^4 -1024*x^5 +2048*x^6)/( (1-8*x)*(1+256*x^4)*(1-8*x+32*x^2)) ).list()
a=A229138_list(30); a[1:] # G. C. Greubel, Dec 21 2019

G.f.: 4*x*(1 -12*x +56*x^2 -128*x^3 +128*x^4 -1024*x^5 +2048*x^6)/((1-8*x)*(1+256*x^4)*(1-8*x+32*x^2)). - Colin Barker, Nov 10 2014

A330619 Array read by rows: T(n,k) is the number of solutions to the equation Sum_{i=1..n} x_i^2 == k (mod 4) with x_i in 0..3, where n >= 0 and 0 <= k <= 3.

Original entry on oeis.org

1, 0, 0, 0, 2, 2, 0, 0, 4, 8, 4, 0, 8, 24, 24, 8, 32, 64, 96, 64, 192, 192, 320, 320, 1024, 768, 1024, 1280, 4608, 3584, 3584, 4608, 18432, 16384, 14336, 16384, 69632, 69632, 61440, 61440, 262144, 278528, 262144, 245760, 1015808, 1081344, 1081344, 1015808, 4063232, 4194304, 4325376, 4194304

Offset: 0

Petros Hadjicostas, Dec 20 2019

Let v(n) = [T(n,0), T(n,1), T(n,2), T(n,3)]' for n >= 0, where ' denotes transpose, and M = [[2,0,0,2], [2,2,0,0], [0,2,2,0], [0,0,2,2]]. We claim that v(n+1) = M*v(n) for n >= 0.
To see why this is the case, let j in 0..3, and consider the expressions 0^2 + j, 1^2 + j, 2^2 + j, and 3^2 + j modulo 4. It can be easily proved that these four numbers contain M[0,j] 0's, M[1,j] 1's, M[2,j] 2's, and M[3,j] 3's. (This is how the transfer matrix M was constructed. The idea is similar to Jianing Song's ideas for sequences A101990, A318609, and A318610.)
It follows that v(n) = M^n * v(0), where v(0) = [1,0,0,0]'.
The minimal polynomial for M is z*(z - 4)*(z^2 - 4*z + 8) = z^4 - 8*z^3 + 24*z^2 - 32*z. Thus, M^4 - 8*M^3 + 24*M^2 - 32*M = 0, and so M^n*v(0) - 8*M^(n-1)*v(0) + 24*M^(n-2)*v(0) - 32*M^(n-3)*v(0) = 0 for n >= 4. This implies v(n) - 8*v(n-1) +  24*v(n-2) - 32*v(n-3) = 0 for n >= 4. Hence each sequence (T(n,k): n >= 0) satisfies the same recurrence b(n) - 8*b(n-1) + 24*b(n-2) - 32*b(n-3) = 0 for n >= 4. (For all k in 0..3, the recurrence is not satisfied for n = 3.)
Clearly, for each k in 0..3, we may find constants c_k, d_k, e_k such that T(n,k) = c_k*(2 + 2*i)^n + d_k*(2 - 2*i)^n + e_k*4^n for n >= 0 (where i = sqrt(-1)). We omit the details.

			Array T(n,k) (with rows n >= 0 and columns 0 <= k <= 3) begins as follows:
     1,    0,    0,    0;
     2,    2,    0,    0;
     4,    8,    4,    0;
     8,   24,   24,    8;
    32,   64,   96,   64;
   192,  192,  320,  320;
  1024,  768, 1024, 1280;
  4608, 3584, 3584, 4608;
  ...
T(n=2,k=0) = 4 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 0 (mod 4) (with x_1, x_2 in 0..3): (0,0), (0,2), (2,0), and (2,2).
T(n=2,k=1) = 8 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 1 (mod 4) (with x_1, x_2 in 0..3): (0,1), (0,3), (1,0), (1,2), (2,1), (2,3), (3,0), and (3,2).
T(n=2,k=2) = 4 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 2 (mod 4) (with x_1, x_2 in 0..3): (1,1), (1,3), (3,1), and (3,3).
T(n=2,k=3) = 0 because we have no solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 3 (mod 4) (with x_1, x_2 in 0..3).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,8,0,0,0,-24,0,0,0,32).

Columns include A228920 (k = 0), A229136 (k = 1).

