A280364 Number of 2 X 2 matrices with all elements in {0,...,n} with permanent = determinant^n.

0, 12, 35, 56, 99, 132, 195, 240, 323, 380, 483, 552, 675, 756, 899, 992, 1155, 1260, 1443, 1560, 1763, 1892, 2115, 2256, 2499, 2652, 2915, 3080, 3363, 3540, 3843, 4032, 4355, 4556, 4899, 5112, 5475, 5700, 6083, 6320, 6723, 6972, 7395, 7656, 8099, 8372, 8835, 9120, 9603, 9900, 10403

Offset: 0

Author

Indranil Ghosh, Jan 01 2017

Keywords

• nonn

Examples

			For n=2, the matrices are [0,0,0,0], [0,0,0,1], [0,0,0,2], [0,0,1,0], [0,0,1,1], [0,0,1,2], [0,0,2,0], [0,0,2,1], [0,0,2,2], [0,1,0,0], [0,1,0,1], [0,1,0,2], [0,1,1,0], [0,1,1,1], [0,1,1,2], [0,2,0,0], [0,2,0,1], [0,2,0,2], [1,0,0,0], [1,0,0,1], [1,0,1,0], [1,0,1,1], [1,0,2,0], [1,0,2,1], [1,1,0,0], [1,1,0,1], [1,1,1,0], [1,2,0,0], [1,2,0,1], [2,0,0,0], [2,0,1,0], [2,0,2,0], [2,1,0,0], [2,1,1,0], [2,2,0,0].
Here each of these matrices M is defined as  M=[a,b,c,d], where a=M[1][1], b=M[1][2], c=M[2][1], d=M[2][2]. There are 35 possibilities. So for n=2, a(n)=35.
		

Links

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

Crossrefs

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

Programs

Mathematica

CoefficientList[Series[x (12 + 23 x - 3 x^2 - 3 x^3 + 3 x^4)/((1 - x)^3*(1 + x)^2), {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)**n==(a*d+b*c):
                        s+=1
    return s
for i in range(0,51):
    print(i, t(i))

Python

def a(n):
    if n==0:
        return 0
    if n==1:
        return 12
    return (((-2*n-1)*a(n-1))//(2*n-1))+8*n**2+10*n+3
for i in range(0,51):
    print(i, a(i))

Formula

a(n) = (((-2*n-1)*a(n-1))/(2*n-1)) + 8*n^2 + 10*n + 3 for n>=2. [Corrected by David Radcliffe, Aug 13 2025]

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

a(n) = 4*n^2 + 8*n + 3 for n>0 and even.

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

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5.

G.f.: x*(12 + 23*x - 3*x^2 - 3*x^3 + 3*x^4) / ((1 - x)^3*(1 + x)^2). (End)