A047929 -id:A047929 - cp's OEIS frontend

A045991 a(n) = n^3 - n^2.

0, 0, 4, 18, 48, 100, 180, 294, 448, 648, 900, 1210, 1584, 2028, 2548, 3150, 3840, 4624, 5508, 6498, 7600, 8820, 10164, 11638, 13248, 15000, 16900, 18954, 21168, 23548, 26100, 28830, 31744, 34848, 38148, 41650, 45360, 49284, 53428, 57798, 62400, 67240, 72324

Offset: 0

N. J. A. Sloane

Number of edges in the line graph of the complete bipartite graph of order 2n, L(K_n,n). - Roberto E. Martinez II, Jan 07 2002
Number of edges of the Cartesian product of two complete graphs K_n X K_n. - Roberto E. Martinez II, Jan 07 2002
That is, number of edges in the n X n rook graph. - Eric W. Weisstein, Jun 20 2017
n such that x^3 + x^2 + n factors over the integers. - James R. Buddenhagen, Apr 19 2005
Also the number of triangles in a 2 X n grid of points and therefore also (n choose 2) * (n choose 1) * 2, or (2n choose 3) - 2*(n choose 3). - Joshua Zucker, Jan 11 2006
Nonnegative X values of solutions to the equation (X-Y)^3-XY=0. To find Y values: b(n)=(n+1)*n^2 (see A011379). I proved that, if(X,Y) is different from (0,0) and m=2, 4, 6, 8, 10, 12,..., then the equation (X-Y)^m-XY=0,... has no solution. - Mohamed Bouhamida, May 10 2006
For n>=1, a(n) is equal to the number of functions f:{1,2,3}->{1,2,...,n} such that for a fixed x in {1,2,3} and a fixed y in {1,2,...,n} we have f(x)<>y. - Aleksandar M. Janjic and Milan Janjic, Mar 13 2007
a(n) equals the coefficient of log(2) in 2F1(n-1,n-1,n+1,-1). - John M. Campbell, Jul 16 2011
Define the infinite square array m(n,k) = (n-k)^2 for 1<=k<=n below the diagonal and m(n,k) = (k+n)(k-n) for 1<=n<=k above the diagonal. Then a(n) = Sum_{k=1..n} m(n,k) + Sum_{r=1..n} m(r,n), the "hook sum" of the terms left from m(n,n) and above m(n,n). - J. M. Bergot, Aug 16 2013
Partial sums of A049451. - Bruno Berselli, Feb 10 2014
Volume of an extruded rectangular brick with side lengths n, n and n-1. - Luciano Ancora, Jun 24 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
R. J. Mathar, On the Diophantine equation (X-Y)^m-XY=0.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
J.S. Seneschal, Oblong cuboid illustration.
Eric Weisstein's World of Mathematics, Edge Count.
Eric Weisstein's World of Mathematics, Rook Graph.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A011379, A047929, A114364 (essentially the same).
Cf. A000217, A000290, A000578, A006002, A049451, A146485.
Cf. A002378, A001105, A011379.
Cf. A028724, A052289.

[n^3-n^2: n in [0..40]]; // Vincenzo Librandi, May 02 2011

A045991:=n->n^3 - n^2: seq(A045991(n), n=0..50); # Wesley Ivan Hurt, Mar 30 2014

Table[n^3 - n^2, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Dec 22 2008 *)
Table[4 Binomial[n, 2] + 6 Binomial[n, 3], {n, 0, 50}] (* Robert G. Wilson v, Mar 25 2012 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 4, 18, 48}, 20] (* Eric W. Weisstein, Jun 20 2017 *)

