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A228591
Determinant of the n X n (0,1)-matrix with (i,j)-entry equal to 1 if and only if i + j is 2 or an odd composite number.
12
1, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, -1, -9, 81, 9, -1225, -2500, 2500, 2500, -225, -121, 841, 19044, -29584, -355216, 1527696, 141376, -40000, -40000, 10000, 59536, -258064, -139876, 935089, 885481, -16384, -1876900, 1710864, 818875456, -22896531856, -23799232900, 66328911936, 158281561, -45320023225
OFFSET
1,19
COMMENTS
Conjecture: a(n) = 0 for no n > 15.
We observe that (-1)^{n*(n-1)/2}*a(n) is always a square. This is a special case of the following general result established by Zhi-Wei Sun.
Theorem: Let M = (m_{i,j}) be an n X n symmetric matrix over a commutative ring. Suppose that the (i,j)-entry m_{i,j} is zero whenever i + j is even and greater than 2. If n is even, then (-1)^{n/2}*det(M) = D(n)^2, where D(n) denotes the determinant |m_{2i,2j-1}|_{i,j = 1,...,n/2}. If n is odd, then (-1)^{(n-1)/2}*det(M) = m_{1,1}*D(n)^2, where D(n) is the determinant |m_{2i,2j+1}|_{i,j = 1,...,(n-1)/2}.
This theorem extends the result mentioned in A069191.
LINKS
MATHEMATICA
a[n_]:=a[n]=Det[Table[If[(i+j==2)||(Mod[i+j, 2]==1&&PrimeQ[i+j]==False), 1, 0], {i, 1, n}, {j, 1, n}]]
Table[a[n], {n, 1, 50}]
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Aug 27 2013
STATUS
approved