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%I #63 Jan 14 2024 16:36:49
%S 0,3,5,10,14,21,27,36,44,55,65,78,90,105,119,136,152,171,189,210,230,
%T 253,275,300,324,351,377,406,434,465,495,528,560,595,629,666,702,741,
%U 779,820,860,903,945,990,1034,1081,1127,1176,1224,1275,1325,1378,1430
%N a(n) = (n^2 - 3*n + 1 + (-1)^n)/2.
%C Let I = I_n be the n X n identity matrix and P = P_n be the incidence matrix of the cycle (1,2,3,...,n).
%C Let T = P^(-1)+I+P.
%C 11000...01
%C 11100....0
%C 01110.....
%C 00111.....
%C ..........
%C 00.....111
%C 10.....011
%C Then a(n) is the number of (0,1) n X n matrices A <= T (i.e., an element of A can be 1 only if T has a 1 at this place) having exactly two 1's in every row and column with per(A) = 4.
%C a(n) is the maximum number m such that m white kings and m black kings can coexist on an n+1 X n+1 chessboard without attacking each other. - _Aaron Khan_, Jul 05 2022
%D V. S. Shevelyov (Shevelev), Extension of the Moser class of four-line Latin rectangles, DAN Ukrainy, 3 (1992), 15-19.
%H G. C. Greubel, <a href="/A176222/b176222.txt">Table of n, a(n) for n = 3..1000</a>
%H Paul Barry, <a href="http://arxiv.org/abs/1205.2565">On sequences with {-1, 0, 1} Hankel transforms</a>, arXiv preprint arXiv:1205.2565 [math.CO], 2012.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F a(n) = (n - t(n))*(n - 3 + t(n))/2, where t(n) = 1-(n mod 2).
%F G.f.: x^4*(3-x)/( (1+x)*(1-x)^3 ). - _R. J. Mathar_, Mar 06 2011
%F From _Bruno Berselli_, Sep 13 2011: (Start)
%F a(n) + a(n+1) = A005563(n-2).
%F a(-n) = A084265(n). (End)
%F a(n) = 1 -2*n +floor(n/2) +floor(n^2/2). - _Wesley Ivan Hurt_, Jun 14 2013
%F From _Wesley Ivan Hurt_, May 25 2015: (Start)
%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4), n>4.
%F a(n) = Sum_{i=(-1)^n..n-2} i. (End)
%F a(n) = A174239(n-2) * A174239(n-1). - _Paul Curtz_, Jul 17 2017
%F With offset 0, this is ceiling(n/2)*(2*floor(n/2)+3). - _N. J. A. Sloane_, Jan 16 2020
%F E.g.f.: (1/2)*((1-x)*exp(x/2) - exp(-x/2))^2. - _G. C. Greubel_, Mar 22 2022
%e For n=5 the reference matrix is:
%e 11001
%e 11100
%e 01110
%e 00111
%e 10011
%e There are 2^(3*n) = 32768 0-1 matrices obtained from removing one or more 1's in it.
%e There are 305 such matrices with permanent 4 and there are 13 such matrices with exactly two 1's in every column and every row.
%e There are 5 matrices having both properties. One of them is:
%e 10001
%e 01100
%e 01100
%e 00011
%e 10010
%e From _Aaron Khan_, Jul 05 2022: (Start)
%e Examples of the sequence when used for kings on a chessboard:
%e .
%e A solution illustrating a(2)=3:
%e +-------+
%e | B B B |
%e | . . . |
%e | W W W |
%e +-------+
%e .
%e A solution illustrating a(3)=5:
%e +---------+
%e | B B B B |
%e | B . . . |
%e | . . . W |
%e | W W W W |
%e +---------+
%e (End)
%p A176222:=n->(n^2-3*n+1+(-1)^n)/2: seq(A176222(n), n=3..100); # _Wesley Ivan Hurt_, May 25 2015
%t Table[(n^2 - 3*n + 1 + (-1)^n)/2, {n, 3, 100}] (* or *) CoefficientList[Series[x (x - 3)/((1 + x)*(x - 1)^3), {x, 0, 30}], x] (* _Wesley Ivan Hurt_, May 25 2015 *)
%t LinearRecurrence[{2,0,-2,1},{0,3,5,10},90] (* _Harvey P. Dale_, Jan 14 2024 *)
%o (Magma) [(n^2-3*n+1+(-1)^n)/2: n in [3..100]]; // _Vincenzo Librandi_, Mar 24 2011
%o (PARI) a(n)=(n^2-3*n+1+(-1)^n)/2 \\ _Charles R Greathouse IV_, Oct 16 2015
%o (Sage) [n*(n-3)/2 + ((n+1)%2) for n in (3..60)] # _G. C. Greubel_, Mar 22 2022
%Y Cf. A000211, A052928, A128209, A250000 (queens on a chessboard), A002620 (rooks on a chessboard), A355509 (knights on a chessboard).
%K nonn,easy
%O 3,2
%A _Vladimir Shevelev_, Apr 12 2010
%E Matrix class definition checked, edited and illustrated by _Olivier Gérard_, Mar 26 2011
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