A005900 -id:A005900 - cp's OEIS frontend

A004006 a(n) = C(n,1) + C(n,2) + C(n,3), or n*(n^2 + 5)/6.

0, 1, 3, 7, 14, 25, 41, 63, 92, 129, 175, 231, 298, 377, 469, 575, 696, 833, 987, 1159, 1350, 1561, 1793, 2047, 2324, 2625, 2951, 3303, 3682, 4089, 4525, 4991, 5488, 6017, 6579, 7175, 7806, 8473, 9177, 9919, 10700, 11521, 12383, 13287, 14234, 15225

Offset: 0

Albert D. Rich (Albert_Rich(AT)msn.com)

3-dimensional analog of centered polygonal numbers.
The Burnside group B(3,n) has order 3^a(n).
Answer to the question: if you have a tall building and 3 plates and you need to find the highest story, a plate thrown from which does not break, what is the number of stories you can handle given n tries? - Leonid Broukhis, Oct 24 2000
Equals row sums of triangle A144329 starting with "1". - Gary W. Adamson, Sep 18 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=4, a(n-1)=-coeff(charpoly(A,x),x^(n-3)). - Milan Janjic, Jan 24 2010
From J. M. Bergot, Aug 03 2011: (Start)
If one formed the 3 X 3 square
  |  n  |  n+1 | n+2 |
  | n+3 |  n+4 | n+5 |
  | n+6 |  n+7 | n+8 |
  and found the sum of the horizontal products n*(n + 1)*(n + 2) + (n + 3)*(n + 4)*(n + 5) + (n + 6)*(n + 7)*(n + 8) and added the sum of the vertical products n*(n + 3)*(n + 6) + (n + 1)*(n + 4)*(n + 7) + (n + 2)*(n + 5)(n + 8) one gets 6*n^3 + 72*n^2 + 318*n + 504. This will give 36 times the values of all the terms in this sequence. (End)
a(n) is divisible by n for n congruent to {1,5} mod 6. (see A007310). - Gary Detlefs, Dec 08 2011
From Beimar Naranjo, Feb 22 2024: (Start)
Number of compositions with at most three parts and sum at most n.
Also the number of compositions with at most one part distinct from 1 and with a sum at most n. (End)
a(n) is the number of strings of length n defined on {0, 1, 2, 3} that contain one 1 and any number of 0's, or two 2's and any number of 0's, or three 3's and any number of 0's. For example, a(6) = 41 since the strings are the 20 permutations of 333000, the 15 permutations of 220000 and the 6 permutations of 100000. - Enrique Navarrete, Jun 18 2025

			G.f. = x + 3*x^2 + 7*x^3 + 14*x^4 + 25*x^5 + 41*x^6 + 63*x^7 + 92*x^8 + ... - _Michael Somos_, Dec 29 2019

W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory, Wiley, 1966, see p. 380.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Michael Boardman, The Egg-Drop Numbers, Mathematics Magazine, 77 (2004), 368-372. [From Parthasarathy Nambi, Sep 30 2009]
F. Javier de Vega, An extension of Furstenberg's theorem of the infinitude of primes, arXiv:2003.13378 [math.NT], 2020.
Oboifeng Dira, A Note on Composition and Recursion, Southeast Asian Bulletin of Mathematics, 2017, Vol. 41 Issue 6, pp. 849-853.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006.
N. Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8.
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Laurent Saloff-Coste, Random walks on finite groups, in Probability on discrete structures, 263-346, Encyclopaedia Math. Sci., 110, Springer, 2004.
Bridget Eileen Tenner, Reduced word manipulation: patterns and enumeration, J. Algebr. Comb. 46, No. 1, 189-217 (2017), w in S_n(231) l(w)=3.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A051576, A055795, A006552. Differences give A000217 + 1 or A000124.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A000292, A001477, A007310, A128834, A144329, A228074.

