A360278 Determinant of the matrix [L(j+k) + d(j,k)]_{1<=j, k<=n}, where L(n) denotes the Lucas number A000032(n), and d(j,k) is 1 or 0 according as j = k or not.

4, 16, 44, 121, 319, 841, 2204, 5776, 15124, 39601, 103679, 271441, 710644, 1860496, 4870844, 12752041, 33385279, 87403801, 228826124, 599074576, 1568397604, 4106118241, 10749957119, 28143753121, 73681302244, 192900153616, 505019158604, 1322157322201, 3461452807999, 9062201101801, 23725150497404, 62113250390416, 162614600673844, 425730551631121, 1114577054219519

Offset: 1

Author

Zhi-Wei Sun, Feb 01 2023

Keywords

• nonn

Comments

Conjecture 1: Let v(0) = 2, v(1) = A, and v(n+1) = A*v(n) + v(n-1) for n > 0. Then A^2*det[v(j+k) + d(j,k)]_{1<=j, k<=n} = v(n+1)^2 - (A^2 + 4)*(n mod 2) for any positive integer n. In particular, a(n) = L(n+1)^2 - 5*(n mod 2) for all n > 0.

Conjecture 2: Let v(0) = 2, v(1) = A, and v(n+1) = A*v(n) - v(n-1) for n > 0. Then det[v(j+k) + d(j,k)]_{1<=j, k<=n} = u(n+1)^2 - n^2 for any positive integer n, where u(0) = 0, u(1) = 1, and u(n+1) = A*u(n) - u(n-1) for all n > 0.

Conjecture 3: Let F(n) denote the Fibonacci number A000045(n). Then, for any positive integer n, we have det[F(j+k) + d(j,k)]_{1<=j, k<=n} = F(n+1)^2 + (n mod 2).

Examples

			a(2) = 16 since the determinant of the 2 X 2 matrix [L(1+1)+1, L(1+2); L(2+1), L(2+2)+1] = [4, 4; 4, 8] is 16.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Han Wang and Zhi-Wei Sun, Evaluations of some Toeplitz-type determinants, arXiv:2206.12317 [math.NT], 2022.
• Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Crossrefs

Cf. A000032, A000045, A027960.

Programs

Magma

A360278:= func< n | Lucas(n+1)^2 - 5*(n mod 2) >;
[A360278(n): n in [1..40]]; // G. C. Greubel, Jun 10 2025

Mathematica

(* First program *)
a[n_]:=a[n]=Det[Table[LucasL[j+k]+Boole[j==k],{j,1,n},{k,1,n}]];
Table[a[n],{n,1,25}]
(* Second program *)
LinearRecurrence[{3, 0, -3, 1}, {4, 16, 44, 121}, 41] (* G. C. Greubel, Jun 10 2025 *)

SageMath

def A360278(n): return lucas_number2(n+1,1,-1)^2 - 5*(n%2)
print([A360278(n) for n in range(1,41)]) # G. C. Greubel, Jun 10 2025

Formula

From G. C. Greubel, Jun 10 2025: (Start)

a(n) = A000032(2*n+2) - 2 - (n mod 2) = A000032(n+1)^2 - 5*(n mod 2).

a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).

a(n) = 1 + Sum_{k=0..n-1} A027960(n, k)*A027960(n, k+1).

G.f.: (1 + x + 4*x^2 - x^3)/((1-x^2)*(1-3*x+x^2)) - 1.

E.g.f.: exp(3*x/2)*( 3*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2) ) - 2*cosh(x) - 3*sinh(x) - 1. (End)