A002415 -id:A002415 - cp's OEIS frontend

A014820 a(n) = (1/3)(n^2 + 2n + 3)*(n+1)^2.

1, 8, 33, 96, 225, 456, 833, 1408, 2241, 3400, 4961, 7008, 9633, 12936, 17025, 22016, 28033, 35208, 43681, 53600, 65121, 78408, 93633, 110976, 130625, 152776, 177633, 205408, 236321, 270600, 308481

Offset: 0

N. J. A. Sloane

a(n) is the number of 4 X 4 pandiagonal magic squares with sum 2n. - Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002
Figurate numbers based on the 4-dimensional regular convex polytope called the 16-cell, hexadecachoron, 4-cross polytope or 4-hyperoctahedron with Schlaefli symbol {3,3,4}. a(n)=(n^2*(n^2+2))/3 if the offset were 1. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 18 2009
If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 7-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Equals binomial transform of [1, 7, 18, 20, 8, 0, 0, 0, ...], where (1, 7, 18, 20, 8) = row 4 of the Chebyshev triangle A081277. Also = row 4 of the array in A142978. - Gary W. Adamson, Jul 19 2008

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Ahmed, J. De Loera and R. Hemmecke, Polyhedral Cones of Magic Cubes and Squares, arXiv:math/0201108 [math.CO], 2002.
Maya Ahmed, Jesús De Loera and Raymond Hemmecke, Polyhedral cones of magic cubes and squares, in Discrete and Computational Geometry, Springer, Berlin, 2003, pp. 25-41.
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Eric Weisstein's World of Mathematics, 16-Cell
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1)

Cf. A000332, A000583, A005900, A069038, A069039, A070212, A081277, A092181, A092182, A092183, A099175, A099193, A099195, A099196, A099197, A142978.

Cf. A061927, A002415, A000290, A039623.

List([0..40], n -> (n+1)^2*((n+1)^2 +2)/3); # G. C. Greubel, Feb 10 2019

[(1/3)*(n^2+2*n+3)*(n+1)^2: n in [0..40]]; // Vincenzo Librandi, May 22 2011

al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n),r=0..d-1); end; [seq(be(4,n),n=0..100)];

LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 33, 96, 225}, 31] (* Jean-François Alcover, Jan 17 2018 *)

a(n)=(n+1)^2*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Apr 17 2012

a <- c(1, 8, 33, 96,225)
for(n in (length(a)+1):30) a[n] <- 5*a[n-1]-10*a[n-2]+10*a[n-3]-5*a[n-4]+a[n-5]
a # Yosu Yurramendi, Sep 03 2013

[((n+1)^2+2)*(n+1)^2/3 for n in range(40)] # G. C. Greubel, Feb 10 2019

Or, a(n-1) = n^2*(n^2+2)/3. - Corrected by R. J. Mathar, Jul 18 2009
From Vladeta Jovovic, Apr 03 2002: (Start)
G.f.: (1+x)^3/(1-x)^5.
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
a(n-1) = C(n+3,4) + 3 C(n+2,4) + 3 C(n+1,4) + C(n,4).
Sum_{n>=0} 1/((1/3*(n^2 + 2*n + 3))*(n+1)^2) = (1/4)*Pi^2 - 3*sqrt(2)*Pi*coth(Pi*sqrt(2))*(1/8) + 3/8 = 1.1758589... - Stephen Crowley, Jul 14 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with n > 4, a(0)=1, a(1)=8, a(2)=33, a(3)=96, a(4)=225. - Yosu Yurramendi, Sep 03 2013
From Bruce J. Nicholson, Jan 23 2019: (Start)
Sum_{i=0..n} a(i) = A061927(n+1).
a(n) = 4*A002415(n+1) + A000290(n+1) = A039623(n+1) + A002415(n+1). (End)
E.g.f.: (3 + 21*x + 27*x^2 + 10*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Feb 10 2019
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 17/3 - 8*log(2) = 1/(8 + 2/(8 + 6/(8 + ... + n*(n-1)/(8 + ...)))). See A142983. - Peter Bala, Mar 06 2024

Formula index corrected by R. J. Mathar, Jul 18 2009

A034827 a(n) = 2*binomial(n,4).

