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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 (list; graph; refs; listen; history; text; internal format)
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).
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.
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
Sequence in context: A118440 A247119 A296439 * A358294 A298616 A366828
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Aug 25 2013
STATUS
approved

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Last modified March 2 10:43 EST 2024. Contains 370466 sequences. (Running on oeis4.)