OFFSET
1,6
COMMENTS
For the (2*n-1) X (2*n-1) determinant with (i,j)-entry equal to 1 or 0 according as i + j is a prime congruent to 1 mod 4 or not, it is easy to see that it vanishes since sum_{i=1}^{2*n-1} (i + tau(i) - 1) is not a multiple of 4 for any permutation tau of {1,...,2n-1}.
Conjecture: a(2*n-1) = 0 for all n > 0, and a(2*n) is nonzero when n > 9.
Zhi-Wei Sun could prove the following related result:
Let m be any positive even integer, and let D(m, n) denote the n X n determinant with (i,j)-entry equal to 1 or 0 according as i + j is a prime congruent to 1 mod m or not. Then (-1)^{n*(n-1)/2}*D(m,n) is always an m-th power. (It is easy to see that D(m,n) = 0 if m does not divide n^2.)
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..150
MATHEMATICA
a[n_]:=a[n]=Det[Table[If[Mod[i+j, 4]==1&&PrimeQ[i+j]==True, 1, 0], {i, 1, 2n}, {j, 1, 2n}]]
Table[a[n], {n, 1, 20}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 25 2013
STATUS
approved