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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.
6

%I #24 Aug 28 2013 02:59:59

%S 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,

%T 84934656,0,12745506816,0,335563778560000,0,1105346784523536,0,

%U 441194850625,0,986262467993856,0,80385880645971214336,0,6387622009837971841

%N 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.

%C 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}.

%C Conjecture: a(2*n-1) = 0 for all n > 0, and a(2*n) is nonzero when n > 9.

%C Zhi-Wei Sun could prove the following related result:

%C 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.)

%H Zhi-Wei Sun, <a href="/A228574/b228574.txt">Table of n, a(n) for n = 1..150</a>

%t a[n_]:=a[n]=Det[Table[If[Mod[i+j,4]==1&&PrimeQ[i+j]==True,1,0],{i,1,2n},{j,1,2n}]]

%t Table[a[n],{n,1,20}]

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

%K nonn

%O 1,6

%A _Zhi-Wei Sun_, Aug 25 2013