A069191 - OEIS

%I #14 Sep 29 2017 20:20:45

%S 1,-1,-1,0,1,-1,-4,4,0,0,0,0,0,0,0,4,4,-9,-25,121,64,-576,-2304,3600,

%T 3136,-256,-144,961,24025,-47089,-345744,1317904,107584,-26896,-30976,

%U 17424,30976,-156025,-76729,485809,478864,-36481,-837225,5776,517198564,-15791440896,-16404230241,45746793225

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

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

%C 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,j<n and the [n/2]X[n/2] Hankel matrix B[i,j]=a(i+j) if n even, a(i+j+1) if n odd. Then det(A) = (-1)^[n/2] det(B)^2 f where f=1 if n even, x if n odd. The proof uses Jacobi's determinant identity. - _Michael Somos_

%C 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

%C 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

%H Alois P. Heinz, <a href="/A069191/b069191.txt">Table of n, a(n) for n = 1..500</a>

%H Anon, <a href="/A069191/a069191.txt">Proof that abs(a(n)) is always a perfect square</a>

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

%Y Cf. A073364.

%K sign

%O 1,7

%A _Santi Spadaro_, Apr 19 2002