List([0..50], n-> n*(n^2+5)/6); # G. C. Greubel, Aug 27 2019

a004006 n = a000292 n + n + 1  -- Reinhard Zumkeller, Mar 31 2012

[n*(n^2+5)/6: n in [0..50]]; // Vincenzo Librandi, May 15 2011

A004006 := proc(n) n*(n^2+5)/6 ;end proc:
seq(A004006(n),n=0..10) ; # R. J. Mathar, Jun 05 2011

Table[Total[Table[Binomial[n,i], {i,3}]], {n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,3,7}, 50] (* Harvey P. Dale, Aug 21 2011 *)

A004006(n):=n*(n^2+5)/6$ makelist(A004006(n),n,0,30); /* Martin Ettl, Jan 08 2013 */

{a(n) = n*(n^2 + 5)/6}; /* Michael Somos, May 04 2007 */

[n*(n^2+5)/6 for n in (0..50)] # G. C. Greubel, Aug 27 2019

G.f.: x*(1-x+x^2)/(1-x)^4.
E.g.f.: x*(1 + x/2 + x^2/6) * exp(x).
a(-n) = -a(n).
a(n) = binomial(n+2,n-1) - binomial(n,n-2). - Zerinvary Lajos, May 11 2006
Euler transform of length 6 sequence [3, 1, 1, 0, 0, -1]. - Michael Somos, May 04 2007
Starting (1, 3, 7, 14, ...) = binomial transform of [1, 2, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson, Apr 24 2008
a(0)=0, a(1)=1, a(2)=3, a(3)=7, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 21 2011
a(n+1) = A000292(n) + n + 1. - Reinhard Zumkeller, Mar 31 2012
a(n) = 2*a(n-1) + (n-1) - a(n-2) with a(0) = 0, a(1) = 1. - Richard R. Forberg, Jan 23 2014
a(n) = Sum_{i=1..n} binomial(n-2i,2). - Wesley Ivan Hurt, Nov 18 2017
a(n) = n + Sum_{k=0..n} k*(n-k). - Gionata Neri, May 19 2018
a(n) = Sum_{k=0..n-1} A000124(k). - Torlach Rush, Aug 05 2018
G.f.: ((1 - x^5)/(1 - x)^5 - 1)/5. - Michael Somos, Dec 29 2019
G.f.: g(f(x)), where g is g.f. of A001477 and f is g.f. of A128834. - Oboifeng Dira, Jun 21 2020
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(5)) + polygamma(0, 1+i*sqrt(5)))/5 = 1.6787729555834452106286261834348972248... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

A006527 a(n) = (n^3 + 2*n)/3.

Original entry on oeis.org

0, 1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531, 28424, 30405

Offset: 0

N. J. A. Sloane

Number of ways to color vertices (or edges) of a triangle using <= n colors, allowing only rotations.
Also: dot_product (1,2,...,n)*(2,3,...,n,1), n >= 0. - Clark Kimberling
Start from triacid and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002
Starting with offset 1 = row sums of triangle A158822 and binomial transform of (1, 3, 4, 2, 0, 0, 0, ...). - Gary W. Adamson, Mar 28 2009
One-ninth of sum of three consecutive cubes: a(n) = ((n-1)^3 + n^3 + (n+1)^3)/9. - Zak Seidov, Jul 22 2013
For n > 2, number of different cubes, formed after splitting a cube in color C_1, by parallel planes in the colors C_2, C_3, ..., C_n in three spatial dimensions (in the order of the colors from a fixed vertex). Generally, in a large hypercube n^d is f(n,d) = C(n+d-1, d) + C(n, d) different small hypercubes. See below for my formula a(n) = f(n,3). - Thomas Ordowski, Jun 15 2014
a(n) is a square for n = 1, 2 & 24; and for no other values up to 10^7 (see M. Gardner). - Michel Marcus, Sep 06 2015
Number of unit tetrahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb. - Jason Pruski, Aug 23 2017

