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A196810 Number of ways to place 2 nonattacking nightriders on an n X n cylindrical board. 3

%I #29 Oct 12 2020 20:02:01

%S 0,4,18,80,200,420,756,1472,2358,3860,5500,8304,11232,15484,21090,

%T 27392,34816,44604,55404,69840,84294,102124,122452,147264,173800,

%U 203476,237762,276752,318304,368340,418500,478208,541398,611524,689780,774576,863136,963148

%N Number of ways to place 2 nonattacking nightriders on an n X n cylindrical board.

%C A nightrider is a fairy chess piece that can move (proportionate to how a knight moves) in any direction.

%H Vincenzo Librandi, <a href="/A196810/b196810.txt">Table of n, a(n) for n = 1..1000</a>

%H S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, <a href="http://arxiv.org/abs/1303.1879">A q-queens problem I. General theory</a>, arXiv:1303.1879 [math.CO], 2013-2014. See <a href="http://people.math.binghamton.edu/zaslav/Tpapers/qqs1.pdf">also</a>. [_N. J. A. Sloane_, Feb 16 2013]

%H V. Kotesovec, <a href="https://oeis.org/wiki/User:Vaclav_Kotesovec">Non-attacking chess pieces</a>

%F G.f.: -(2*x^2*(2 + 17*x + 96*x^2 + 384*x^3 + 1203*x^4 + 3100*x^5 + 6917*x^6 + 13670*x^7 + 24466*x^8 + 39974*x^9 + 60206*x^10 + 83709*x^11 + 107667*x^12 + 128088*x^13 + 141070*x^14 + 143882*x^15 + 136037*x^16 + 119239*x^17 + 96892*x^18 + 72808*x^19 + 50428*x^20 + 31926*x^21 + 18321*x^22 + 9388*x^23 + 4223*x^24 + 1622*x^25 + 514*x^26 + 127*x^27 + 22*x^28 + 2*x^29))/((-1+x)^5*(1+x)^3*(1+x^2)^3*(1+x+x^2)^3*(1+x+x^2+x^3+x^4)^3).

%F Recurrence: a(n) = a(n-32) + 4*a(n-31) + 10*a(n-30) + 17*a(n-29) + 20*a(n-28) + 11*a(n-27) - 15*a(n-26) - 54*a(n-25) - 90*a(n-24) - 99*a(n-23) - 63*a(n-22) + 18*a(n-21) + 116*a(n-20) + 188*a(n-19) + 194*a(n-18) + 123*a(n-17) - 123*a(n-15) - 194*a(n-14) - 188*a(n-13) - 116*a(n-12) - 18*a(n-11) + 63*a(n-10) + 99*a(n-9) + 90*a(n-8) + 54*a(n-7) + 15*a(n-6) - 11*a(n-5) - 20*a(n-4) - 17*a(n-3) - 10*a(n-2) - 4*a(n-1).

%F Explicit formula: a(n) = -n/4+(572*n^2)/225-(3*n^3)/2+n^4/2+(-1)^n*(n/4+n^2/2)+1/2*n^2*cos((n*Pi)/2)+16/25*n^2*cos((4*n*Pi)/5)+4/9*n^2*cos((4*n*Pi)/3)+16/25*n^2*cos((8*n*Pi)/5).

%F Chaiken et al. give a 4th degree quasi-polynomial formula. - _N. J. A. Sloane_, Feb 16 2013

%F Note that cited formula is for normal chessboard (not cylindrical), see sequence A172141. - _Vaclav Kotesovec_, Dec 09 2013

%t Table[(143*n^2)/30-(79*n^3)/15+n^4/2+16/5*n^2*Floor[n/5]+n^2*Floor[n/4]+4/3*n^2*Floor[n/3]+(n+2*n^2)*Floor[n/2]+8/5*n^2*Floor[(1+n)/5]+n^2*Floor[(1+n)/4]+2/3*n^2*Floor[(1+n)/3]+8/5*n^2*Floor[(2+n)/5]+8/5*n^2*Floor[(3+n)/5],{n,1,100}]

%Y Cf. A172141, A196812.

%K nonn,easy

%O 1,2

%A _Vaclav Kotesovec_, Oct 06 2011

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