A004320 a(n) = n*(n+1)*(n+2)^2/6.

0, 3, 16, 50, 120, 245, 448, 756, 1200, 1815, 2640, 3718, 5096, 6825, 8960, 11560, 14688, 18411, 22800, 27930, 33880, 40733, 48576, 57500, 67600, 78975, 91728, 105966, 121800, 139345, 158720, 180048, 203456, 229075, 257040, 287490, 320568, 356421, 395200

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Consider the set B(n) = {1,2,3,...n}. Let a(0) = 0. Then a(n) = Sum [ b(i)^2 - b(j)^2] for all i, j = 1 to n, b(i) belongs to B(n). E.g., a(3) = (3^2-1^2) + (3^2-2^2) + (2^2-1^2) = 16. - Amarnath Murthy, Jun 01 2001

Partial sums of A016061. - J. M. Bergot, Jun 18 2013

For n >= 3, a(n-2) is the number of permutations of n symbols that 3-commute with an n-cycle (see A233440 for definition). - Luis Manuel Rivera Martínez, Feb 24 2014

a(n) is the sum of all pairs with repetitions allowed drawn from the set of triangular numbers from A000217(0) to A000217(n). This is similar to A027480 but uses triangular numbers instead of the integers. Example for n=2: 0+1, 0+3, 1+1, 1+3, 3+3 gives sum of 16 = a(2). - J. M. Bergot, Mar 23 2016

From Mircea Dan Rus, Jul 29 2020: (Start)

a(n) is the number of lattice rectangles (squares included) inside half of an Aztec diamond of order n. This shape is obtained by stacking n rows of consecutive unit lattice squares, with the centers of rows vertically aligned and consisting successively of 2n, 2n-2,..., 4, 2 squares. See below the representation for n=6.

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(End)

a(n-1) = (n+1)*binomial(n+1, 3) is the number of certain rectangles (squares included) in an n X n square filled with 1 X 1 squares. Divide the n X n square, for n >= 2, into two complementary staircases by the boundary consisting of 2*n length 1 edges. For n = 1 there is no boundary. See a A000332 figure in the Mircea Dan Rus comment for the staircase with basis length n = 4. The complementary staircase is upside down with basis length n-1 = 3. Then a(n-1) is the number of rectangles in the n X n square which have at least one border link in their interior. This counting is based on the binomial identity given in the formula section, using A096948 (for n=m), A000332(n+3) and A000332(n+2). - Wolfdieter Lang, Sep 22 2020

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Teofil Bogdan and Mircea Dan Rus, Counting the lattice rectangles inside Aztec diamonds and square biscuits, arXiv:2007.13472 [math.CO], 2020.
• Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A016061, A047929, A233440, A000332, A096948, A000217.

Programs

Magma

[n*(n+1)*(n+2)^2/6: n in [0..40] ]; // Vincenzo Librandi, Aug 19 2011

Maple

[seq ((n+2)*(binomial(n+2,3)), n=0..45)]; # Zerinvary Lajos, May 26 2006

Mathematica

Table[n (n + 1) (n + 2)^2/6, {n, 0, 40}] (* Wesley Ivan Hurt, Oct 28 2014 *)
LinearRecurrence[{5,-10,10,-5,1},{0,3,16,50,120,245},40] (* Harvey P. Dale, Nov 09 2022 *)

PARI

a(n)=n*(n+1)*(n+2)^2/6 \\ Charles R Greathouse IV, Jun 18 2013

Formula

G.f.: x*(3+x)/(1-x)^5. - Paul Barry, Feb 27 2003

a(n) = (n+2)*A000292(n). - Zerinvary Lajos, May 26 2006

a(n) = A047929(n+2)/6. - Zerinvary Lajos, May 09 2007

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Wesley Ivan Hurt, Oct 28 2014

a(n) = 3*A000332(n+3) + A000332(n+2). - Mircea Dan Rus, Jul 29 2020

Sum_{n>=1} 1/a(n) = Pi^2/2 - 9/2. - Jaume Oliver Lafont, Jul 13 2017

a(n-1) = T(n)^2 - (s(n) + s(n-1)), with T(n) = binomial(n+1, 2) = A000217(n) and s(n) = binomial(n+3, 4) = A000332(n+3), for n >= 1. See a comment above, and the formula by Mircea Dan Rus. - Wolfdieter Lang, Sep 22 2020

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/4 + 12*log(2) - 21/2. - Amiram Eldar, Jan 28 2022

E.g.f.: exp(x)*x*(18 + 30*x + 11*x^2 + x^3)/6. - Stefano Spezia, Mar 04 2023

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