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Determinant of rank n matrix of 1..n^2 filled successively along antidiagonals.
6

%I #23 Aug 14 2021 15:36:41

%S 1,-2,-5,6,13,-12,-25,20,41,-30,-61,42,85,-56,-113,72,145,-90,-181,

%T 110,221,-132,-265,156,313,-182,-365,210,421,-240,-481,272,545,-306,

%U -613,342,685,-380,-761,420,841,-462,-925,506,1013,-552,-1105,600,1201,-650,-1301,702,1405,-756,-1513

%N Determinant of rank n matrix of 1..n^2 filled successively along antidiagonals.

%C The sorted absolute values of this sequence begin: 1, 2, 5, 6, 12, 13, 20, 25, 30, 41, 42, 56, 61, 72, 85, 90, 110, 113, 132, 145, 156, 181, 182, ...

%D On comp.soft-sys.math.mathematica (Mar 20 2002), _Mark R. Diamond_ asked for a nice way to generate matrices 'along the diagonals' of the form (see example). The coding is from _Wouter Meeussen_.

%H Stefano Spezia, <a href="/A069480/b069480.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,-3,0,-3,0,-1).

%F a(n) = n*(n+2)/4 when n == 0 (mod 4), (n^2+1)/2 when n == 1 (mod 4), -n*(n+2)/4 when n == 2 (mod 4), -(n^2+1)/2, n == 3 (mod 4).

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

%F From _Stefano Spezia_, Apr 25 2021: (Start)

%F E.g.f.: ((2 - x)*x*cos(x) + (2 - 3*x - 2*x^2)*sin(x))/4.

%F a(n) = (n*(n + 2)*cos(n*Pi/2) + 2*(n^2 + 1)*sin(n*Pi/2))/4.

%F a(n) = -3*a(n-2) - 3*a(n-4) - a(n-6) for n > 6. (End)

%e a(4) = 6 = 4*6/4 = |1 3 6 10 | 2 5 9 13 | 4 8 12 15 | 7 11 14 16|.

%t f[n_] := Table[(i + j - 1)((i + j - 1) - 1)/2 + 1 + (j - 1) - Mod[i + j - 1, n, 1]^2 Quotient[i + j - 1, n, 1], {i, n}, {j, n}]; Table[ Det[ f[n]], {n, 1, 50}]

%t LinearRecurrence[{0,-3,0,-3,0,-1},{1,-2,-5,6,13,-12},60] (* _Harvey P. Dale_, Aug 14 2021 *)

%K easy,sign

%O 1,2

%A _Robert G. Wilson v_ and _Marc LeBrun_, Mar 26 2002