Original entry on oeis.org

0, 0, 0, 0, 2, 10, 30, 70, 140, 252, 420, 660, 990, 1430, 2002, 2730, 3640, 4760, 6120, 7752, 9690, 11970, 14630, 17710, 21252, 25300, 29900, 35100, 40950, 47502, 54810, 62930, 71920, 81840, 92752, 104720, 117810, 132090, 147630, 164502, 182780

Offset: 0

N. J. A. Sloane

Also number of ways to insert two pairs of parentheses into a string of n-4 letters (allowing empty pairs of parentheses). E.g., there are 30 ways for 2 letters. Cf. A002415.
2,10,30,70, ... gives orchard crossing number of complete graph K_n. - Ralf Stephan, Mar 28 2003
If Y is a 2-subset of an n-set X then, for n>=4, a(n-1) is the number of 4-subsets and 5-subsets of X having exactly one element in common with Y. - Milan Janjic, Dec 28 2007
Middle column of table on p. 6 of Feder and Garber. - Jonathan Vos Post, Apr 23 2009
Number of pairs of non-intersecting lines when each of n points around a circle is joined to every other point by straight lines. A pair of lines is considered non-intersecting if the lines do not intersect in either the interior or the boundary of a circle. - Melvin Peralta, Feb 05 2016
From a(2), convolution of the oblong numbers (A002378) with the nonnegative numbers (A001477). - Bruno Berselli, Oct 24 2016
Also the number of 3-cycles in the n-triangular honeycomb bishop graph. - Eric W. Weisstein, Aug 10 2017

Charles Jordan, Calculus of Finite Differences, Chelsea, 1965, p. 449.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Bruno Berselli, Table of n, a(n) for n = 0..1000
M. Aganagic, A. Klemm and C. Vafa, Disk Instantons, Mirror Symmetry and the Duality Web, arXiv:hep-th/0105045, 2001.
Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
Elie Feder and David Garber, The Orchard crossing number of an abstract graph, arXiv:math/0303317 [math.CO], 2003-2009.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), pp. 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Eric Weisstein's World of Mathematics, Graph Cycle.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

A diagonal of A088617.
Cf. A033487, A050534, A060008.
Partial sums of A007290.
Cf. A001477, A002378.
Cf. A051843 (4-cycles in the triangular honeycomb bishop graph), A290775 (5-cycles), A290779 (6-cycles).

[2*Binomial(n,4): n in [0..40]]; // Vincenzo Librandi, Oct 20 2013

[seq(binomial(n,4)*2,n=0..40)]; # Zerinvary Lajos, Jul 18 2006

CoefficientList[Series[2 x^4/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Oct 20 2013 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 0, 0, 0, 2}, 50] (* Harvey P. Dale, Jun 09 2016 *)
Table[2 Binomial[n, 4], {n, 0, 40}] (* Bruno Berselli, Oct 24 2016 *)
2 Binomial[Range[0, 20], 4] (* Eric W. Weisstein, Aug 10 2017 *)

a(n)=2*binomial(n,4) \\ Charles R Greathouse IV, Jun 23 2015

a(n) = A096338(2*n-6) = 2*A000332(n), n>2. - R. J. Mathar, Nov 08 2010
G.f.: 2*x^4/(1-x)^5. - Colin Barker, Feb 29 2012
a(n) = Sum_{k=1..n-3} ( Sum_{i=1..k} i*(2*k-n+4) ). - Wesley Ivan Hurt, Sep 26 2013
E.g.f.: x^4*exp(x)/12. - G. C. Greubel, Feb 23 2017
From Amiram Eldar, Jul 19 2022: (Start)
Sum_{n>=4} 1/a(n) = 2/3.
Sum_{n>=4} (-1)^n/a(n) = 16*log(2) - 32/3. (End)

A006011 a(n) = n^2*(n^2 - 1)/4.

Original entry on oeis.org

0, 0, 3, 18, 60, 150, 315, 588, 1008, 1620, 2475, 3630, 5148, 7098, 9555, 12600, 16320, 20808, 26163, 32490, 39900, 48510, 58443, 69828, 82800, 97500, 114075, 132678, 153468, 176610, 202275, 230640, 261888, 296208, 333795, 374850, 419580, 468198

Offset: 0

N. J. A. Sloane, Simon Plouffe

Products of two consecutive triangular numbers (A000217).
a(n) is the number of Lyndon words of length 4 on an n-letter alphabet. A Lyndon word is a primitive word that is lexicographically earliest in its cyclic rotation class. For example, a(2)=3 counts 1112, 1122, 1222. - David Callan, Nov 29 2007
For n >= 2 this is the second rightmost column of A163932. - Johannes W. Meijer, Oct 16 2009
Partial sums of A059270. - J. M. Bergot, Jun 27 2013
Using the integers, triangular numbers, and squares plot the points (A001477(n),A001477(n+1)), (A000217(n), A000217(n+1)), and (A000290(n),A000290(n+1)) to create the vertices of a triangle. One-half the area of this triangle = a(n). - J. M. Bergot, Aug 01 2013
a(n) is the Wiener index of the triangular graph T(n+1). - Emeric Deutsch, Aug 26 2013