M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
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..5000
B. Babcock and A. van Tuyl, Revisiting the spreading and covering numbers, arXiv preprint arXiv:1109.5847 [math.AC], 2011.
Richard A. Brualdi and Geir Dahl, Alternating Sign Matrices and Hypermatrices, and a Generalization of Latin Square, arXiv:1704.07752 [math.CO], 2017. See p. 8.
Peter Esser?, Guenter Stertenbrink, Triangles with Mac Mahon's pieces, digest of 14 messages in polyforms Yahoo group, Apr 14 - May 2, 2002.
Th. Grüner, A. Kerber, R. Laue and M. Meringer, Mathematics for Combinatorial Chemistry
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
polyforms list, Triangles with MacMahon's pieces.
Taskcentre, McMahon's Triangles 2
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Column 1 of triangle A094414. Row 6 of the array in A107735.
Cf. A002264, A005843, A054602, A059100, A135184, A158822, A177342.
Cf. A000292 (unoriented), A000292(n-2) (chiral), A000290 (achiral) triangle colorings.
Row 2 of A324999 (simplex vertices and facets) and A327083 (simplex edges and ridges).

a006527 n = n * (n ^ 2 + 2) `div` 3  -- Reinhard Zumkeller, Jan 06 2014

[(n^3 + 2*n)/3: n in [0..50]]; // Vincenzo Librandi, May 15 2011

A006527:=z*(1+z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
with(combinat):seq(lcm(fibonacci(4,n),fibonacci(2,n))/3,n=0..42); # Zerinvary Lajos, Apr 20 2008

Table[ (n^3 + 2*n)/3, {n, 0, 45} ]
LinearRecurrence[{4,-6,4,-1},{0,1,4,11},46] (* or *) CoefficientList[ Series[(x+x^3)/(x-1)^4,{x,0,49}],x] (* Harvey P. Dale, Jun 13 2011 *)

a(n)=n*(n^2+2)/3 \\ Charles R Greathouse IV, Jul 25 2011

a(0)=0, a(1)=1, a(2)=4, a(3)=11; for n > 3, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Jun 13 2011
From Paul Barry, Mar 13 2003: (Start)
a(n) = 2*binomial(n+1, 3) + binomial(n, 1).
G.f.: x*(1+x^2)/(1-x)^4. (End)
a(n) = A000292(n) + A000292(n-2). - Alexander Adamchuk, May 20 2006
a(n) = n*A059100(n)/3. - Lekraj Beedassy, Feb 06 2007
a(n) = A054602(n)/3. - Zerinvary Lajos, Apr 20 2008
a(n) = ( n + Sum_{i=1..n} A177342(i) )/(n+1), with n > 0. - Bruno Berselli, May 19 2010
a(n) = A002264(A000578(n) + A005843(n)). - Reinhard Zumkeller, Jun 16 2011
a(n) = binomial(n+2, 3) + binomial(n, 3). - Thomas Ordowski, Jun 15 2014
a(n) = A000292(n) - A000292(-n). - Bruno Berselli, Sep 22 2016
E.g.f.: (x/3)*(3 + 3*x + x^2)*exp(x). - G. C. Greubel, Sep 01 2017
From Robert A. Russell, Oct 20 2020: (Start)
a(n) = 1*C(n,1) + 2*C(n,2) + 2*C(n,3), where the coefficient of C(n,k) is the number of oriented triangle colorings using exactly k colors.
a(n) = 2*A000292(n) - A000290(n) = 2*A000292(n-2) + A000290(n). (End)
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i*sqrt(2)) + polygamma(0, 1+i*sqrt(2)))/4 = 1.45245201414472469745354677573358867... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

More terms from Alexander Adamchuk, May 20 2006
Corrected and replaced 5th formula from Harvey P. Dale, Jun 13 2011
Deleted an erroneous comment. - N. J. A. Sloane, Dec 10 2018

A063496 a(n) = (2n - 1)(8n^2 - 8n + 3)/3.

Original entry on oeis.org

1, 19, 85, 231, 489, 891, 1469, 2255, 3281, 4579, 6181, 8119, 10425, 13131, 16269, 19871, 23969, 28595, 33781, 39559, 45961, 53019, 60765, 69231, 78449, 88451, 99269, 110935, 123481, 136939, 151341, 166719, 183105, 200531, 219029, 238631, 259369, 281275, 304381

Offset: 1

N. J. A. Sloane, Aug 01 2001

Number of potential flows in a 2 X 2 matrix with integer velocities in -n..n, i.e., number of 2 X 2 matrices with adjacent elements differing by no more than n, counting matrices differing by a constant only once. - R. H. Hardin, Feb 27 2002
Number of ordered quadruples (a,b,c,d), -(n-1) <= a,b,c,d <= n-1, such that a+b+c+d = 0. - Benoit Cloitre, Jun 14 2003
If Y and Z are 2-blocks of a (2n+1)-set X then a(n-1) is the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007
Equals binomial transform of [1, 18, 48, 32, 0, 0, 0, ...]. - Gary W. Adamson, Jul 19 2008

Harry J. Smith, Table of n, a(n) for n = 1..1000
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
Milan Janjic, Two Enumerative Functions
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Cf. A003215, A010006, A142992, A142993, A142994.

