A281194 Number of 2 X 2 matrices with all terms in {-n,..,0,..,n} and (sum of terms) = determinant.

1, 31, 111, 271, 479, 831, 1167, 1711, 2239, 2975, 3631, 4687, 5407, 6655, 7759, 9135, 10367, 12127, 13231, 15375, 16991, 19135, 20879, 23471, 25215, 27999, 30319, 33167, 35359, 39167, 41039, 44975, 47615, 50975, 54511, 58767, 61791, 66239, 69391

Offset: 0

Author

Indranil Ghosh, Jan 17 2017

Keywords

• nonn

Examples

			For n = 3, few of the possible matrices are [-3,-3,-3,0], [-3,-3,-1,1], [-3,-3,1,2], [-3,-3,3,3], [-3,-2,-1,1], [-3,-2,3,2], [-3,-1,-3,1], [-3,-1,-2,1], [-3,-1,-1,1], [-3,-1,0,1], [-3,-1,1,1], [-3,-1,2,1], [-3,-1,3,1], [-3,0,-1,1], [2,0,0,2], [2,0,1,3], [2,1,-3,-3], [2,1,-2,-1], [2,1,-1,1], [3,3,0,3],...There are 271 possibilities.
Here each of the 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]. So, a(3) = 271.
		

Links

• David Radcliffe, Table of n, a(n) for n = 0..10000 (terms 0..186 from Indranil Ghosh).

Crossrefs

Cf. A280588, A280934, A209981.

Programs

PARI

a(n)=sum(a=-n,n, sum(d=-n,n, my(t=a*d+a+d); sum(b=-n,n, if(b==-1, if(t==-1, 2*n+1, 0), my(c=(t-b)/(b+1)); denominator(c)==1 && c<=n && c>=-n)))) \\ Charles R Greathouse IV, Jan 17 2017

Python

def t(n):
    s=0
    for a in range(-n, n+1):
        for b in range(-n, n+1):
            for c in range(-n, n+1):
                for d in range(-n, n+1):
                    if (a+b+c+d)==(a*d-b*c):
                        s+=1
    return s
for i in range(0, 187):
    print(f"{i} {t(i)}")