			From _Bruno Berselli_, Aug 29 2014: (Start)
After the zeros, the sequence is provided by the row sums of the triangle:
   3;
   4, 14;
   5, 16, 39;
   6, 18, 42,  84;
   7, 20, 45,  88, 155;
   8, 22, 48,  92, 160, 258;
   9, 24, 51,  96, 165, 264, 399;
  10, 26, 54, 100, 170, 270, 406, 584;
  11, 28, 57, 104, 175, 276, 413, 592, 819;
  12, 30, 60, 108, 180, 282, 420, 600, 828, 1110; etc.,
where T(r,c) = c*(c^2+r+1), with r = row index, c = column index, r >= c > 0. (End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Miguel Azaola and Francisco Santos, The number of triangulations of the cyclic polytope C(n,n-4), Discrete Comput. Geom., Vol. 27 (2002), pp. 29-48 (see Prop. 4.2(a)).
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926. (Annotated scanned copy)
Eric Weisstein's World of Mathematics, Triangular Graph.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Thrice A002415. Row 4 of A074650.
Cf. A000217, A000290, A000537, A001477, A002415, A008911, A047928, A059270, A083374, A126274, A163932, A228317
A column of A124428.

[n^2*(n^2-1)/4: n in [0..40]]; // Vincenzo Librandi, Sep 14 2011

A006011 := proc(n)
    n^2*(n^2-1)/4 ;
end proc: # R. J. Mathar, Nov 29 2015

Table[n^2 (n^2 - 1)/4, {n, 0, 38}]
Binomial[Range[20]^2, 2]/2 (* Eric W. Weisstein, Sep 08 2017 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 3, 18, 60, 150}, 20] (* Eric W. Weisstein, Sep 08 2017 *)
CoefficientList[Series[-3 x (1 + x)/(-1 + x)^5, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 08 2017 *)
Join[{0},Times@@@Partition[Accumulate[Range[0,40]],2,1]] (* Harvey P. Dale, Aug 08 2025 *)

a(n)=binomial(n^2,2)/2 \\ Charles R Greathouse IV, Jun 27 2013

G.f.: 3*(1 + x) / (1 - x)^5.
a(n) = (n-1)*n/2 * n*(n+1)/2 = A000217(n-1)*A000217(n) = 1/2*(n^2-1)*n^2/2 = 1/2*A000217(n^2-1). - Alexander Adamchuk, Apr 13 2006
a(n) = 3*A002415(n) = A047928(n-1)/4 = A083374(n-1)/2 = A008911(n)*3/2. - Zerinvary Lajos, May 09 2007
a(n) = (A126274(n) - A000537(n+1))/2. - Enrique Pérez Herrero, Mar 11 2013
Ceiling(sqrt(a(n)) + sqrt(a(n-1)))/2 = A000217(n). - Richard R. Forberg, Aug 14 2013
a(n) = Sum_{i=1..n-1} i*(i^2+n) for n > 1 (see Example section). - Bruno Berselli, Aug 29 2014
Sum_{n>=2} 1/a(n) = 7 - 2*Pi^2/3 = 0.42026373260709425411... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n^2+n) - A000217(n)*A000217(n+1). - Charlie Marion, Feb 15 2020
Sum_{n>=2} (-1)^n/a(n) = Pi^2/3 - 3. - Amiram Eldar, Nov 02 2021
E.g.f.: exp(x)*x^2*(6 + 6*x + x^2)/4. - Stefano Spezia, Mar 12 2024

A047659 Number of ways to place 3 nonattacking queens on an n X n board.

Original entry on oeis.org

0, 0, 0, 0, 24, 204, 1024, 3628, 10320, 25096, 54400, 107880, 199400, 348020, 579264, 926324, 1431584, 2148048, 3141120, 4490256, 6291000, 8656860, 11721600, 15641340, 20597104, 26797144, 34479744, 43915768, 55411720, 69312516, 86004800, 105919940

Offset: 0

Paul Zimmermann

Lucas mentions that the number of ways of placing p <= n non-attacking queens on an n X n chessboard is given by a polynomial in n of degree 2p and attribute the result to Mantel, professor in Delft. Cf. Stanley, exercise 15.