a(n)=n^2*(n-1) \\ Charles R Greathouse IV, Jul 17 2011

[n^2*(n-1) for n in range(0, 40)] # Zerinvary Lajos, Dec 03 2009

G.f.: 2*x^2*(x+2)/(-1+x)^4 = 6/(-1+x)^4+10/(-1+x)^2+14/(-1+x)^3+2/(-1+x). - R. J. Mathar, Nov 19 2007
a(n) = floor(n^5/(n^2+n+1)). - Gary Detlefs, Feb 10 2010
a(n) = 4*binomial(n,2) + 6*binomial(n,3). - Gary Detlefs, Mar 25 2012
a(n+1) = 2*A006002(n). - R. J. Mathar, Oct 31 2012
a(n) = (A000217(n-1)+A000217(n))*(A000217(n-1)-A000217(n-2)). - J. M. Bergot, Oct 31 2012
From Wesley Ivan Hurt, May 19 2015: (Start)
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4).
a(n) = Sum_{k=0..n-1} Sum_{i=n-k-1..n+k-1} i. (End)
Sum_{n>=2} 1/a(n) = 2 - Pi^2/6. - Daniel Suteu, Feb 06 2017
Sum_{n>=2} (-1)^n/a(n) = Pi^2/12 + 2*log(2) - 2. - Amiram Eldar, Jul 05 2020
E.g.f.: exp(x)*x^2*(2 + x). - Stefano Spezia, May 20 2021
Product_{n>=2} (1 - 1/a(n)) = A146485. - Amiram Eldar, Nov 22 2022
From J.S. Seneschal, Jun 21 2024: (Start)
a(n) = (n-1)*A000290(n).
a(n) = n*A002378(n-1).
a(n) = A011379(n) - A001105(n). (End)

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

Original entry on oeis.org

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

N. J. A. Sloane

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.
              
           ||_|_
         ||_|||_
       ||_|||_||
     ||_|||_|||_|_
   ||_|||_|||_|||_
  |||_|||_|||_|||_|
(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

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).

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

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

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

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 *)

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

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

A214943 T(n,k) = Number of squarefree words of length n in a (k+1)-ary alphabet.

Original entry on oeis.org

2, 3, 2, 4, 6, 2, 5, 12, 12, 0, 6, 20, 36, 18, 0, 7, 30, 80, 96, 30, 0, 8, 42, 150, 300, 264, 42, 0, 9, 56, 252, 720, 1140, 696, 60, 0, 10, 72, 392, 1470, 3480, 4260, 1848, 78, 0, 11, 90, 576, 2688, 8610, 16680, 15960, 4848, 108, 0, 12, 110, 810, 4536, 18480, 50190, 80040

Offset: 1

R. H. Hardin, Jul 30 2012

Table starts
.2..3...4....5.....6.....7......8......9.....10......11......12......13......14
.2..6..12...20....30....42.....56.....72.....90.....110.....132.....156.....182
.2.12..36...80...150...252....392....576....810....1100....1452....1872....2366
.0.18..96..300...720..1470...2688...4536...7200...10890...15840...22308...30576
.0.30.264.1140..3480..8610..18480..35784..64080..107910..172920..265980..395304
.0.42.696.4260.16680.50190.126672.281736.569520.1068210.1886280.3169452.5108376
Empirical: row n is a polynomial of degree n
Coefficients for rows 1-12, highest power first:
...1...1
...1...1...0
...1...1...0...0
...1...1..-1..-1...0
...1...1..-2..-1...1...0
...1...1..-3..-2...2...1...0
...1...1..-4..-3...5...2..-2...0
...1...1..-5..-4...8...4..-4..-1...0
...1...1..-6..-5..12...8..-9..-4...2...0
...1...1..-7..-6..17..12.-17..-7...6...0...0
...1...1..-8..-7..23..17.-28.-13..10...2...2...0
...1...1..-9..-8..30..23.-45.-23..25...3..-2...4...0
Terms in column k are multiples of k+1 due to symmetry. - Michael S. Branicky, May 20 2021

			Some solutions for n=6 k=4
..0....1....1....0....4....4....4....0....2....2....1....2....1....4....1....1
..2....0....4....4....3....0....0....4....1....3....4....0....0....2....0....3
..1....4....2....1....2....3....2....1....0....4....3....2....2....1....2....1
..4....3....4....2....3....1....4....2....4....1....2....4....4....3....4....4
..1....0....3....0....0....4....2....3....2....0....1....3....0....4....2....3
..0....2....1....3....1....0....3....1....4....4....0....0....1....3....0....1

R. H. Hardin, Table of n, a(n) for n = 1..245
A. M. Shur, Growth of Power-Free Languages over Large Alphabets, CSR 2010, LNCS vol. 6072, 350-361.
A. M. Shur, Numerical values of the growth rates of power-free languages, arXiv:1009.4415 [cs.FL], 2010.

Cf. A006156 (column 2), A051041 (column 3), A214939 (column 4).