[(2*n-1)*(8*n^2-8*n+3)/3: n in [1..40]]; // Wesley Ivan Hurt, May 09 2014

A063496:=n->(2*n-1)*(8*n^2-8*n+3)/3; seq(A063496(n), n=1..40); # Wesley Ivan Hurt, May 09 2014

Table[(2*n - 1)*(8*n^2 - 8*n + 3)/3, {n, 40}] (* Wesley Ivan Hurt, May 09 2014 *)
LinearRecurrence[{4,-6,4,-1}, {1,19,85,231}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(8*n^2 - 8*n + 3)/3 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-3+6*x+24*x^2+16*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017

From Peter Bala, Jul 18 2008: (Start)
The following remarks about the C_3 lattice assume the sequence offset is 0.
Partial sums of A010006. So this sequence is the crystal ball sequence for the C_3 lattice - row 3 of A142992. The lattice C_3 consists of all integer lattice points v = (a,b,c) in Z^3 such that a + b + c is even, equipped with the taxicab type norm ||v|| = (1/2) * (|a| + |b| + |c|).
The crystal ball sequence of C_3 gives the number of lattice points v in C_3 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].
For example, a(1) = 19 because the origin has norm 0 and the 18 lattice points in Z^3 of norm 1 (as defined above) are +-(2,0,0), +-(0,2,0), +-(0,0,2), +-(1,1,0), +-(1,0,1), +-(0,1,1), +-(1,-1,0), +-(1,0,-1) and +-(0,1,-1). These 18 vectors form a root system of type C_3.
O.g.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4 = x/(1 - x) * T(3, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
2*log(2) = 4/3 + Sum_{n >= 1} 1/(n*a(n)*a(n+1)). (End)
a(n+1) = (1/Pi) * Integral_{x=0..Pi} (sin((n+1/2)*x)/sin(x/2))^4. - Yalcin Aktar, Nov 02 2011, corrected by R. J. Mathar, Dec 01 2011
From G. C. Greubel, Dec 01 2017: (Start)
G.f.: x*(1 + 15*x + 15*x^2 + x^3)/(1 - x)^4.
E.g.f.: (-3 + 6*x + 24*x^2 + 16*x^3)*exp(x)/3 + 1. (End)
a(n) = A005900(2n-1). - Ivan N. Ianakiev, Mar 27 2022
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k+1)) = 1/(19 - 3/(27 - 60/(43 - 315/(67 - ... -n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*3^2))))).
E.g.f.: exp(x)*(1 + 18*x + 48*x^2/2! + 32*x^3/3!). Note that -T(6, i*sqrt(x)) = 1 + 18*x + 48*x^2 + 32*x^3, where T(n, x) denotes the n-th Chebyshev polynomial of the first kind. See A008310. (End)

A006331 a(n) = n(n+1)(2*n+1)/3.

Original entry on oeis.org

0, 2, 10, 28, 60, 110, 182, 280, 408, 570, 770, 1012, 1300, 1638, 2030, 2480, 2992, 3570, 4218, 4940, 5740, 6622, 7590, 8648, 9800, 11050, 12402, 13860, 15428, 17110, 18910, 20832, 22880, 25058, 27370, 29820, 32412, 35150, 38038, 41080, 44280