E. Landau, Naturwissenschaftliche Wochenschrift (Aug. 2 1896).
R. P. Stanley, Enumerative Combinatorics, vol. I, exercise 15 in chapter 4 (and its solution) asks one to show the existence of a rational generating function for the number of ways of placing k non-attacking queens on an n X n chessboard.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem I. General theory, Jan 26 2013. - _N. J. A. Sloane_, Feb 16 2013
S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv:1609.00853 [math.CO], Sep 03 2016.
Christopher R. H. Hanusa, T Zaslavsky, S Chaiken, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 11.
Edmund Landau, Ueber das Achtdamenproblem und seine Verallgemeinerung, Naturwissenschaftliche Wochenschrift, Aug 02 1896.
Edouard Lucas, Récréations mathématiques, Gauthier-Villars, Paris, 1882-1894, Vol. I, p. 228.
Antal Pinter, Numerical solution of the k=3 Queens problem, 2011, P(3) at p.8-9.
I. Rivin, I. Vardi and P. Zimmermann, The n-queens problem, Amer. Math. Monthly, 101 (1994), 629-639.
Wenxi Wang, Muhammad Usman, Alyas Almaawi, Kaiyuan Wang, Kuldeep S. Meel, Sarfraz Khurshid, A Study of Symmetry Breaking Predicates and Model Counting, National University of Singapore (2020).
Index entries for linear recurrences with constant coefficients, signature (5,-8,0,14,-14,0,8,-5,1).

Cf. A036464, A061994, A108792, A176186, A178721.

Column k=3 of A348129.

[(3*(2*n-1)*(-1)^n +4*n^6 -40*n^5 +158*n^4 -300*n^3 +264*n^2 -86*n +3)/24: n in [0..35]]; // Vincenzo Librandi, Sep 21 2015

f:=n-> n^6/6 - 5*n^5/3 + 79*n^4/12 - 25*n^3/2 + 11*n^2 - 43*n/12 + 1/8 + (-1)^n*(n/4 - 1/8); [seq(f(n),n=1..40)]; # N. J. A. Sloane, Feb 16 2013

Table[If[EvenQ[n],n (n-2)^2 (2n^3-12n^2+23n-10)/12,(n-1)(n-3) (2n^4- 12n^3+25n^2-14n+1)/12],{n,0,30}] (* or *) LinearRecurrence[ {5,-8,0,14,-14,0,8,-5,1},{0,0,0,0,24,204,1024,3628,10320},30] (* Harvey P. Dale, Nov 06 2011 *)

a(n)=if(n%2, (n - 1)*(n - 3)*(2*n^4 - 12*n^3 + 25*n^2 - 14*n + 1), n*(n - 2)^2*(2*n^3 - 12*n^2 + 23*n - 10))/12 \\ Charles R Greathouse IV, Feb 09 2017

a(n) = n(n - 2)^2(2n^3 - 12n^2 + 23n - 10)/12 if n is even and (n - 1)(n - 3)(2n^4 - 12n^3 + 25n^2 - 14n + 1)/12 if n is odd (Landau, 1896).
a(n) = 5a(n - 1) - 8a(n - 2) + 14a(n - 4) - 14a(n - 5) + 8a(n - 7) - 5a(n - 8) + a(n - 9) for n >= 9.
G.f.: 4(9*x^4 + 35*x^3 + 49*x^2 + 21*x + 6)*x^4/((1 - x)^7*(1 + x)^2).
a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=24, a(5)=204, a(6)=1024, a(7)=3628, a(8)=10320, a(n) = 5*a(n-1)-8*a(n-2)+14*a(n-4)-14*a(n-5)+8*a(n-7)- 5*a(n-8)+ a(n-9). - Harvey P. Dale, Nov 06 2011
a(n) = n^6/6 - 5*n^5/3 + 79*n^4/12 - 25*n^3/2 + 11*n^2 - 43*n/12 + 1/8 + (-1)^n*(n/4 - 1/8) [Chaiken et al.]. - N. J. A. Sloane, Feb 16 2013
a(n) = (3*(2*n-1)*(-1)^n +4*n^6 -40*n^5 +158*n^4 -300*n^3 +264*n^2 -86*n +3)/24. - Antal Pinter, Oct 03 2014
E.g.f.: (exp(2*x)*(3 - 6*x^2 + 8*x^3 + 18*x^4 + 20*x^5 + 4*x^6) -3 - 6*x) / (24*exp(x)). - Vaclav Kotesovec, Feb 15 2015
For n>3, a(n) = A179058(n) -4*(n-2)*A000914(n-2) -2*(n-2)*A002415(n-1) + 2*A008911(n-1) +8*(A001752(n-4) +A007009(n-3)). - Antal Pinter, Sep 20 2015
In general, for m <= n, n >= 3, the number of ways to place 3 nonattacking queens on an m X n board is n^3/6*(m^3 - 3*m^2 + 2*m) - n^2/2*(3*m^3 - 9*m^2 + 6*m) + n/6*(2*m^4 + 20*m^3 - 77*m^2 + 58*m) - 1/24*(39*m^4 - 82*m^3 - 36*m^2 + 88*m) + 1/16*(2*m - 4*n + 1)*(1 + (-1)^(m+1)) + 1/2*(1 + abs(n - 2*m + 3) - abs(n - 2*m + 4))*(1/24*((n - 2*m + 11)^4 - 42*(n - 2*m + 11)^3 + 656*(n - 2*m + 11)^2 - 4518*(n - 2*m + 11) + 11583) - 1/16*(4*m - 2*n - 1)*(1 + (-1)^(n+1))) [Panos Louridas, idee & form 93/2007, pp. 2936-2938]. - Vaclav Kotesovec, Feb 20 2016

The formula given in the Rivin et al. paper is wrong.