Cf. A101990, A228921, A229138, A318609, A318610, A330607, A330635.

with(LinearAlgebra);
v := proc(n) local M, v0;
  M := Matrix([[2, 0, 0, 2], [2, 2, 0, 0], [0, 2, 2, 0], [0, 0, 2, 2]]);
  v0 := Matrix([[1], [0], [0], [0]]); if n = 0 then v0; else  MatrixMatrixMultiply(MatrixPower(M, n), v0); end if;
end proc;
seq(seq(v(n)[k, 1], k = 1 .. 4), n = 0 .. 10);

a(n) = ([2,0,0,2; 2,2,0,0; 0,2,2,0; 0,0,2,2]^n*mattranspose([1, 0, 0, 0]));
for(n=0, 30, print1(a(n), ", "));  /* after Michel Marcus's program for A101990 */

Vec((1 - 2*x^4 + 2*x^5)*(1 - 4*x^4 + 4*x^8 + 4*x^10) / ((1 - 2*x^2)*(1 + 2*x^2)*(1 - 4*x^4 + 8*x^8)) + O(x^50)) \\ Colin Barker, Dec 21 2019

T(n,k) = 8*T(n-1,k) - 24*T(n-2,k) + 32*T(n-3,k) for n >= 4 with initial conditions for T(1,k), T(2,k), and T(3,k) (for each value of k in 0..3) given in the example below. (The recurrence is not true for n = 3.)
T(n,k) = 4*T(n-1,k) - 8*T(n-2,k) + 2^(2*n-3) for n >= 3.
T(n,k) ~ 4^(n-1) for each k in 0..3.
Sum_{k = 0..3} T(n,k) = 4^n.
v(n+1) = M*v(n) and v(n) = M^n * [1,0,0,0]' for n >= 0, where M = [[2,0,0,2], [2,2,0,0], [0,2,2,0], [0,0,2,2]] and v(n) = [T(n,0), T(n,1), T(n,2), T(n,3)]'.
From Colin Barker, Dec 21 2019: (Start)
If we consider this array as a single sequence (a(n): n >= 0), then:
G.f.: (1 - 2*x^4 + 2*x^5)*(1 - 4*x^4 + 4*x^8 + 4*x^10) / ((1 - 2*x^2)*(1 + 2*x^2)*(1 - 4*x^4 + 8*x^8)).
a(n) = 8*a(n-4) - 24*a(n-8) + 32*a(n-12) for n > 15. (End)

A330635 Array read by rows: T(n,k) is the number of solutions to the equation Sum_{i=1..n} x_i^2 == k (mod 6) with x_i in 0..5, where n >= 0 and 0 <= k <= 5.

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 1, 2, 0, 1, 2, 0, 2, 8, 8, 2, 8, 8, 36, 24, 48, 36, 24, 48, 264, 192, 192, 264, 192, 192, 1296, 1440, 1152, 1296, 1440, 1152, 7200, 8064, 8064, 7200, 8064, 8064, 46656, 44928, 48384, 46656, 44928, 48384, 286848, 276480, 276480, 286848, 276480, 276480

Offset: 0

Petros Hadjicostas, Dec 21 2019

Let v(n) = [T(n,0), T(n,1), T(n,2), T(n,3), T(n,4), T(n,5)]' for n >= 0, where ' denotes transpose, and M = [[1, 0, 2, 1, 0, 2], [2, 1, 0, 2, 1, 0], [0, 2, 1, 0, 2, 1], [1, 0, 2, 1, 0, 2], [2, 1, 0, 2, 1, 0], [0, 2, 1, 0, 2, 1]]. We claim that v(n+1) = M*v(n) for n >= 0.
To see why this is the case, let j in 0..5, and consider the expressions 0^2 + j, 1^2 + j, 2^2 + j, 3^2 + j, 4^2 + j, and 5^2 + j modulo 6. It can be easily proved that these six numbers contain M[0,j] 0's, M[1,j] 1's, M[2,j] 2's, M[3,j] 3's, M[4,j] 4's, and M[5,j] 5's. (This is how the transfer matrix M was constructed. The idea is similar to Jianing Song's ideas for sequences A101990, A318609, and A318610.)
It follows that v(n) = M^n * v(0), where v(0) = [1,0,0,0,0,0]'.
The minimal polynomial for M is z*(z - 6)*(z^2 + 12) = z^4 - 6*z^3 + 12*z^2 - 72*z. Thus, M^4 - 6*M^3 + 12*M^2 - 72*M = 0, and so M^n*v(0) - 6*M^(n-1)*v(0) + 12*M^(n-2)*v(0) - 72*M^(n-3)*v(0) = 0 for n >= 4. This implies v(n) - 6*v(n-1) + 12*v(n-2) - 72*v(n-3) = 0 for n >= 4. Hence each sequence (T(n,k): n >= 0) satisfies the same recurrence b(n) - 6*b(n-1) + 12*b(n-2) - 72*b(n-3) = 0 for n >= 4.
Clearly, for each k in 0..5, we may find constants c_k, d_k, e_k such that T(n,k) = c_k*(2*sqrt(3)*i)^n + d_k*(-2*sqrt(3)*i)^n + e_k*6^n for n >= 0, where i = sqrt(-1). We omit the details.

