A069191 Determinant of n X n matrix defined by m(i,j)=1 if i+j is a prime, m(i,j)=0 otherwise.

1, -1, -1, 0, 1, -1, -4, 4, 0, 0, 0, 0, 0, 0, 0, 4, 4, -9, -25, 121, 64, -576, -2304, 3600, 3136, -256, -144, 961, 24025, -47089, -345744, 1317904, 107584, -26896, -30976, 17424, 30976, -156025, -76729, 485809, 478864, -36481, -837225, 5776, 517198564, -15791440896, -16404230241, 45746793225

Offset: 1

Author

Santi Spadaro, Apr 19 2002

Keywords

• sign

Comments

Abs(a(n)) is always a perfect square.

A general result for Hankel determinants: Given any sequence a(0),a(1),... of numbers, let the n X n Hankel matrix A[i,j]=x if i=j=0, 0 if i+j even and a(((i+j)-1)/2) otherwise where 0<=i,jMichael Somos

Conjecture: Let M(n) be the n X n matrix with (i,j)-entry equal to 1 or 0 according as i + j is prime or not. For each n = 24, 25, ..., the characteristic polynomial of M(n) is irreducible over the field of rational numbers, and the n eigenvalues can be listed as lambda(1), ..., lambda(n) such that lambda(1) > -lambda(2) > lambda(3) > -lambda(4) > .... > (-1)^n*lambda(n) > 0. - Zhi-Wei Sun, Aug 25 2013

That (-1)^(n(n-1)/2)*a(n) is a perfect square is a special case of a general theorem mentioned in A228591. - Zhi-Wei Sun, Aug 25 2013

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..500
• Anon, Proof that abs(a(n)) is always a perfect square

Crossrefs

Cf. A073364.

Programs

Mathematica

f[n_] := Det[ Table[ If[ PrimeQ[i + j], 1, 0], {i, 1, n}, {j, 1, n}]]; Table[ f[n], {n, 1, 45}]