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%I #34 May 26 2017 11:21:27
%S 1,-1,-1,1,0,-1,-1,1,1,-1,0,1,1,-1,-4,16,0,-64,-64,64,0,0,0,0,0,-64,
%T -64,64,0,-16,-4,1,1,-1,0,4,16,-64,-144,324,0,-81,-9,1,4,-16,0,64,64,
%U -64,0,0,0,0,0,262144,4194304,-67108864,0,1073741824
%N Determinant of the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j is a Sophie Germain prime or not.
%C If p > 3 and 2*p+1 are both prime, then p == -1 (mod 6). If tau is a permutation of {1,...,n}, and i + tau(i) is a Sophie Germain prime for each i = 1,...,n, then n*(n+1) = sum_{i=1}^n (i + tau(i)) is congruent to -n or 2 - (n - 1) or 3 - (n - 1) or 3 + 3 - (n - 2) modulo 6, which is impossible when n == -1 (mod 6). Therefore a(6*n-1) = 0 for all n > 0.
%C Note also that (-1)^{n*(n-1)/2}*a(n) is always a square in view of the comments in A228591.
%C Conjecture: a(n) is nonzero if n is greater than 55 and not congruent to 5 modulo 6.
%C Zhi-Wei Sun also had some similar conjectures. For example, if b(n) denotes the n X n determinant with (i,j)-entry equal to 1 or 0 according as i + j and 2*(i + j) - 1 are both prime or not, then b(n) is nonzero when n is greater than 125 and not congruent to 3 modulo 6. (Just like a(6*n-1) = 0, we can show that b(6*n-3) = 0.)
%H Zhi-Wei Sun, <a href="/A228559/b228559.txt">Table of n, a(n) for n = 1..300</a>
%e a(1) = 1 since 1 + 1 is a Sophie Germain prime.
%t a[n_]:=a[n]=Det[Table[If[PrimeQ[i+j]==True&&PrimeQ[2(i+j)+1]==True,1,0],{i,1,n},{j,1,n}]]
%t Table[a[n],{n,1,20}]
%t Det/@Table[If[AllTrue[{i+j,2(i+j)+1},PrimeQ],1,0],{n,60},{i,n},{j,n}] (* The program uses the AllTrue function from Mathematica version 10 *) (* _Harvey P. Dale_, May 26 2017 *)
%Y Cf. A005384, A228591, A069191, A071524, A228552, A228557, A228561, A228574, A228578, A228615, A228616, A228548, A228549.
%K sign
%O 1,15
%A _Zhi-Wei Sun_, Aug 25 2013
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