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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.
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%I #8 Feb 01 2023 10:23:38

%S 4,16,44,121,319,841,2204,5776,15124,39601,103679,271441,710644,

%T 1860496,4870844,12752041,33385279,87403801,228826124,599074576,

%U 1568397604,4106118241,10749957119,28143753121,73681302244,192900153616,505019158604,1322157322201,3461452807999,9062201101801,23725150497404,62113250390416,162614600673844,425730551631121,1114577054219519

%N 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.

%C 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.

%C 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.

%C 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).

%H Han Wang and Zhi-Wei Sun, <a href="http://arxiv.org/abs/2206.12317">Evaluations of some Toeplitz-type determinants</a>, arXiv:2206.12317 [math.NT], 2022.

%e 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.

%t a[n_]:=a[n]=Det[Table[LucasL[j+k]+Boole[j==k],{j,1,n},{k,1,n}]];

%t Table[a[n],{n,1,25}]

%Y Cf. A000032, A000045.

%K nonn

%O 1,1

%A _Zhi-Wei Sun_, Feb 01 2023