A280344 Number of 2 X 2 matrices with all elements in {0,...,n} with determinant = permanent^n.

0, 12, 30, 56, 90, 132, 182, 240, 306, 380, 462, 552, 650, 756, 870, 992, 1122, 1260, 1406, 1560, 1722, 1892, 2070, 2256, 2450, 2652, 2862, 3080, 3306, 3540, 3782, 4032, 4290, 4556, 4830, 5112, 5402, 5700, 6006, 6320, 6642, 6972, 7310, 7656, 8010, 8372, 8742, 9120, 9506, 9900, 10302

Offset: 0

Author

Indranil Ghosh, Jan 01 2017

Keywords

• nonn

Comments

For n>0, a(n) is the perimeter of a primitive Pythagorean triangle. - Torlach Rush, Jul 11 2019

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..995

Crossrefs

Same as both A002939 and A118239 without A002939(1) = 2.

Cf. A016754, A280343.

Programs

Mathematica

Table[Boole[n != 0] 2 # (2 # - 1) &[n + 1], {n, 0, 50}] (* or *)
CoefficientList[Series[2 x (6 - 3 x + x^2)/(1 - x)^3, {x, 0, 50}], x] (* Michael De Vlieger, Jan 01 2017 *)

Python

def t(n):
    s=0
    for a in range(0,n+1):
        for b in range(0,n+1):
            for c in range(0,n+1):
                for d in range(0,n+1):
                    if (a*d-b*c)==(a*d+b*c)**n:
                        s+=1
    return s
for i in range(0,41):
    print(i, t(i))

Formula

a(0) = A002939(0) = 0; a(n) = A002939(n+1), for n>=1.

a(n) = (((n-2)*a(n-1))/(n-4)) - (6*(3*(n-1)+1)/(n-4)) for n>=4.

Conjectures from Colin Barker, Jan 01 2017: (Start)

a(n) = 2 + 6*n + 4*n^2 for n>0.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>3.

G.f.: 2*x*(6 - 3*x + x^2) / (1 - x)^3.

(End)

From Torlach Rush, Jul 11 2019: (Start)

a(n) = (2*n+1)*(2*n+2), n>0.

a(n) = 2*((n+1)^2 + ((n+1)*n)), n>0.

(End)