

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.


9



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
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OFFSET

1,42


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).
ZhiWei 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.


LINKS



EXAMPLE

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


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}]


CROSSREFS

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


KEYWORD

sign


AUTHOR



STATUS

approved