Entry improved by comments from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 30 2001

A173121 a(n) = sinh(2arccosh(n))^2 = 4n^2*(n^2 - 1).

Original entry on oeis.org

0, 0, 48, 288, 960, 2400, 5040, 9408, 16128, 25920, 39600, 58080, 82368, 113568, 152880, 201600, 261120, 332928, 418608, 519840, 638400, 776160, 935088, 1117248, 1324800, 1560000, 1825200, 2122848, 2455488, 2825760, 3236400, 3690240

Offset: 0

Artur Jasinski, Feb 10 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A002415, A047928, A001079, A037270, A071253, A108741, A132592, A146311-A146313, A173115.

[4*n^2*(n^2-1): n in [0..40]]; // Vincenzo Librandi, Jun 15 2011

Table[4 n^2*(n^2 - 1), {n, 0, 30}] (* or *) Table[Round[N[Sinh[2 ArcCosh[n]]^2, 100]], {n, 0, 50}]
LinearRecurrence[{5,-10,10,-5,1},{0,0,48,288,960},40] (* Harvey P. Dale, Jul 22 2015 *)

a(n)=4*n^2*(n^2-1) \\ Charles R Greathouse IV, Jul 01 2013

a(n) = 48*A002415(n) = 4*A047928(n).
G.f.: 48*x^2*(1+x)/(1-x)^5. - Colin Barker, Mar 22 2012
From Amiram Eldar, Jul 26 2022: (Start)
Sum_{n>=2} 1/a(n) = (21 - 2*Pi^2)/48.
Sum_{n>=2} (-1)^n/a(n) = (Pi^2 - 9)/48. (End)

A001249 Squares of tetrahedral numbers: a(n) = binomial(n+3,n)^2.

Original entry on oeis.org

1, 16, 100, 400, 1225, 3136, 7056, 14400, 27225, 48400, 81796, 132496, 207025, 313600, 462400, 665856, 938961, 1299600, 1768900, 2371600, 3136441, 4096576, 5290000, 6760000, 8555625, 10732176, 13351716, 16483600, 20205025, 24601600, 29767936, 35808256

Offset: 0

N. J. A. Sloane

Total area of all square and rectangular regions from an n+1 X n+1 grid. E.g., n = 2, there are 9 individual squares, 4 2 X 2's and 1 3 X 3, total area 9 + 16 + 9 = 34. The rectangular regions include 6 2 X 1's, 6 1 X 2's, 3 3 X 1's, 3 1 X 3's, 2 3 X 2's and 2 2 X 3's, total area 12 + 12 + 9 + 9 + 12 + 12 = 66, hence a(2) = 34 + 66 = 100. - Jon Perry, Jul 29 2003 [Index/grid size adjusted by Rick L. Shepherd, Jun 27 2017]
Number of 3 X 3 submatrices of an n+3 X n+3 matrix. - Rick L. Shepherd, Jun 27 2017
The inverse binomial transform gives row n=2 of A087107. - R. J. Mathar, Aug 31 2022

T. D. Noe, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000290, A000292, A006542, A033455, A108674 (first diffs.), A086020 (partial sums).
Third column of triangle A008459.
Cf. A000579, A001303, A040977.

A001249 := proc(n) binomial(n+3,n)^2 end proc: seq(A001249(n),n=0..10) ; # Zerinvary Lajos, May 17 2006

Table[Binomial[n + 3, 3]^2, {n, 0, 100}] (* T. D. Noe, Jun 26 2012 *)

