A280844 Number of 2 X 2 matrices with entries in {-n,..,0,..,n} with no entries repeated having permanent = trace^n.
0, 0, 0, 4, 16, 20, 16, 36, 56, 60, 72, 76, 80, 100, 112, 100, 136, 124, 152, 172, 192, 196, 224, 196, 232, 236, 264, 252, 288, 276, 288, 308, 344, 332, 344, 332, 384, 388, 416, 404, 456, 428, 456, 444, 496, 468, 512, 468, 536, 556, 648
Offset: 0
Keywords
Examples
For n = 5, the possible matrices are [-3,-5,-1,2], [-3,-1,-5,2], [-3,1,5,2], [-3,5,1,2], [-2,-4,-1,2], [-2,-1,-4,2], [-2,1,4,2], [-2,4,1,2], [-1,1,3,2], [-1,3,1,2], [2,-5,-1,-3], [2,-4,-1,-2], [2,-1,-5,-3], [2,-1,-4,-2], [2,1,3,-1], [2,1,4,-2], [2,1,5,-3], [2,3,1,-1], [2,4,1,-2] and [2,5,1,-3]. Here each of the matrices 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 20 possibilities. So, for n = 5, a(n) = 20.
Links
- David Radcliffe, Table of n, a(n) for n = 0..10000 (terms 0..103 from Indranil Ghosh).
Programs
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Python
def t(n): s=0 for a in range(-n,n+1): for b in range(-n,n+1): if a!=b: for c in range(-n,n+1): if a!=c and b!=c: for d in range(-n,n+1): if d!=a and d!=b and d!=c: if (a*d+b*c)==(a+d)**n: s+=1 return s for i in range(0,24): print(t(i), end=', ')
Comments