			Array T(n,k) (with rows n >= 0 and columns 0 <= k <= 5) begins as follows:
     1,    0,    0,    0,    0,    0;
     1,    2,    0,    1,    2,    0;
     2,    8,    8,    2,    8,    8;
    36,   24,   48,   36,   24,   48;
   264,  192,  192,  264,  192,  192;
  1296, 1440, 1152, 1296, 1440, 1152;
  7200, 8064, 8064, 7200, 8064, 8064;
  ...
T(n=2,k=0) = 2 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 0 (mod 6) (with x_1, x_2 in 0..5): (0,0) and (3,3).
T(n=2,k=1) = 8 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 1 (mod 6) (with x_1, x_2 in 0..5): (0,1), (0,5), (1,0), (2,3), (3,2), (3,4), (4,3), and (5,0).
T(n=2,k=2) = 8 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 2 (mod 6) (with x_1, x_2 in 0..5): (1,1), (1,5), (2,2), (2,4), (4,2), (4,4), (5,1), and (5,5).
T(n=2,k=3) = 2 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 3 (mod 6) (with x_1, x_2 in 0..5): (0,3) and (3,0).
T(n=2,k=4) = 8 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 4 (mod 6) (with x_1, x_2 in 0..5): (0,2), (0,4), (1,3), (2,0), (3,1), (3,5), (4,0), and (5,3).
T(n=2,k=5) = 8 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 5 (mod 6) (with x_1, x_2 in 0..5): (1,2), (1,4), (2,1), (2,5), (4,1), (4,5), (5,2), and (5,4).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,6,0,0,0,0,0,-12,0,0,0,0,0,72).

Cf. A101990, A228920, A228921, A229136, A229138, A318609, A318610, A330607, A330619.

with(LinearAlgebra);
v := proc(n) local M, v0;
   M := Matrix([[1,0,2,1,0,2],[2,1,0,2,1,0],[0,2,1,0,2,1],[1,0,2,1,0,2],[2,1,0,2,1,0],[0,2,1,0,2,1]]);
   v0 := Matrix([[1], [0], [0], [0], [0], [0]]);
   if n = 0 then v0; else MatrixMatrixMultiply(MatrixPower(M, n), v0); end if;
end proc;
seq(seq(v(n)[k, 1], k = 1..6), n = 0..10);

Vec((1 - 5*x^6 + 2*x^7 + x^9 + 2*x^10 + 8*x^12 - 4*x^13 + 8*x^14 - 4*x^15 - 4*x^16 + 8*x^17 - 36*x^18 + 36*x^21) / ((1 - 6*x^6)*(1 + 12*x^12)) + O(x^60)) \\ Colin Barker, Jan 17 2020

T(n,k) = T(n, k+3) for k = 0,1,2 and n >= 0.
T(n,k) = 6*T(n-1,k) - 12*T(n-2,k) + 72*T(n-3,k) for n >= 4 and each k in 0..5. (This is not true for k = 0 or 3 when n = 3 because of the presence of z in the minimal polynomial for M.)
Sum_{k=0..5} T(n,k) = 6^n.
T(n,k) = -12*T(n-2,k) + 8*6^(n-2) for n >= 3 and k = 0..5.
T(n,k) ~ 6^(n-1) for each k in 0..5.
From Colin Barker, Jan 12 2020: (Start)
If we consider this array as a single sequence (a(n): n >= 0), then:
a(n) = 6*a(n-6) - 12*a(n-12) + 72*a(n-18) for n > 21.
G.f.: (1 - 5*x^6 + 2*x^7 + x^9 + 2*x^10 + 8*x^12 - 4*x^13 + 8*x^14 - 4*x^15 - 4*x^16 + 8*x^17 - 36*x^18 + 36*x^21) / ((1 - 6*x^6)*(1 + 12*x^12)).
(End)

cp's OEIS Frontend

A318609 a(1) = 2, a(2) = 4, a(3) = 6; for n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + 9*a(n-3).

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A071304 a(n) = (1/2) * (number of n X n 0..4 matrices M with MM' mod 5 = I, where M' is the transpose of M and I is the n X n identity matrix).

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A318610 a(1) = 0, a(2) = 4, a(3) = 12; for n > 3, a(n) = 3*a(n-1) - 3*a(n-2) + 9*a(n-3).

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A229136 Number of solutions to Sum_{i=1..n} x_i^2 == 1 (mod 4) with x_i in 0..3.

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A229138 Number of solutions to Sum_{i=1...n} x_i^2 == 1 (mod 8) with x_i in 0..7.

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A330619 Array read by rows: T(n,k) is the number of solutions to the equation Sum_{i=1..n} x_i^2 == k (mod 4) with x_i in 0..3, where n >= 0 and 0 <= k <= 3.

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A330635 Array read by rows: T(n,k) is the number of solutions to the equation Sum_{i=1..n} x_i^2 == k (mod 6) with x_i in 0..5, where n >= 0 and 0 <= k <= 5.

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A318609 a(1) = 2, a(2) = 4, a(3) = 6; for n > 3, a(n) = 3a(n-1) - 3a(n-2) + 9*a(n-3).

A318610 a(1) = 0, a(2) = 4, a(3) = 12; for n > 3, a(n) = 3a(n-1) - 3a(n-2) + 9*a(n-3).