Offset: 0

N. J. A. Sloane

Triangles in rhombic matchstick arrangement of side n.
Maximum accumulated number of electrons at energy level n. - Scott A. Brown, Feb 28 2000
Let M_n denote the n X n matrix M_n(i,j)=i^2+j^2; then the characteristic polynomial of M_n is x^n - a(n)x^(n-1) - .... - Michael Somos, Nov 14 2002
Convolution of odds (A005408) and evens (A005843). - Graeme McRae, Jun 06 2006
a(n) is the number of non-monotonic functions with domain {0,1,2} and codomain {0,1,...,n}. - Dennis P. Walsh, Apr 25 2011
For any odd number 2n+1, find Sum_{aJ. M. Bergot, Jul 16 2011
a(n) gives the number of (n+1) X (n+1) symmetric (0,1)-matrices containing three ones (see [Cameron]). - L. Edson Jeffery, Feb 18 2012
a(n) is the number of 4-tuples (w,x,y,z) with all terms in {0,...,n} and |w - x| < y. - Clark Kimberling, Jun 02 2012
Partial sums of A001105. - Omar E. Pol, Jan 12 2013
Total number of square diagonals (of any size) in an n X n square grid. - Wesley Ivan Hurt, Mar 24 2015
Number of diagonal attacks of two queens on (n+1) X (n+1) chessboard. - Antal Pinter, Sep 20 2015
a(n) is the minimum value obtainable by partitioning either the set {x in the natural numbers | 1 <= x <= 2n} or the set {x in the natural numbers | 0 <= x <= 2n+1} into pairs, taking the product of all such pairs, and taking the sum of all such products. - Thomas Anton, Oct 21 2020
a(n) is the irregularity of the n-th power of a path of length at least 3*n.  (The irregularity of a graph is the sum of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 16 2023
a(n) is the maximum possible total number of inversions in all rows and all columns of a Latin square of order n+1. - Ivaylo Kortezov, Jun 28 2025

			For n=2, a(2)=10 since there are 10 non-monotonic functions f from {0,1,2} to {0,1,2}, namely, functions f = <f(1),f(2),f(3)> given by <0,1,0>, <0,2,0>, <0,2,1>, <1,0,1>, <1,0,2>, <1,2,0>, <1,2,1>, <2,0,1>, <2,0,2>, and <2,1,2>. - _Dennis P. Walsh_, Apr 25 2011
Let n=4, 2*n+1 = 9. Since 9 = 1+8 = 3+6 = 5+4 = 7+2, a(4) = 1*8 + 3*6 + 5*4 + 7*2 = 60. - _Vladimir Shevelev_, May 11 2012

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
J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., Vol. 2 (1931), pp. 355-359.
J. L. Bailey, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., Vol. 2 (1931), pp. 355-359. [Annotated scanned copy]
Rowan Beckworth, Basic atomic information.
Allan Bickle and Zhongyuan Che, Irregularities of Maximal k-degenerate Graphs, Discrete Applied Math. 331 (2023) 70-87.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
P. Cameron, T. Prellberg and D. Stark, Asymptotics for incidence matrix classes, Electron. J. Combin. 13 (2006), #R85, p. 11.
Jose Manuel Garcia Calcines, Luis Javier Hernandez Paricio, and Maria Teresa Rivas Rodriguez, Semi-simplicial combinatorics of cyclinders and subdivisions, arXiv:2307.13749 [math.CO], 2023. See p. 25.
N. S. S. Gu, H. Prodinger and S. Wagner, Bijections for a class of labeled plane trees, Eur. J. Combinat., Vol. 31 (2010), pp. 720-732, doi|10.1016/j.ejc.2009.10.007, Theorem 2 at n=3.
JBMO 2025, 29th Junior Balkan Mathematical Olympiad, Problem 4, author: Boris Mihov
Germain Kreweras, Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, Vol. 6 (1965), circa p. 82.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Dennis Walsh, Notes on finite monotonic and non-monotonic functions.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

A row of A132339.
Cf. A002378, A048147, A098077, A163102.
Cf. A005900, A084980, A077415, A112524, A188475, A100178, A035597, A096000.
Cf. A002378, A046092, A028896 (irregularities of maximal k-degenerate graphs).

a006331 n = sum $ zipWith (*) [2*n-1, 2*n-3 .. 1] [2, 4 ..]
-- Reinhard Zumkeller, Feb 11 2012