a(n)=binomial(n+3,3)^2 \\ Charles R Greathouse IV, Sep 24 2015

From R. J. Mathar, Aug 19 2008: (Start)
a(n) = (A000292(n+1))^2.
O.g.f.: (1+x)(x^2+8x+1)/(1-x)^7. (End)
a(n) = C(n+4, 3)*C(n+4, 4)/(n+4) + A001303(n) = C(n+4, 3)*C(n+3, 3)/4 + A001303(n) = C(n+4, 6) + 3*C(n+5, 6) + C(n+6,6) + A001303(n). - Gary Detlefs, Aug 07 2013
-n^2*a(n) + (n+3)^2*a(n-1) = 0. - R. J. Mathar, Aug 15 2013
a(n) = 9*A040977(n-1) + A000579(n+6) + A000579(n+3). - R. J. Mathar, Aug 15 2013
a(n) = (n+3)*C(n+2, 2)*C(n+3, 3)/3. - Gary Detlefs, Jan 06 2014
a(n) = A000290(n+1)*A000290(n+2)*A000290(n+3)/36. - Bruno Berselli, Nov 12 2014
G.f. 2F1(4,4;1;x). - R. J. Mathar, Aug 09 2015
E.g.f.: exp(x)*(1 + 15*x + 69*x^2/2! + 147*x^3/3! + 162*x^4/4! +  90*x^5/5! +  20*x^6/6!). Computed from the o.g.f with the formulas (23) - (25) of the W. Lang link given in A060187. - Wolfdieter Lang, Jul 27 2017
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=0} 1/a(n) = 9*Pi^2 - 351/4.
Sum_{n>=0} (-1)^n/a(n) = 63/4 - 3*Pi^2/2. (End)
a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7). - Wesley Ivan Hurt, Aug 29 2022
a(n) = a(n-1)+A000217(n+1)*A000330(n+1). - J. M. Bergot, Aug 29 2022
a(n) = A002415(n+2)^2 - 20*A006857(n-1). - Yasser Arath Chavez Reyes, Nov 08 2024

A047928 a(n) = n(n-1)^2(n-2).

Original entry on oeis.org

0, 12, 72, 240, 600, 1260, 2352, 4032, 6480, 9900, 14520, 20592, 28392, 38220, 50400, 65280, 83232, 104652, 129960, 159600, 194040, 233772, 279312, 331200, 390000, 456300, 530712, 613872, 706440, 809100, 922560, 1047552, 1184832

Offset: 2

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A002378, A002415, A006011, A008911, A083374.

[ n*(n-1)^2*(n-2): n in [2..40]]; // Vincenzo Librandi, May 02 2011

seq(floor(n^6/(n^2+1)),n=1..25); # Gary Detlefs, Feb 11 2010

f[n_]:=n*(n-1)^2*(n-2); f[Range[2,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2011 *)
LinearRecurrence[{5,-10,10,-5,1},{0,12,72,240,600},40] (* or *) CoefficientList[Series[-((12 x (1+x))/(-1+x)^5),{x,0,40}],x] (* Harvey P. Dale, Jul 31 2021 *)

a(n)=n^4 - 4*n^3 + 5*n^2 - 2*n \\ Charles R Greathouse IV, May 02 2011

a(n) = A002378(n)*A002378(n+1). - Zerinvary Lajos, Apr 11 2006
a(n) = 12*A002415(n+1) = 2*A083374(n) = 4*A006011(n+1) = 6*A008911(n+1). - Zerinvary Lajos, May 09 2007
a(n) = floor((n-1)^6/((n-1)^2+1)). - Gary Detlefs, Feb 11 2010
From Amiram Eldar, Nov 05 2020: (Start)
Sum_{n>=3} 1/a(n) = 7/4 - Pi^2/6.
Sum_{n>=3} (-1)^(n+1)/a(n) = Pi^2/12 - 3/4. (End)
G.f.: -12*x*(1+x)/(-1+x)^5. - Harvey P. Dale, Jul 31 2021
a(n) = (n-1)^4 - (n-1)^2. - Katherine E. Stange, Mar 31 2022

A047819 a(n) = Product_{i=1..n} ((i+3)(i+4)(i+5))/(i(i+1)(i+2)).

Original entry on oeis.org

1, 20, 175, 980, 4116, 14112, 41580, 108900, 259545, 572572, 1184183, 2318680, 4331600, 7768320, 13441968, 22535064, 36729945, 58373700, 90684055, 138003404, 206108980, 302588000, 437287500, 622849500, 875343105, 1215006156, 1667110095

Offset: 0

N. J. A. Sloane

Number of tilings of a <3,n,3> hexagon.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 18 2005
Partial sums of A107891. - Peter Bala, Sep 21 2007
Determinant of the 3 X 3 matrix with rows [C(n+3,3) C(n+3,4) C(n+3,5)], [C(n+4,3) C(n+4,4) C(n+4,5)], and [C(n+5,3) C(n+5,4) C(n+5,5)]. - J. M. Bergot, Sep 10 2013

			G.f. = 1 + 20*x + 175*x^2 + 980*x^3 + 4116*x^4 + 14112*x^5 + 41580*x^6 + ...

O. D. Anderson, Find the next sequence, J. Rec. Math., 8 (No. 4, 1975-1976), 241.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 232, # 2 and p. 105, eq.(ii), K(0a(2,5,n))).

