A035287 Number of ways to place a non-attacking white and black rook on n X n chessboard.

0, 4, 36, 144, 400, 900, 1764, 3136, 5184, 8100, 12100, 17424, 24336, 33124, 44100, 57600, 73984, 93636, 116964, 144400, 176400, 213444, 256036, 304704, 360000, 422500, 492804, 571536, 659344, 756900, 864900, 984064, 1115136, 1258884

Offset: 1

Author

Erich Friedman

Keywords

• nonn
• easy

Comments

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

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

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

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
• Leo Tavares, Illustration: Square of squares
• Wolfram Research, Hypergeometric Function 3F2, The Wolfram Functions site. [From _Johannes W. Meijer_, Jul 21 2009]
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A002378.

Cf. A000290.

Programs

Magma

[n^2 * (n-1)^2: n in [1..40]]; // Vincenzo Librandi, May 21 2011

Mathematica

Table[(n - 1)^2 n^2, {n, 30}] (* Alonso del Arte, May 20 2011 *)

Python

for n in range(100):
  print(((n+1)*n)**2) # John H. Chakkour, Dec 14 2019

Sage

[n^2*(n-1)^2 for n in range(1, 35)] # Zerinvary Lajos, Dec 03 2009

Formula

a(n) = n^2 * (n-1)^2.

a(n) = A002378(n-1)^2. - Zerinvary Lajos, Apr 11 2006

From Stephen Crowley, Jul 19 2009: (Start)

a(n) = n!*(2*n+1) / lim_{x->0} (d^n/dx^n) (polylog(2,x)*(1-1/x));

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]

a(n) = 4*A000537(n-1) = 2*A163102(n-1). - Omar E. Pol, Nov 29 2011

G.f.: 4*x^2*(1+4*x+x^2)/(1-x)^5. - Colin Barker, Apr 04 2012

a(n) = 4*A000217(n-1)^2. - J. M. Bergot, Nov 01 2012

E.g.f.: x^2*(2 + 4*x + x^2)*exp(x). - Ilya Gutkovskiy, May 24 2016

Sum_{n>=2} (-1)^n/a(n) = 3 - 4*log(2). - Amiram Eldar, Jul 02 2020

Product_{n>=2} (1 - 1/a(n)) = -cos(sqrt(5)*Pi/2)*cosh(sqrt(3)*Pi/2)/Pi^2. - Amiram Eldar, Jan 29 2021

(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

a(n) = A000290(n-1) * A000290(n). - Leo Tavares, Dec 03 2022

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