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%I #27 Nov 28 2024 16:58:09
%S 0,-1,4,-12,-64,-208,1088,-960,-8192,-6400,58368,-27648,-344064,
%T -151552,1982464,-638976,-10485760,-3211264,54788096,-13369344,
%U -272629760,-63963136,1346371584,-264241152,-6442450944,-1224736768,30668750848
%N Determinant of the n X n matrix whose element (i,j) equals |i-j| (mod 4).
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (0,-4,0,32,0,128,0,-256,0,-1024).
%F a(4n+1) = (-1)*2^(4*n+2)*n*(3*n-2), a(4n+2) = (-1)*2^(4*n)*(12*n+1), a(4n+3) = 2^(4*n+2)*(12*n^2+4*n+1), a(4n+4) = (-1)*2^(4*n+2)*(12*n+3). 2*abs(a(n-1)) = A087982(n) for n=2, 3, 4, 5. - _Benoit Cloitre_, Nov 07 2003
%F G.f.: -x^2*(2304*x^8 + 2304*x^7 + 1280*x^6 - 704*x^5 + 224*x^4 + 48*x^3 + 16*x^2 - 4*x + 1) / ((2*x-1)^2*(2*x+1)^2*(4*x^2+1)^3). - _Colin Barker_, Sep 29 2014
%F E.g.f.: -9/4 + ((3/2)*x^2 - 15/8)*sin(2*x) + ((11/4)*x + 1/2)*cos(2*x) + ((3/4)*x + 5/8)*exp(-2*x) + (-(3/4)*x + 9/8)*exp(2*x). - _Robert Israel_, Sep 29 2014
%p seq(LinearAlgebra:-Determinant(Matrix(n,n,(i,j) -> abs(i-j) mod 4)), n=1..100); # _Robert Israel_, Sep 29 2014
%t Table[ Det[ Table[ Mod[ Abs[i - j], 4], {i, 1, n}, {j, 1, n}]], {n, 1, 30}]
%t LinearRecurrence[{0,-4,0,32,0,128,0,-256,0,-1024},{0,-1,4,-12,-64,-208,1088,-960,-8192,-6400},30] (* _Harvey P. Dale_, Nov 28 2024 *)
%o (PARI) vector(30, n, matdet(matrix(n, n, i, j, abs(i-j)%4))) \\ _Colin Barker_, Sep 29 2014
%Y Cf. A071768 (with mod 3).
%K sign,easy
%O 1,3
%A _Robert G. Wilson v_, Jun 04 2002