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A163102 a(n) = n^2*(n+1)^2/2.

%I A163102 #108 Aug 20 2025 00:27:05
%S A163102 0,2,18,72,200,450,882,1568,2592,4050,6050,8712,12168,16562,22050,
%T A163102 28800,36992,46818,58482,72200,88200,106722,128018,152352,180000,
%U A163102 211250,246402,285768,329672,378450,432450,492032,557568,629442,708050,793800,887112,988418
%N A163102 a(n) = n^2*(n+1)^2/2.
%C A163102 Row sums of triangle A163282.
%C A163102 Also, the number of nonattacking placements of 2 rooks on an (n+1) X (n+1) board. - _Thomas Zaslavsky_, Jun 26 2010
%C A163102 If P_{k}(n) is the n-th k-gonal number, then a(n) = P_{s}(n+1)*P_{t}(n+1) - P_{s+1}(n+1)*P_{t-1}(n+1) for s=t+1. - _Bruno Berselli_, Sep 05 2014
%C A163102 Subsequence of A000982, see formula. - _David James Sycamore_, Jul 31 2018
%C A163102 Number of edges in the (n+1) X (n+1) rook complement graph. - _Freddy Barrera_, May 02 2019
%C A163102 Number of paths from (0,0) to (n+2,n+2) consisting of exactly three forward horizontal steps and three upward vertical steps. - _Greg Dresden_ and Snezhana Tuneska, Aug 24 2023
%D A163102 Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-queens problem, in preparation. - _Thomas Zaslavsky_, Jun 26 2010
%H A163102 Vincenzo Librandi, <a href="/A163102/b163102.txt">Table of n, a(n) for n = 0..1000</a>
%H A163102 Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, <a href="http://arxiv.org/abs/1609.00853">A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks)</a>, arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
%H A163102 A. Umar, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Umar/umar2.html">Combinatorial Results for Semigroups of Orientation-Preserving Partial Transformations</a>, J. Int. Seq., Vol. 14 (2011), Article 11.7.5.
%H A163102 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RookComplementGraph.html">Rook Complement Graph</a>.
%H A163102 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A163102 a(n) = 2*A000537(n) = A035287(n+1)/2. - _Omar E. Pol_, Nov 29 2011
%F A163102 G.f.: 2*x*(1+4*x+x^2)/(1-x)^5. - _R. J. Mathar_, Nov 30 2011
%F A163102 Let t(n) = A000217(n). Then a(n) = (t(n-1)*(t(n)+t(n+1)) + t(n)*(t(n-1)+t(n+1)) + t(n+1)*(t(n-1)+t(n)))/3. - _J. M. Bergot_, Jun 21 2012
%F A163102 a(n) = A000982(n*(n+1)). - _David James Sycamore_, Jul 31 2018
%F A163102 From _Amiram Eldar_, Nov 02 2021: (Start)
%F A163102 Sum_{n>=1} 1/a(n) = 2*Pi^2/3 - 6.
%F A163102 Sum_{n>=1} (-1)^(n+1)/a(n) = 6 - 8*log(2). (End)
%F A163102 Another identity: ..., a(4) = 200 = 1*(2+4+6+8) + 3*(4+6+8) + 5*(6+8) + 7*(8), a(5) = 450 = 1*(2+4+6+8+10) + 3*(4+6+8+10) + 5*(6+8+10) + 7*(8+10) + 9*(10) = 30+84+120+126+90, and so on. - _J. M. Bergot_, Aug 25 2022
%F A163102 From _Elmo R. Oliveira_, Aug 14 2025: (Start)
%F A163102 E.g.f.: x*(2 + x)*(2 + 6*x + x^2)*exp(x)/2.
%F A163102 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
%F A163102 a(n) = A254371(n)/4 = A060300(n)/8. (End)
%t A163102 CoefficientList[Series[2*x*(1+4*x+x^2)/(1-x)^5,{x,0,40}],x] (* _Vincenzo Librandi_, Mar 26 2012 *)
%o A163102 (Magma) [n^2*(n+1)^2/2: n in [0..40]]; // _Vincenzo Librandi_, Mar 26 2012
%o A163102 (PARI) a(n)=n^2*(n+1)^2/2 \\ _Charles R Greathouse IV_, Oct 07 2015
%o A163102 (GAP) List([0..40],n->(n*(n+1))^2/2); # _Muniru A Asiru_, Aug 02 2018
%Y A163102 Column k=2 of A144084.
%Y A163102 Cf. A000217, A000537, A000982, A035287.
%Y A163102 Cf. A006002, A099903, A163274, A163275, A163276, A163277, A163282.
%Y A163102 Cf. A060300, A254371.
%K A163102 nonn,easy,changed
%O A163102 0,2
%A A163102 _Omar E. Pol_, Jul 24 2009