A008911 a(n) = n^2*(n^2 - 1)/6.

0, 0, 2, 12, 40, 100, 210, 392, 672, 1080, 1650, 2420, 3432, 4732, 6370, 8400, 10880, 13872, 17442, 21660, 26600, 32340, 38962, 46552, 55200, 65000, 76050, 88452, 102312, 117740, 134850, 153760, 174592, 197472, 222530, 249900, 279720, 312132

Offset: 0

Author

N. J. A. Sloane, R. K. Guy

Keywords

• nonn
• easy

Comments

Number of equilateral triangles in rhombic portion of side n+1 in hexagonal lattice.

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.

Sum of squared distances on n X n board between n queens each on its own row and column. - Zak Seidov, Sep 04 2002

For queens "each on its column and row" the sum of squared distances does not depend on configuration - while sum of distances does.

Number of cycles of length 3 in the bishop's graph associated with an (n+1) X (n+1) chessboard. - Anton Voropaev (anton.n.voropaev(AT)gmail.com), Feb 01 2009

a(n) is number of ways to place 3 queens on an (n+1) X (n+1) chessboard so that they diagonally attack each other exactly 3 times. The maximal possible attack number, p=binomial(k,2)=3 for k=3 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form the corresponding complete graph. - Antal Pinter, Dec 27 2015

From a(1), convolution of the oblong numbers (A002378) with the odd numbers (A005408). - Bruno Berselli, Oct 24 2016

Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, p and |q-p|. - Wesley Ivan Hurt, Apr 15 2018

Examples

			a(2)=2 because on 2 X 2 board queens "each on its column and row" may take only two angular cells, then squared distance is 1^2+1^2=2. a(3)=12 because on 3 X 3 board queens "each on its column and row" make only two essentially distinct configurations: {1,2,3}, {1,3,2} and in both cases the sum of three squared distances is 12.
G.f.: 2*x^2 + 12*x^3 + 40*x^4 + 100*x^5 + 210*x^6 + 392*x^7 + 672*x^8 + ...
		

References

• James Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 6).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Gabriele Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2
• James Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics, 1999.
• James Propp, Updated article, 2009.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A002415, A006011, A047928, A083374.

Cf. A002378, A005408.

Convolution of the oblong numbers with the even numbers: A033488.

Programs

GAP

List([0..40], n-> n^2*(n^2-1)/6); # G. C. Greubel, Sep 13 2019

Magma

[n^2*(n^2-1)/6: n in [0..40]]; // Vincenzo Librandi, Sep 14 2011

Maple

A008911 := n->n^2*(n^2-1)/6; seq(A008911(n), n=0..40);

Mathematica

a[m_]:= m^2(m^2-1)/6;
Binomial[Range[0,40]^2, 2]/3 (* G. C. Greubel, Sep 13 2019 *)

PARI

{a(n) = n^2*(n^2-1)/6};

Sage

[n^2*(n^2-1)/6 for n in (0..40)] # G. C. Greubel, Sep 13 2019

Formula

G.f.: 2*x^2*(1+x)/(1-x)^5.

a(n) = 2*A002415(n) = A047928(n-1)/6 = A083374(n-1)/3 = A006011(n)*2/3. - Zerinvary Lajos, May 09 2007

a(n) = n*binomial(n+1,3). - Martin Renner, Apr 03 2011

a(n+1) = (n+1)*A000292(n). - Tom Copeland, Sep 13 2011

From G. C. Greubel, Sep 13 2019: (Start)

a(n) = binomial(n^2,2)/3.

E.g.f.: x^2*(6 + 6*x + x^2)*exp(x)/6. (End)

From Amiram Eldar, Nov 02 2021: (Start)

Sum_{n>=2} 1/a(n) = 21/2 - Pi^2.

Sum_{n>=2} (-1)^n/a(n) = (Pi^2 - 9)/2. (End)

a(n) = Sum_{j=0..n-1} binomial(n,2) + binomial(n,3). - Detlef Meya, Jan 20 2024