A228549 -id:A228549 - cp's OEIS frontend

A228552 Square root of the absolute value of A069191(n).

1, 1, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 0, 2, 2, 3, 5, 11, 8, 24, 48, 60, 56, 16, 12, 31, 155, 217, 588, 1148, 328, 164, 176, 132, 176, 395, 277, 697, 692, 191, 915, 76, 22742, 125664, 128079, 213885, 7371, 171654, 89678, 114902, 149465, 353497, 144573, 388325, 198676, 1738118, 1311164, 222898

Offset: 1

Zhi-Wei Sun, Aug 25 2013

nonn

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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..400

Cf. A000040, A069191, A228616, A228615, A228548, A228549.

a[n_]:=a[n]=Sqrt[Abs[Det[Table[If[PrimeQ[i+j]==True,1,0],{i,1,n},{j,1,n}]]]]
Table[a[n],{n,1,20}]

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.

Original entry on oeis.org

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

Zhi-Wei Sun, Aug 25 2013

sign

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.

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

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

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

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}]

A228559 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.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Aug 25 2013

sign

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.)

			a(1) = 1 since 1 + 1 is a Sophie Germain prime.

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

Cf. A005384, A228591, A069191, A071524, A228552, A228557, A228561, A228574, A228578, A228615, A228616, A228548, A228549.

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 *)

A228548 Determinant of the n X n matrix with (i,j)-entry equal to A008683(i+j-1) for all i,j = 1..n.

Original entry on oeis.org

1, -2, 3, 3, -7, -5, 12, -19, -52, -52, -20, 73, -919, 6209, 2206, -1869, -8835, -4021, 23202, -122489, -174347, 1106682, 1114088, 388318, -7528057, 55753005, 81020413, -530178192, -6348221604, 101952770365, -371734984964, -16091176203501, 90823940064758, 163339092651834, -3480231557696967

Offset: 1

Zhi-Wei Sun, Aug 25 2013

sign

Conjecture: a(n) is always nonzero. Moreover, |a(n)|^(1/n) tends to infinity.

We have verified that a(n) is nonzero for all n = 1..500.

			a(1) = 1 since Moebius(1+1-1) = 1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..200

Cf. A008683, A228549, A069191, A228552.

a[n_]:=a[n]=Det[Table[MoebiusMu[i+j-1],{i,1,n},{j,1,n}]]
Table[a[n],{n,1,10}]

a(n) = matdet(matrix(n, n, i, j, moebius(i+j-1))); \\ Michel Marcus, Apr 14 2023

cp's OEIS Frontend

A228552 Square root of the absolute value of A069191(n).

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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.

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A228559 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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A228548 Determinant of the n X n matrix with (i,j)-entry equal to A008683(i+j-1) for all i,j = 1..n.

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