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A004320 a(n) = n*(n+1)*(n+2)^2/6. 16

%I #94 Jan 20 2024 20:22:38

%S 0,3,16,50,120,245,448,756,1200,1815,2640,3718,5096,6825,8960,11560,

%T 14688,18411,22800,27930,33880,40733,48576,57500,67600,78975,91728,

%U 105966,121800,139345,158720,180048,203456,229075,257040,287490,320568,356421,395200

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

%C 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

%C Partial sums of A016061. - _J. M. Bergot_, Jun 18 2013

%C 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

%C 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

%C From _Mircea Dan Rus_, Jul 29 2020: (Start)

%C 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.

%C _ _

%C _|_|_|_

%C _|_|_|_|_|_

%C _|_|_|_|_|_|_|_

%C _|_|_|_|_|_|_|_|_|_

%C _|_|_|_|_|_|_|_|_|_|_|_

%C |_|_|_|_|_|_|_|_|_|_|_|_|

%C (End)

%C 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

%H Vincenzo Librandi, <a href="/A004320/b004320.txt">Table of n, a(n) for n = 0..10000</a>

%H Teofil Bogdan and Mircea Dan Rus, <a href="https://arxiv.org/abs/2007.13472">Counting the lattice rectangles inside Aztec diamonds and square biscuits</a>, arXiv:2007.13472 [math.CO], 2020.

%H Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv preprint arXiv:1406.3081 [math.CO], 2014.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: x*(3+x)/(1-x)^5. - _Paul Barry_, Feb 27 2003

%F a(n) = (n+2)*A000292(n). - _Zerinvary Lajos_, May 26 2006

%F a(n) = A047929(n+2)/6. - _Zerinvary Lajos_, May 09 2007

%F 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

%F a(n) = 3*A000332(n+3) + A000332(n+2). - _Mircea Dan Rus_, Jul 29 2020

%F Sum_{n>=1} 1/a(n) = Pi^2/2 - 9/2. - _Jaume Oliver Lafont_, Jul 13 2017

%F 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

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

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

%F a(n) = Sum_{j=0..n+1} binomial(n+1,2) + binomial(n+1,3). - _Detlef Meya_, Jan 20 2024

%p [seq ((n+2)*(binomial(n+2,3)), n=0..45)]; # _Zerinvary Lajos_, May 26 2006

%t Table[n (n + 1) (n + 2)^2/6, {n, 0, 40}] (* _Wesley Ivan Hurt_, Oct 28 2014 *)

%t LinearRecurrence[{5,-10,10,-5,1},{0,3,16,50,120,245},40] (* _Harvey P. Dale_, Nov 09 2022 *)

%o (Magma) [n*(n+1)*(n+2)^2/6: n in [0..40] ]; // _Vincenzo Librandi_, Aug 19 2011

%o (PARI) a(n)=n*(n+1)*(n+2)^2/6 \\ _Charles R Greathouse IV_, Jun 18 2013

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

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

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