A228557 Determinant of the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j and i + j + 2 are twin primes or not.

0, -1, 0, 1, 0, -1, 0, 1, 0, 0, 0, 0, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -4, 0, 64, 0, -324, 0, 81, 0, -1, 0, 16, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Zhi-Wei Sun, Aug 25 2013

Keywords

• sign

Comments

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.

Examples

			a(1) = 0 since {2, 4} is not a twin prime pair.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A001359, A006512, A228615, A069191, A071524, A228591, A228552, A228548, A228549, A228559, A228561, A228574, A228578.

Programs

Mathematica

a[n_]:=a[n]=Det[Table[If[PrimeQ[i+j]==True&&PrimeQ[i+j+2]==True,1,0],{i,1,n},{j,1,n}]]
Table[a[n],{n,1,100}]