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A228559
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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.
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8
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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, -64, 64, 0, -16, -4, 1, 1, -1, 0, 4, 16, -64, -144, 324, 0, -81, -9, 1, 4, -16, 0, 64, 64, -64, 0, 0, 0, 0, 0, 262144, 4194304, -67108864, 0, 1073741824
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OFFSET
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1,15
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COMMENTS
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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.
Note also that (-1)^{n*(n-1)/2}*a(n) is always a square in view of the comments in A228591.
Conjecture: a(n) is nonzero if n is greater than 55 and not congruent to 5 modulo 6.
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.)
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LINKS
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EXAMPLE
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a(1) = 1 since 1 + 1 is a Sophie Germain prime.
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MATHEMATICA
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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}]]
Table[a[n], {n, 1, 20}]
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 *)
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CROSSREFS
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Cf. A005384, A228591, A069191, A071524, A228552, A228557, A228561, A228574, A228578, A228615, A228616, A228548, A228549.
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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