|
|
A004320
|
|
a(n) = n*(n+1)*(n+2)^2/6.
|
|
13
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
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 MartÃnez, 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.
_ _
_|_|_|_
_|_|_|_|_|_
_|_|_|_|_|_|_|_
_|_|_|_|_|_|_|_|_|_
_|_|_|_|_|_|_|_|_|_|_|_
|_|_|_|_|_|_|_|_|_|_|_|_|
(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).
|
|
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
|
|
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 *)
|
|
PROG
|
(MAGMA) [n*(n+1)*(n+2)^2/6: n in [0..40] ]; // Vincenzo Librandi, Aug 19 2011
(PARI) a(n)=n*(n+1)*(n+2)^2/6 \\ Charles R Greathouse IV, Jun 18 2013
|
|
CROSSREFS
|
Cf. A016061, A047929, A233440, A000332, A096948, A000217.
Sequence in context: A172482 A212564 A222843 * A089363 A000574 A041233
Adjacent sequences: A004317 A004318 A004319 * A004321 A004322 A004323
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
N. J. A. Sloane
|
|
STATUS
|
approved
|
|
|
|