Cf. A002378 (row 2), A011379 (row 3), A047929(n+1) (row 4).

from itertools import product
def T(n, k):
  if n == 1: return k+1
  symbols = "".join(chr(48+i) for i in range(k+1))
  squares = ["".join(u)*2 for r in range(1, n//2 + 1)
    for u in product(symbols, repeat = r)]
  words = ("0" + "".join(w) for w in product(symbols, repeat=n-1))
  return (k+1)*sum(all(s not in w for s in squares) for w in words)
def atodiag(maxd): # maxd antidiagonals
  return [T(n, d+1-n) for d in range(1, maxd+1) for n in range(1, d+1)]
print(atodiag(11)) # Michael S. Branicky, May 20 2021

From Arseny Shur, Apr 26 2015: (Start)
Let L_k be the limit lim T(n,k)^{1/n}, which exists because T(n,k) is a submultiplicative sequence for any k. Then L_k=k-1/k-1/k^3-O(1/k^5) (Shur, 2010).
Exact values of L_k for small k, rounded up to several decimal places:
L_2=1.30176..., L_3=2.6215080..., L_4=3.7325386... (for L_5,...,L_14 see Shur arXiv:1009.4415).
Empirical observation: for k=2 the O-term in the general formula is slightly bigger than 2/k^5, and for k=3,...,14 this O-term is slightly smaller than 2/k^5.
(End)

A054026 a(n) is the number of sets of natural numbers [a,b,c,d,e] that can be produced with the numbers [0..n] such that the values of all the distinct parenthesized expressions of a-b-c-d-e are different.

Original entry on oeis.org

0, 0, 0, 0, 300, 1296, 4116, 9984, 21384, 40800, 72600, 120960, 192660, 294000, 434700, 623616, 873936, 1197504, 1611504, 2131200, 2778300, 3571920, 4538820, 5702400, 7095000, 8744736, 10690056, 12964224, 15612324, 18673200, 22199100, 26234880, 30840480, 36067200, 41983200, 48646656, 56134476

Offset: 0

Asher Auel, Jan 27 2000

There are 14 ways to put parentheses in the expression a - b - c - d - e: ((a - (b - c)) - d) - e, (((a - b) - c) - d) - e, ((a - b) - (c - d)) - e, etc. This sequence describes how many sets of natural numbers [a,b,c,d,e] can be produced with the numbers {0,1,2,3,...,n} such that the values of all the distinct expressions are different.

It can be shown that in the set of expressions obtained this way, for any number of variables, a is always positive, b is always negative, and the other variables appear with every possible combination of signs. Therefore, the valid k-tuples of numbers in [0..n] are precisely those such that every subset of {c,d,e,...}, including the empty subset, has a distinct sum. For 5 variables, there are n*(n-1)*(n-2) ways to choose distinct, nonzero values for c, d, and e. For each k, there are floor((n-1)/2) ways to choose distinct numbers x and y in [0..n] such that x + y = k. Summing over all k in [0..n], allowing arbitrary permutations of {x,y,k}, and allowing a and b to be any value gives the formula below. - Charlie Neder, Jan 13 2019

			For example, no such sets can be produced with only 0's, only 0's and 1's, only 0's and 1's and 2's, only 1's and 2's and 3's; with {0,1,2,3,4}, 300 such sets can be produced.

Charlie Neder, Table of n, a(n) for n = 0..1000
Index entries for sequences related to parenthesizing
Index entries for linear recurrences with constant coefficients, signature (3,0,-8,6,6,-8,0,3,-1).

Cf. A045991 (similar for a - b - c), A047929 (similar for a - b - c - d).

LinearRecurrence[{3,0,-8,6,6,-8,0,3,-1},{0,0,0,0,300,1296,4116,9984,21384},40] (* Harvey P. Dale, May 25 2025 *)

a(n) = (1+n)^2*(3*(-1)^n+4*n^3-18*n^2+20*n-3)/4; \\ Jinyuan Wang, Jun 27 2020

a(n) = (n+1)^2 * (n*(n-1)*(n-2) - 6*A002620(n-1)). - Charlie Neder, Jan 13 2019

a(9)-a(36) from Charlie Neder, Jan 13 2019

Incorrect formula removed by Jinyuan Wang, Jun 27 2020

cp's OEIS Frontend

A045991 a(n) = n^3 - n^2.

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

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A214943 T(n,k) = Number of squarefree words of length n in a (k+1)-ary alphabet.

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A054026 a(n) is the number of sets of natural numbers [a,b,c,d,e] that can be produced with the numbers [0..n] such that the values of all the distinct parenthesized expressions of a-b-c-d-e are different.

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