A303260 Determinant of n X n matrix A[i,j] = (j - i - 1 mod n) + [i=j], i.e., the circulant having (n, 0, 1, ..., n-2) as first row.

1, 1, 4, 28, 273, 3421, 52288, 941578, 19505545, 456790123, 11931215316, 343871642632, 10840081272265, 371026432467913, 13702802011918048, 543154131059225686, 23000016472483168305, 1036227971225610466711, 49492629462587441963140, 2497992686980609418282548, 132849300060919364474261281

Offset: 0

Author

M. F. Hasler, Apr 23 2018

Keywords

• nonn

Comments

It is remarkable that for odd n, this determinant has its base n+1 digits equal to the middle row: e.g., a(9) = 456790123 is the determinant of the circulant matrix having [4,5,6,7,9,0,1,2,3] as middle row.

a(0) = 1 is (by convention) the determinant of a 0 X 0 matrix.

Examples

			a(5) = 3421 is the determinant of the matrix
   ( 5 0 1 2 3 )
   ( 3 5 0 1 2 )
   ( 2 3 5 0 1 )  and 3421 = 23501[6], i.e., written in base 6.
   ( 1 2 3 5 0 )
   ( 0 1 2 3 5 ).
		

Links

• Max Alekseyev, Illustration for a(9) = 456790123 = A219324(20).
• N. I. Belukhov, Solution to Problem 14.7 (in Russian), Matematicheskoe Prosveshchenie 15 (2011), pp. 241-244.
• Wikipedia, Circulant matrix.

Crossrefs

Cf. A081131(n+1) = determinant of the circulant matrix C(n) defined in formula, A070896 (signed variant).

See also A219324.

Programs

PARI

a(n)=matdet(matrix(n,n,i,j,(j-i-1)%n+(i==j)))

Python

from sympy import Matrix
def A303260(n): return Matrix(n,n, lambda i,j:(j-i-1) % n + (i==j)).det() # Chai Wah Wu, Oct 18 2021

Formula

a(n) = det(I(n) + C(n)), where I(n) is the n X n identity matrix and C(n) is the circulant having (n-1, ..., 0) as first column.