A228574 Determinant of the 2*n X 2*n matrix with (i,j)-entry equal to 1 or 0 according as i + j is a prime congruent to 1 mod 4 or not.

0, 1, 0, 1, 0, 16, 0, 1, 0, 1, 0, 6561, 0, 0, 0, 0, 0, 0, 0, 6561, 0, 456976, 0, 65536, 0, 84934656, 0, 12745506816, 0, 335563778560000, 0, 1105346784523536, 0, 441194850625, 0, 986262467993856, 0, 80385880645971214336, 0, 6387622009837971841

Offset: 1

Author

Zhi-Wei Sun, Aug 25 2013

Keywords

• nonn

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

Crossrefs

Cf. A002144, A069191, A228591, A228552, A228557, A228559, A228561, A228615, A228616.

Programs

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}]