A080322 Determinant of the n X n tridiagonal matrix M with the elements on the diagonals equal to 1, except M(n,n-1)=M(n-1,n)=n.

-3, -9, -1, 24, 36, 1, -63, -81, -1, 120, 144, 1, -195, -225, -1, 288, 324, 1, -399, -441, -1, 528, 576, 1, -675, -729, -1, 840, 900, 1, -1023, -1089, -1, 1224, 1296, 1, -1443, -1521, -1, 1680, 1764, 1, -1935, -2025, -1, 2208, 2304, 1, -2499, -2601, -1, 2808, 2916, 1, -3135, -3249, -1, 3480, 3600, 1, -3843

Offset

2,1

Links

Iain Fox, Table of n, a(n) for n = 2..10000

Index entries for linear recurrences with constant coefficients, signature (3,-6,7,-6,3,-1).

Formula

a(n+2) = A010892(n+1) - (n+2)^2*A010892(n).

G.f.: -x^2*(3-8x^2+6x^3-3x^4+x^5)/(1-x+x^2)^3. [corrected by Iain Fox, Feb 26 2018]

From Iain Fox, Feb 26 2018: (Start)

a(n) = 3*a(n-1) - 6*a(n-2) + 7*a(n-3) - 6*a(n-4) + 3*a(n-5) - a(n-6).

a(3*k+1) = (-1)^k, where k is an integer >= 1. (End)

E.g.f.: - x - exp(x/2)*(3*x^2*cos(sqrt(3)*x/2) + sqrt(3)*(x^2 + 2*x - 2)*sin(sqrt(3)*x/2))/3. - Stefano Spezia, Feb 09 2025

Mathematica

CoefficientList[Series[-(3-8x^2+6x^3-3x^4+x^5)/(1-x+x^2)^3, {x, 0, 70}], x]

Prog

(PARI) a(n) = matdet(matrix(n, n, i, j, if(abs(i-j) <= 1, if((i==n || j==n) && i!=j, n, 1), 0))) \\ Iain Fox, Feb 26 2018
(PARI) first(n) = Vec(-x^2*(3 - 8*x^2 + 6*x^3 - 3*x^4 + x^5)/(1 - x + x^2)^3 + O(x^(n+2))) \\ Iain Fox, Feb 26 2018

Crossrefs

Cf. A010892.

Sequence in context: A243526 A329214 A390310 * A126179 A304249 A128727

Adjacent sequences: A080319 A080320 A080321 * A080323 A080324 A080325

Keyword

easy,sign

Author

Mario Catalani (mario.catalani(AT)unito.it), Feb 14 2003

Status

approved