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According to the comments of A069191, a(n) should be always integral. Note that a(2*n) is the absolute value of A228616(n) by the comments of A228591. We conjecture that a(n) > 0 for all n > 15.

a(n)=permanent(m), where the n X n matrix m is defined by m(i,j) = 1 or 0, depending on whether i+j is prime or composite respectively. - T. D. Noe, Oct 16 2007

Conjecture: a(n) = 0 for no n > 15.

We observe that (-1)^{n*(n-1)/2}*a(n) is always a square. This is a special case of the following general result established by Zhi-Wei Sun.

Theorem: Let M = (m_{i,j}) be an n X n symmetric matrix over a commutative ring. Suppose that the (i,j)-entry m_{i,j} is zero whenever i + j is even and greater than 2. If n is even, then (-1)^{n/2}*det(M) = D(n)^2, where D(n) denotes the determinant |m_{2i,2j-1}|{i,j = 1,...,n/2}. If n is odd, then (-1)^{(n-1)/2}*det(M) = m{1,1}*D(n)^2, where D(n) is the determinant |m_{2i,2j+1}|_{i,j = 1,...,(n-1)/2}.

This theorem extends the result mentioned in A069191.

Conjecture: a(n) is nonzero for any n > 35.

Clearly this conjecture implies the twin prime conjecture.

Clearly p is odd if p and p + 2 are twin primes. If sigma is a permutation of {1,...,n}, and i + sigma (i) and i + sigma(i) + 2 are twin primes for all i = 1,...,n, then we must have sum_{i=1}^n (i + sigma(i)) == n (mod 2) and hence n is even. Therefore a(n) = 0 if n is odd.

By the general result mentioned in A228591, (-1)^n*a(2*n) equals the square of A228615(n).

Zhi-Wei Sun made the following general conjecture:

Let d be any positive even integer, and let D(d,n) be the n X n determinant with (i,j)-entry eual to 1 or 0 according as i + j and i + j + d are both prime or not. Then D(d,2*n) is nonzero for large n.

Note that when n is odd we have D(d,n) = 0 (just like a(n) = 0). Also, the conjecture implies de Polignac's conjecture that there are infinitely many primes p such that p and p + d are both prime.

Conjecture: a(n) is nonzero for any n > 7.

Clearly this conjecture implies that there are infinitely many primes.

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.)

Conjecture: a(n) is nonzero for each n > 28.

This implies that there are infinitely many primes p with 4*p^2 + 1 also prime. Note also that (-1)^{n*(n-1)/2}*a(n) is always a square in view of the comments in A228591.

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.)