[n*(n+1)*(2*n+1)/3: n in [0..40]]; // Vincenzo Librandi, Aug 15 2011

A006331 := proc(n)
    n*(n+1)*(2*n+1)/3 ;
end proc:
seq(A006331(n),n=0..80) ; # R. J. Mathar, Sep 27 2013

Table[n(n+1)(2n+1)/3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,2,10,28},50] (* Harvey P. Dale, Apr 12 2013 *)

PARI
```
a(n)=if(n<0,0,n*(n+1)*(2*n+1)/3)
```

G.f.: 2*x*(1 + x)/(1 - x)^4. - Simon Plouffe (in his 1992 dissertation)
a(n) = 2*binomial(n+1,3) + 2*binomial(n+2,3).
a(n) = 2*A000330(n) = A002492(n)/2.
a(n) = Sum_{i=0..n} T(i,n-i), array T as in A048147. - N. J. A. Sloane, Dec 11 1999
From the formula for the sum of squares of positive integers 1^2 + 2^2 + 3^2 + ... + n^2 = n*(n+1)(2*n+1)/6, if we multiply both sides by 2 we get Sum_{k=0..n} 2*k^2 = n*(n+1)*(2*n+1)/3, which is an alternative formula for this sequence. - Mike Warburton, Sep 08 2007
10*a(n) = A016755(n) - A001845(n); since A016755 are odd cubes and A001845 centered octahedral numbers, 10*a(n) are the "odd cubes without their octahedral contents." - Damien Pras, Mar 19 2011
a(n) = sum(a*b), where the summing is over all unordered partitions 2*n+1=a+b. - Vladimir Shevelev, May 11 2012
a(n) = binomial(2*n+2, 3)/2. - Ronan Flatley, Dec 13 2012
a(n) = A000292(n) + A002411(n). - Omar E. Pol, Jan 11 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3, with a(0)=0, a(1)=2, a(2)=10, a(3)=28. - Harvey P. Dale, Apr 12 2013
a(n) = A208532(n+1,2). - Philippe Deléham, Dec 05 2013
Sum_{n>0} 1/a(n) = 9 - 12*log(2). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A000292(n-1) + (n+1)*A000217(n). - J. M. Bergot, Sep 02 2015
a(n) = 2*(A000332(n+3) - A000332(n+1)). - Antal Pinter, Sep 20 2015
From Bruno Berselli, May 17 2018: (Start)
a(n) = n*A002378(n) - Sum_{k=0..n-1} A002378(k) for n>0, a(0)=0. Also:
A163102(n) = n*a(n) - Sum_{k=0..n-1} a(k) for n>0, A163102(0)=0. (End)
a(n) = A005900(n) - A000290(n) = A096000(n) - A000578(n+1) = A000578(n+1) - A084980(n+1) = A000578(n+1) - A077415(n)-1 = A112524(n) + 1 = A188475(n) - 1 = A061317(n) - A100178(n) = A035597(n+1) - A006331(n+1). - Bruce J. Nicholson, Jun 24 2018
E.g.f.: (1/3)*exp(x)*x*(6 + 9*x + 2*x^2). - Stefano Spezia, Jan 05 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*Pi - 9. - Amiram Eldar, Jan 04 2022

A006325 4-dimensional analog of centered polygonal numbers.

Original entry on oeis.org

0, 0, 1, 7, 26, 70, 155, 301, 532, 876, 1365, 2035, 2926, 4082, 5551, 7385, 9640, 12376, 15657, 19551, 24130, 29470, 35651, 42757, 50876, 60100, 70525, 82251, 95382, 110026, 126295, 144305, 164176, 186032, 210001, 236215, 264810, 295926

Offset: 0

Albert Rich (Albert_Rich(AT)msn.com)

