A280914 Number of 2 X 2 matrices with all terms in {-n,...,0,...,n} and (sum of terms) = permanent.

1, 21, 52, 172, 268, 428, 588, 812, 1004, 1324, 1580, 1900, 2252, 2668, 2988, 3532, 3916, 4460, 5004, 5548, 6028, 6764, 7308, 8044, 8716, 9548, 10156, 11116, 11852, 12620, 13548, 14444, 15244, 16524, 17228, 18380, 19340, 20588, 21548

Offset: 0

Author

Indranil Ghosh, Jan 18 2017

Keywords

• nonn

Comments

a(n) mod 4 = 0 for n > 1.

Examples

			For n = 4, a few of the possible matrices are [-4,-3,-2,3], [-4,-3,3,-1], [-4,-2,-3,3], [-4,-2,2,0], [-3,4,-1,-1], [-3,4,3,2], [-2,-4,0,2], [-2,-4,3,-3], [-1,4,1,0], [-1,4,3,3], [0,-4,0,4], [0,-4,1,-1], [0,-3,0,3], [1,2,3,0], [1,2,3,1], [1,2,3,2], [1,2,3,3], [1,2,3,4], [1,3,2,-4], [1,3,2,-3], [2,-1,0,1],... There are 268 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] and d = M[2][2]. So, a(4) = 268.
		

Links

• David Radcliffe, Table of n, a(n) for n = 0..10000 (terms 0..168 from Indranil Ghosh).

Crossrefs

Cf. A280588, A280934, A281194.

Programs

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(0,169):
    print(f"{i} {t(i)}")