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%I #126 Feb 28 2023 09:42:13
%S 0,4,36,144,400,900,1764,3136,5184,8100,12100,17424,24336,33124,44100,
%T 57600,73984,93636,116964,144400,176400,213444,256036,304704,360000,
%U 422500,492804,571536,659344,756900,864900,984064,1115136,1258884
%N Number of ways to place a non-attacking white and black rook on n X n chessboard.
%C a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n} such that for fixed different x_1, x_2 in {1,2,3,4} and fixed y_1, y_2 in {1,2,...,n} we have f(x_1) <> y_1 and f(x_2) <> y_2. - _Milan Janjic_, Apr 17 2007
%C The third differences of certain values of the hypergeometric function 3F2 lead to this sequence, i.e., 3F2([1,n+1,n+1], [n+2,n+2], z=1) - 3*3F2([1,n+2,n+2], [n+3,n+3], z=1) + 3*3F2([1,n+3,n+3], [n+4,n+4], z=1) - 3F2([1,n+4,n+4], [n+5,n+5], z=1) = (1/((n+2)*(n+3)))^2 with n = -1, 0, 1, 2, ... See also A162990. - _Johannes W. Meijer_, Jul 21 2009
%C a(n) is the denominator (m*n)^2 of the term (1/m^2 - 1/n^2) = (2*n-1)/(m*n)^2, n = m+1, m > 0 in the Rydberg formula, while A005408 is the numerator 2n-1. So the quotient A005408/A035287 simulates the hydrogen spectral series of all hydrogen-like elements. - _Freimut Marschner_, Aug 10 2013
%H Vincenzo Librandi, <a href="/A035287/b035287.txt">Table of n, a(n) for n = 1..1000</a>
%H Milan Janjic, <a href="https://pmf.unibl.org/wp-content/uploads/2017/10/enumfun.pdf">Enumerative Formulas for Some Functions on Finite Sets</a>.
%H Leo Tavares, <a href="/A035287/a035287.jpg">Illustration: Square of squares</a>
%H Wolfram Research, <a href="http://functions.wolfram.com/webMathematica/FunctionEvaluation.jsp?name=Hypergeometric3F2">Hypergeometric Function 3F2</a>, The Wolfram Functions site. [From _Johannes W. Meijer_, Jul 21 2009]
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = n^2 * (n-1)^2.
%F a(n) = A002378(n-1)^2. - _Zerinvary Lajos_, Apr 11 2006
%F From _Stephen Crowley_, Jul 19 2009: (Start)
%F a(n) = n!*(2*n+1) / lim_{x->0} (d^n/dx^n) (polylog(2,x)*(1-1/x));
%F Sum_{n >= 2} 1/a(n) = 2*zeta(2) - 3 = A145426. (End) [Comment from _Jianing Song_, Dec 31 2022: Note that polylog(2,x)*(1-1/x) = -1 + Sum_{n>=1} ((2*n+1)/(n^2*(n+1)^2))*x^n, so (d^n/dx^n) (polylog(2,x)*(1-1/x)) = n!*(2*n+1)/(n^2*(n+1)^2) for n >= 1. - _Jianing Song_, Dec 31 2022]
%F a(n) = 4*A000537(n-1) = 2*A163102(n-1). - _Omar E. Pol_, Nov 29 2011
%F G.f.: 4*x^2*(1+4*x+x^2)/(1-x)^5. - _Colin Barker_, Apr 04 2012
%F a(n) = 4*A000217(n-1)^2. - _J. M. Bergot_, Nov 01 2012
%F E.g.f.: x^2*(2 + 4*x + x^2)*exp(x). - _Ilya Gutkovskiy_, May 24 2016
%F Sum_{n>=2} (-1)^n/a(n) = 3 - 4*log(2). - _Amiram Eldar_, Jul 02 2020
%F Product_{n>=2} (1 - 1/a(n)) = -cos(sqrt(5)*Pi/2)*cosh(sqrt(3)*Pi/2)/Pi^2. - _Amiram Eldar_, Jan 29 2021
%F (n^2)^2 + (n^2+1)^2 + ... + (n^2 + n)^2 + a(n) = (n^2 + n + 1)^2 + ... + (n^2 + 2*n)^2. - _Charlie Marion_, Jun 18 2022
%F a(n) = A000290(n-1) * A000290(n). - _Leo Tavares_, Dec 03 2022
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n >= 6. - _Jianing Song_, Dec 30 2022
%t Table[(n - 1)^2 n^2, {n, 30}] (* _Alonso del Arte_, May 20 2011 *)
%o (Sage) [n^2*(n-1)^2 for n in range(1, 35)] # _Zerinvary Lajos_, Dec 03 2009
%o (Magma) [n^2 * (n-1)^2: n in [1..40]]; // _Vincenzo Librandi_, May 21 2011
%o (Python)
%o for n in range(100):
%o print(((n+1)*n)**2) # _John H. Chakkour_, Dec 14 2019
%Y Cf. A002378.
%Y Cf. A000290.
%K nonn,easy
%O 1,2
%A _Erich Friedman_
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