%I #6 Jan 10 2017 19:43:28
%S 0,0,0,4,16,20,16,36,56,60,72,76,80,100,112,100,136,124,152,172,192,
%T 196,224,196,232,236,264,252,288,276,288,308,344,332,344,332,384,388,
%U 416,404,456,428,456,444,496,468,512,468,536,556,648
%N Number of 2 X 2 matrices with entries in {-n,..,0,..,n} with no entries repeated having permanent = trace^n.
%C a(n) is also equal to the number of 2 X 2 matrices with entries in {-n,..,0,..n} with no elements repeated having determinant = trace^n except a(4). For permanent = trace^n, a(4) = 16 but for determinant = trace^n, a(4) = 24.
%C a(n) mod 4 = 0.
%H Indranil Ghosh, <a href="/A280844/b280844.txt">Table of n, a(n) for n = 0..103</a>
%e For n = 5, the possible matrices are [-3,-5,-1,2], [-3,-1,-5,2],
%e [-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],
%e [2,-1,-5,-3], [2,-1,-4,-2], [2,1,3,-1], [2,1,4,-2], [2,1,5,-3],
%e [2,3,1,-1], [2,4,1,-2] and [2,5,1,-3].
%e 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.
%o (Python)
%o def t(n):
%o s=0
%o for a in range(-n,n+1):
%o for b in range(-n,n+1):
%o if a!=b:
%o for c in range(-n,n+1):
%o if a!=c and b!=c:
%o for d in range(-n,n+1):
%o if d!=a and d!=b and d!=c:
%o if (a*d+b*c)==(a+d)**n:
%o s+=1
%o return s
%o for i in range(0,104):
%o print str(i)+" "+str(t(i))
%Y Cf. A278933, A278902, A279018.
%K nonn
%O 0,4
%A _Indranil Ghosh_, Jan 09 2017