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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.

%I A360278 #17 Jun 16 2025 00:42:57
%S A360278 4,16,44,121,319,841,2204,5776,15124,39601,103679,271441,710644,
%T A360278 1860496,4870844,12752041,33385279,87403801,228826124,599074576,
%U A360278 1568397604,4106118241,10749957119,28143753121,73681302244,192900153616,505019158604,1322157322201,3461452807999,9062201101801,23725150497404,62113250390416,162614600673844,425730551631121,1114577054219519
%N 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.
%C A360278 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 A360278 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 A360278 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 A360278 G. C. Greubel, <a href="/A360278/b360278.txt">Table of n, a(n) for n = 1..1000</a>
%H A360278 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.
%H A360278 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-3,1).
%F A360278 From _G. C. Greubel_, Jun 10 2025: (Start)
%F A360278 a(n) = A000032(2*n+2) - 2 - (n mod 2) = A000032(n+1)^2 - 5*(n mod 2).
%F A360278 a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).
%F A360278 a(n) = 1 + Sum_{k=0..n-1} A027960(n, k)*A027960(n, k+1).
%F A360278 G.f.: (1 + x + 4*x^2 - x^3)/((1-x^2)*(1-3*x+x^2)) - 1.
%F A360278 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)
%e A360278 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 A360278 (* First program *)
%t A360278 a[n_]:=a[n]=Det[Table[LucasL[j+k]+Boole[j==k],{j,1,n},{k,1,n}]];
%t A360278 Table[a[n],{n,1,25}]
%t A360278 (* Second program *)
%t A360278 LinearRecurrence[{3, 0, -3, 1}, {4, 16, 44, 121}, 41] (* _G. C. Greubel_, Jun 10 2025 *)
%o A360278 (Magma)
%o A360278 A360278:= func< n | Lucas(n+1)^2 - 5*(n mod 2) >;
%o A360278 [A360278(n): n in [1..40]]; // _G. C. Greubel_, Jun 10 2025
%o A360278 (SageMath)
%o A360278 def A360278(n): return lucas_number2(n+1,1,-1)^2 - 5*(n%2)
%o A360278 print([A360278(n) for n in range(1,41)]) # _G. C. Greubel_, Jun 10 2025
%Y A360278 Cf. A000032, A000045, A027960.
%K A360278 nonn
%O A360278 1,1
%A A360278 _Zhi-Wei Sun_, Feb 01 2023