|
Search: id:A069191
|
|
|
| A069191 |
|
Determinant of n X n matrix defined by m(i,j)=1 if i+j is a prime, m(i,j)=0 otherwise. |
|
+0 4
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
1,7
|
|
|
COMMENT
|
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,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
|
|
LINKS
|
Anon, Proof that abs(a(n)) is always a perfect square
|
|
MATHEMATICA
|
f[n_] := Det[ Table[ If[ PrimeQ[i + j], 1, 0], {i, 1, n}, {j, 1, n}]]; Table[ f[n], {n, 1, 45}]
|
|
CROSSREFS
|
Sequence in context: A013335 A079324 A002298 this_sequence A098445 A143266 A133845
Adjacent sequences: A069188 A069189 A069190 this_sequence A069192 A069193 A069194
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
Santi Spadaro (spados(AT)katamail.com), Apr 19 2002
|
|
|
Search completed in 0.002 seconds
|