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Paolo Aluffi, Degrees of projections of rank loci, arXiv:1408.1702 [math.AG], 2014. ["After compiling the results of many explicit computations, we noticed that many of the numbers d_{n,r,S} appear in the existing literature in contexts far removed from the enumerative geometry of rank conditions; we owe this surprising (to us) observation to perusal of [Slo14]."]
O. D. Anderson, Find the next sequence, J. Rec. Math., 8 (No. 4, 1975-1976), 241. [Annotated scanned copy]
Harald Helfgott and Ira M. Gessel, Enumeration of tilings of diamonds and hexagons with defects, arXiv:math/9810143 [math.CO], 1998.
J. M. Landsberg and L. Manivel, The sextonions and E7 1/2, Adv. Math. 201 (2006), 143-179. [Th. 7.2(ii), case a=2].
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 25.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Third row of array A103905.

Cf. A002415, A107891, A107915, A133708.

a:=n->(n+1)*(n+2)^2*(n+3)^3*(n+4)^2*(n+5)/8640: seq(a(n),n=0..30); # Emeric Deutsch, Jun 18 2005

a[n_] :=(n + 1)*(n + 2)^2*(n + 3)^3*(n + 4)^2*(n + 5)/8640;
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 19 2018, after Emeric Deutsch *)

{a(n) = if( n<0, 0, binomial(n+5, 5) * binomial(n+4, 3) * (n+3) / 12)}; /* Michael Somos, Nov 14 2002 */

{a(n) = my(s=sign(n+3)); n=abs(n+3)-3; -s/8 * polcoeff( charpoly( matrix(n+3, n+3, i, j, (i-j)^2)), n)}; /* Michael Somos, Nov 14 2002 */

G.f.: (1 + 10*x + 20*x^2 + 10*x^3 + x^4) / (1 - x)^10. - Michael Somos, Nov 14 2002
a(n) = C(n+3,n+2)*C(n+4,n+1)*C(n+5,n)/12. - Zerinvary Lajos, May 29 2007
a(n-3) = (1/24)*Sum_{1 <= x_1, x_2, x_3 <= n} (det V(x_1,x_2,x_3))^2 = (1/24)*Sum_ {1 <= i,j,k <= n} ((i-j)(i-k)(j-k))^2, where V(x_1,x_2,x_3) is the Vandermonde matrix of order 3. - Peter Bala, Sep 21 2007
a(n) = -a(-6-n) for all n in Z. - Michael Somos, Dec 26 2016
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 5195/2 - 2160*zeta(3).
Sum_{n>=0} (-1)^n/a(n) = 17205/2 - 9600*log(2) - 1620*zeta(3). (End)

A080852 Square array of 4D pyramidal numbers, read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 5, 10, 1, 6, 15, 20, 1, 7, 20, 35, 35, 1, 8, 25, 50, 70, 56, 1, 9, 30, 65, 105, 126, 84, 1, 10, 35, 80, 140, 196, 210, 120, 1, 11, 40, 95, 175, 266, 336, 330, 165, 1, 12, 45, 110, 210, 336, 462, 540, 495, 220, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 286

Offset: 0

Paul Barry, Feb 21 2003

The first row contains the tetrahedral numbers, which are really three-dimensional, but can be regarded as degenerate 4D pyramidal numbers. - N. J. A. Sloane, Aug 28 2015

			Array, n >= 0, k >= 0, begins
1 4 10 20  35  56 ...
1 5 15 35  70 126 ...
1 6 20 50 105 196 ...
1 7 25 65 140 266 ...
1 8 30 80 175 336 ...

Index to sequences related to pyramidal numbers

Rows include A000292, A000332, A002415, A001296, A002418, A002419, A051740, A051797.
Cf. A057145, A080851, A180266, A055796 (antidiagonal sums).
See A257200 for another version of the array.

vector(vector(poly_coeff(Taylor((1+kx)/(1-x)^5,x,11),x,n),n,0,11),k,-1,10)

VECTOR(VECTOR(comb(k+3,3)+comb(k+3,4)n, k, 0, 11), n, 0, 11)

A080852 := proc(n,k)
        binomial(k+4,4)+(n-1)*binomial(k+3,4) ;
end proc:
seq( seq(A080852(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Oct 01 2021

T[n_, k_] := Binomial[k+3, 3] + Binomial[k+3, 4]n;
Table[T[n-k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 05 2023 *)

T(n, k) = binomial(k + 4, 4) + (n-1)*binomial(k + 3, 4), corrected Oct 01 2021.
T(n, k) = T(n - 1, k) + C(k + 3, 4) = T(n - 1, k) + k(k + 1)(k + 2)(k + 3)/24.
G.f. for rows: (1 + nx)/(1 - x)^5, n >= -1.
T(n,k) = sum_{j=0..k} A080851(n,j). - R. J. Mathar, Jul 28 2016