If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-4) is equal to the number of 6-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
For n>0, a(n+1) is the n-th antidiagonal sum of A213751. - Clark Kimberling, Jun 20 2012
This sequence is the case m=n-1, k=n+3 of b(m,k) = m*(m+1)*((k-2)*m-(k-5))/6, which is the m-th k-gonal pyramidal number. - Luciano Ancora, Apr 11 2015
Starting (1, 7, 26, ...), this is the binomial transform of (1, 6, 13, 12, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 31 2015
Also starting (1, 7, 26, ...), this appears to be the number of magic labelings of the cycle-of-loops graph LOOP X C_4 having magic sum n, where LOOP is the 1-vertex, 1-loop-edge graph. - David J. Seal, Sep 13 2017
The conjecture by David J. Seal is true and easily proved using MacMahon's Omega operators via the "Omega" package for Mathematica authored by Axel Riese (obtaining (up to an offset) the generating function listed in the formula section below). See the second Mathematica program in which the edges of LOOP X C_4 are indexed as in the example below. The Omega package can be downloaded from the link provided in the article by G. E. Andrews et al. - L. Edson Jeffery, Oct 15 2017

			A representation of the LOOP X C_4 graph, with edges and loops indexed as shown, as used in the second Mathematica program below:
.             3         1
.              O_______O
.              |   2   |
.              |4     0|
.              |_______|
.              O   6   O
.             5         7

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
G. E. Andrews, P. Paule and A. Riese, MacMahon's partition analysis III. The Omega package.
David Galvin and Courtney Sharpe, Independent set sequence of linear hyperpaths, arXiv:2409.15555 [math.CO], 2024. See p. 7.
Milan Janjic, Two Enumerative Functions
Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
J. K. Merikoski, R. Kumar and R. A. Rajput, Upper bounds for the largest eigenvalue of a bipartite graph, Electronic Journal of Linear Algebra ISSN 1081-3810, A publication of the International Linear Algebra Society, Volume 26, pp. 168-176, April 2013.
Richard P. Stanley, Examples of Magic Labelings, Unpublished Notes, 1973 [Cached copy, with permission]
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A005900, A169938, A213751, A288486, A019298, A061907, A244497, A244879, A244873, A244880.
Cf. A000027, A000217, A019298, A244497, A244879, A244873, A244880, A293310, A293309 (magic labelings of LOOP X C_k, for k = 1..3,5..10).
Cf. A293311, A293312.

[n*(n-1)*(n^2-n+1)/6: n in [0..40]]; // Vincenzo Librandi, May 22 2011

Table[n*(n-1)*(n^2-n+1)/6, {n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Apr 22 2011 *)
<< Omega.m; n = 4; cond = {}; Do[AppendTo[cond, Sum[a[Mod[2*k - j, 2*n]], {j, 0, 2}] == a[2*n]], {k, 0, n - 1}]; f = OEqSum[Product[x[i]^a[i], {i, 0, 2*n}], cond, u][[1]]; Do[f = OEqR[f, Subscript[u, k]], {k, n}];
(* Generating function: *)
f = Factor[f /. {x[2*n] -> x} /. {x[_] -> 1}]
(* This sequence (with initial zeros dropped): *)
CoefficientList[Series[f, {x, 0, 35}], x] (* L. Edson Jeffery, Oct 15 2017 *)

a(n)=n*(n-1)*(n^2-n+1)/6 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = n*(n-1)*(n^2-n+1)/6.
a(n) = ((n^5 - (n-1)^5) - (n^1 - (n-1)^1))/30 = (n^5 - (n-1)^5 - 1)/30. - Xavier Acloque, Jan 25 2003
The partial sums of the octahedral numbers: a(n+1) = Sum_{i=0..n} A005900(i). - Jonathan Vos Post, Mar 14 2006
G.f.: -x^2*(x+1)^2/(x-1)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009
a(n) = Sum_{i=1..n-1} Sum_{j=1..n-1} min(i,j)^2. - Enrique Pérez Herrero, Jan 15 2013 [Which is just rephrasing the partial sum formula with the Murthy formula in A005900. - R. J. Mathar, Jun 14 2014]
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4. - Yosu Yurramendi, Sep 03 2013
From Daniel Poveda Parrilla, Sep 09 2017: (Start)
a(n) = A169938(n-1)/6.
a(n+1) = A288486(n)/24. (End)
Sum_{n>=2} 1/a(n) = 12 - 2 * sqrt(3) * tanh(sqrt(3)*Pi/2). - Amiram Eldar, Jun 28 2020
E.g.f.: exp(x)*x^2*(3 + 4*x + x^2)/6. - Stefano Spezia, