A229136 -id:A229136 - cp's OEIS frontend

A071303 1/2 times the number of n X n 0..3 matrices M with MM' mod 4 = I, where M' is the transpose of M and I is the n X n identity matrix.

1, 8, 192, 12288, 1966080, 1509949440, 5411658792960

Offset: 1

R. H. Hardin, Jun 11 2002

It seems that a(n) = n! * 2^(binomial(n+1,2) - 1) for n = 1, 2, 3, 4, 5, while for n = 6, a(n) is twice this number. The number n! * 2^(binomial(n+1,2) - 1) appears in Proposition 6.1 in Eriksson and Linusson (2000) as an upper bound to the number of three-dimensional permutation arrays of size n (see column k = 3 of A330490). - Petros Hadjicostas, Dec 16 2019

a(7) = 7! * 2^30. - Sean A. Irvine, Jul 11 2024

			From _Petros Hadjicostas_, Dec 16 2019: (Start)
For n = 2, here are the 2*a(2) = 16 2 x 2 matrices M with elements in {0,1,2,3} that satisfy MM'  mod 4 = I:
(a) With 1 = det(M) mod 4:
  [[1,0],[0,1]]; [[0,1],[3,0]]; [[0,3],[1,0]]; [[1,2],[2,1]];
  [[2,1],[3,2]]; [[2,3],[1,2]]; [[3,0],[0,3]]; [[3,2],[2,3]].
These form the abelian group SO(2, Z_n). See the comments for sequence A060968.
(b) With 3 = det(M) mod 4:
  [[0,1],[1,0]]; [[0,3],[3,0]]; [[1,0],[0,3]];  [[1,2],[2,3]];
  [[2,1],[1,2]]; [[2,3],[3,2]]; [[3,0],[0,1]];  [[3,2],[2,1]].
Note that, for n = 3, we have 2*a(3) = 2*192 = 384 = A264083(4). (End)

Kimmo Eriksson and Svante Linusson, A combinatorial theory of higher-dimensional permutation arrays, Adv. Appl. Math. 25(2) (2000a), 194-211; see Proposition 6.1 (p. 210).
Sean A. Irvinne, Java program (github)
Jianing Song, Structure of the group SO(2,Z_n).
László Tóth, Counting solutions of quadratic congruences in several variables revisited, arXiv:1404.4214 [math.NT], 2014.
László Tóth, Counting Solutions of Quadratic Congruences in Several Variables Revisited, J. Int. Seq. 17 (2014), #14.11.6.

Cf. A003920, A060968, A071302, A071304, A071305, A071306, A071307, A071308, A071309, A071310, A071900, A087784, A208895, A229136, A264083, A330490.

a(7) from Sean A. Irvine, Jul 11 2024

A228920 Number of solutions to Sum_{i=1..n} x_i^2 == 0 (mod 4) with x_i in 0..3.

Original entry on oeis.org

2, 4, 8, 32, 192, 1024, 4608, 18432, 69632, 262144, 1015808, 4063232, 16515072, 67108864, 270532608, 1082130432, 4311744512, 17179869184, 68585259008, 274341036032, 1098437885952, 4398046511104, 17600775979008, 70403103916032, 281543696187392

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 (8,-24,32).

Cf. A101990, A228921, A229136, A318609, A318610.

Column k = 0 of A330619.

Table[((2 + 2I)^n + (2 - 2I)^n + 4^n)/4, {n, 1, 30}]

a(n)=((2+2*I)^n+(2-2*I)^n+4^n)/4 \\ Charles R Greathouse IV, Sep 15 2013

Vec(-2*x*(12*x^2-6*x+1)/((4*x-1)*(8*x^2-4*x+1)) + O(x^100)) \\ Colin Barker, Nov 10 2014

a(n) = ((2+2i)^n + (2-2i)^n + 4^n)/4. - Charles R Greathouse IV, Sep 15 2013

G.f.: -2*x*(12*x^2-6*x+1) / ((4*x-1)*(8*x^2-4*x+1)). - Colin Barker, Nov 10 2014

a(13)-a(25) from Charles R Greathouse IV, Sep 15 2013

A228921 Number of solutions to Sum_{i=1..n} x_i^2 == 0 (mod 8) with x_i in 0..7.

Original entry on oeis.org

2, 8, 32, 128, 3072, 32768, 294912, 2392064, 17825792, 134217728, 1040187392, 8313110528, 67645734912, 549755813888, 4432406249472, 35461397479424, 282574488338432, 2251799813685248, 17979214137393152, 143833163343331328, 1151795604700004352, 9223372036854775808

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. A228920, A229138, A229136.

a[n_]:= a[n]=16 a[n-1]-96 a[n-2] + 256 a[n-3]-256 a[n-4]+4096a[n-5]-24576a[n-6]+ 65536 a[n-7];Do[a[i] = {2, 8, 32, 128, 3072, 32768, 294912}[[i]], {i, 1, 7}];Array[a,33]

a(n)=my(v=vector(8,i,i==1)); for(i=1,n,v+=[2*v[8]+v[5], 2*v[1]+v[6], 2*v[2]+v[7], 2*v[3]+v[8], 2*v[4]+v[1], 2*v[5]+v[2], 2*v[6]+v[3], 2*v[7]+v[4]]); v[1]<Charles R Greathouse IV, Sep 15 2013

Vec(-2*x*(28672*x^6-9216*x^5+1280*x^4-64*x^3+48*x^2-12*x+1)/((8*x-1)*(32*x^2-8*x+1)*(256*x^4+1)) + O(x^100)) \\ Colin Barker, Nov 10 2014

G.f.: -2*x*(28672*x^6-9216*x^5+1280*x^4-64*x^3+48*x^2-12*x+1) / ((8*x-1)*(32*x^2-8*x+1)*(256*x^4+1)). - Colin Barker, Nov 10 2014

a(10)-a(22) from Charles R Greathouse IV, Sep 15 2013

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

A330607 Array read by rows: T(n,k) is the number of solutions to the equation Sum_{i=1..n} x_i^2 == k (mod 5) with x_i in 0..4, where n >= 0 and 0 <= k <= 4.

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 2, 0, 0, 2, 9, 4, 4, 4, 4, 25, 30, 20, 20, 30, 145, 120, 120, 120, 120, 625, 650, 600, 600, 650, 3225, 3100, 3100, 3100, 3100, 15625, 15750, 15500, 15500, 15750, 78625, 78000, 78000, 78000, 78000, 390625, 391250, 390000, 390000, 391250, 1955625, 1952500, 1952500, 1952500, 1952500

Offset: 0

Petros Hadjicostas, Dec 20 2019

Let v(n) = [T(n,0), T(n,1), T(n,2), T(n,3), T(n,4)]' for n >= 0, where ' denotes transpose, and M = [[1,2,0,0,2], [2,1,2,0,0], [0,2,1,2,0], [0,0,2,1,2], [2,0,0,2,1]]. We claim that v(n+1) = M*v(n) for n >= 0.
To see why this is the case, let j in 0..4, and consider the expressions 0^2 + j, 1^2 + j, 2^2 + j, 3^2 + j, and 4^2 + j modulo 5. It can be easily proved that these five numbers contain M[0,j] 0's, M[1,j] 1's, M[2,j] 2's, M[3,j] 3's, and M[4,j] 4'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]'.
The minimal polynomial for M is (z - 5)*(z^2 - 5) = z^3 - 5*z^2 - 5*z + 25. Thus, M^3 - 5*M^2 - 5*M + 25*I = 0, and so M^n*v(0) - 5*M^(n-1)*v(0) - 5*M^(n-2)*v(0) + 25*M^(n-3)*v(0) = 0 for n >= 3. This implies v(n) - 5*v(n-1) - 5*v(n-2) + 25*v(n-3) = 0 for n >= 3. Hence each sequence (T(n,k): n >= 0) satisfies the same recurrence b(n) - 5*b(n-1) - 5*b(n-2) + 25*b(n-3) = 0 for n >= 3.
Clearly, for each k in 0..4, we may find constants c_k, d_k, e_k such that T(n,k) = c_k*sqrt(5)^n + d_k*(-sqrt(5))^n + e_k*5^n for n >= 0. We omit the details.

			Array T(n,k) (with rows n >= 0 and columns 0 <= k <= 4) begins as follows:
     1,    0,    0,    0,    0;
     1,    2,    0,    0,    2;
     9,    4,    4,    4,    4;
    25,   30,   20,   20,   30;
   145,  120,  120,  120,  120;
   625,  650,  600,  600,  650;
  3225, 3100, 3100, 3100, 3100;
  ...
T(n=2,k=0) = 9 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 0 (mod 5) (with x_1, x_2 in 0..4): (0,0), (1,2), (1,3), (2,1), (2,4), (3,1), (3,4), (4,2), and (4,3).
T(n=2,k=1) = 4 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 1 (mod 5) (with x_1, x_2 in 0..4): (0,1), (0,4), (1,0), and (4,0).
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 5) (with x_1, x_2 in 0..4): (1,1), (1,4), (4,1), and (4,4).
T(n=2,k=3) = 4 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 3 (mod 5) (with x_1, x_2 in 0..4): (2,2), (2,3), (3,2), and (3,3).
T(n=2,k=4) = 4 because we have the following solutions (x_1, x_2) to the equation x_1^2 + x_2^2 == 4 (mod 5) (with x_1, x_2 in 0..4): (0,2), (0,3), (2,0), and (3,0).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,5,0,0,0,0,5,0,0,0,0,-25).

Cf. A071304, A101990, A228920, A228921, A229136, A229138, A318609, A318610.

with(LinearAlgebra);
v := proc(n) local M, v0;
   M := Matrix([[1, 2, 0, 0, 2], [2, 1, 2, 0, 0], [0, 2, 1, 2, 0], [0, 0, 2, 1, 2], [2, 0, 0, 2, 1]]); v0 := Matrix([[1], [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 .. 5), n = 0 .. 10);

Vec((1 - 4*x^5 + 2*x^6 + 2*x^9 - x^10 - 6*x^11 + 4*x^12 + 4*x^13 - 6*x^14) / ((1 - 5*x^5)*(1 - 5*x^10)) + O(x^50)) \\ Colin Barker, Dec 21 2019

T(n,k) = 5*T(n-1,k) + 5*T(n-2,k) - 25*T(n-3,k) for n >= 3 with initial conditions for T(0,k), T(1,k), and T(2,k) (for each value of k in 0..4) given in the example below.
T(n,k) = 5*T(n-2,k) + 4*5^(n-2) for n >= 2.
T(n,k=1) = T(n,k=4) and T(n,k=2) = T(n,k=3).
T(n,k) ~ 5^(n-1) for each k in 0..4.
Sum_{k = 0..4} T(n,k) = 5^n.
v(n+1) = M*v(n) and v(n) = M^n * [1,0,0,0,0]' for n >= 0, where M = [[1,2,0,0,2], [2,1,2,0,0], [0,2,1,2,0], [0,0,2,1,2], [2,0,0,2,1]] and v(n) = [T(n,0), T(n,1), T(n,2), T(n,3), T(n,4)]'.
Conjecture: T(n,k=1) = A071304(n)/A071304(n-1) for n >= 2.
From Colin Barker, Dec 21 2019: (Start)
If we consider the array as a single sequence (a(n): n >= 1), then:
G.f.: (1 - 4*x^5 + 2*x^6 + 2*x^9 - x^10 - 6*x^11 + 4*x^12 + 4*x^13 - 6*x^14) / ((1 - 5*x^5)*(1 - 5*x^10)).
a(n) = 5*a(n-5) + 5*a(n-10) - 25*a(n-15) for n > 14. (End)

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

A071303 1/2 times the number of n X n 0..3 matrices M with MM' mod 4 = I, where M' is the transpose of M and I is the n X n identity matrix.

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A228920 Number of solutions to Sum_{i=1..n} x_i^2 == 0 (mod 4) with x_i in 0..3.

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A228921 Number of solutions to Sum_{i=1..n} x_i^2 == 0 (mod 8) with x_i in 0..7.

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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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A330607 Array read by rows: T(n,k) is the number of solutions to the equation Sum_{i=1..n} x_i^2 == k (mod 5) with x_i in 0..4, where n >= 0 and 0 <= k <= 4.

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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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