A101089 Second partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 18, 116, 470, 1449, 3724, 8400, 17172, 32505, 57838, 97812, 158522, 247793, 375480, 553792, 797640, 1125009, 1557354, 2120020, 2842686, 3759833, 4911236, 6342480, 8105500, 10259145, 12869766, 16011828, 19768546, 24232545, 29506544

Offset: 1

Cecilia Rossiter, Dec 14 2004

a(n) is the n-th antidiagonal sum of the convolution array A213553. - Clark Kimberling, Jun 17 2012
a(n-1)/n^5 is the "retention" of water on a 3 X 3 random surface of n levels - see Knecht et al., 2012, Schrenk et al., 2014. - Robert M. Ziff, Mar 08 2014
The general formula for the second partial sums of m-th powers is: b(n,m) = (n+1)*F(m) - F(m+1), where F(m) is the m-th Faulhaber’s polynomial. - Luciano Ancora, Jan 26 2015

			a(7) = 8400 = 1*(8-1)^4 + 2*(8-2)^4 + 3*(8-3)^4 + 4*(8-4)^4 + 5*(8-5)^4 + 6*(8-6)^4 + 7*(8-7)^4. - _Bruno Berselli_, Jan 31 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Luciano Ancora, Recurrence relation for the second partial sums of m-th powers.
Luciano Ancora, Second partial sums of the m-th powers.
Craig L. Knecht, Walter Trump, Daniel ben-Avraham and Robert M. Ziff, Retention Capacity of Random Surfaces, Phys. Rev. Lett., Vol. 108 (2012), 045703.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 13.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev., Vol. 19, No. 1 (1977), pp. 90-99. MR0428678 (55 #1698). See Table 1. - _N. J. A. Sloane_, Mar 23 2014
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013), Article 13.5.7.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Dead link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
K. J. Schrenk, N. A. M. Araújo, R. M. Ziff and H. J. Herrmann Retention Capacity of Correlated Surfaces, arXiv:1403.2082 [cond-mat.stat-mech], 2014.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Partial sums of A000538.

Cf. A000583, A002415, A101090, A201126, A213553.

List([1..40], n-> (n+1)^2*(2*(n+1)^4-5*(n+1)^2+3)/60); # G. C. Greubel, Jul 31 2019

[(1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1): n in [1..40]]; // Vincenzo Librandi, Mar 24 2014

f:=n->(2*n^6-5*n^4+3*n^2)/60;
[seq(f(n),n=0..50)]; # N. J. A. Sloane, Mar 23 2014

a[n_] := n(n+1)^2(n+2)(2n(n+2) -1)/60; Table[a[n], {n, 40}]
CoefficientList[Series[(1+x)*(1+10*x+x^2)/(1-x)^7, {x,0,40}], x] (* Vincenzo Librandi, Mar 24 2014 *)
Nest[Accumulate[#]&,Range[30]^4,2] (* Harvey P. Dale, Aug 13 2024 *)

a(n)=n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 \\ Charles R Greathouse IV, Mar 18 2014

[n*(n+1)^2*(n+2)*(2*n*(n+2)-1)/60 for n in range(1,40)] # Danny Rorabaugh, Apr 20 2015

a(n) = (1/60)*n*(n+1)^2*(n+2)*(2*n*(n+2)-1).
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^7. - Colin Barker, Apr 04 2012
a(n) = Sum_{i=1..n} i*(n+1-i)^4, by the definition. - Bruno Berselli, Jan 31 2014
a(n) = 2*a(n-1) - a(n-2) + n^4. - Luciano Ancora, Jan 08 2015
Sum_{n>=1} 1/a(n) = 85/3 + 10*Pi^2/3 - 20*sqrt(2/3)*Pi*cot(sqrt(3/2)*Pi). - Amiram Eldar, Jan 26 2022
a(n) = (1/2)*Sum_{1 <= i, j <= n+1} (i - j)^4 - Peter Bala, Jun 11 2024

Edited by Ralf Stephan, Dec 16 2004

cp's OEIS Frontend

A014820 a(n) = (1/3)*(n^2 + 2*n + 3)*(n+1)^2.

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A034827 a(n) = 2*binomial(n,4).

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A006011 a(n) = n^2*(n^2 - 1)/4.

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A047659 Number of ways to place 3 nonattacking queens on an n X n board.

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A173121 a(n) = sinh(2*arccosh(n))^2 = 4*n^2*(n^2 - 1).

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A001249 Squares of tetrahedral numbers: a(n) = binomial(n+3,n)^2.

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A014820 a(n) = (1/3)(n^2 + 2n + 3)*(n+1)^2.

A173121 a(n) = sinh(2arccosh(n))^2 = 4n^2*(n^2 - 1).

A047928 a(n) = n(n-1)^2(n-2).

A047819 a(n) = Product_{i=1..n} ((i+3)(i+4)(i+5))/(i(i+1)(i+2)).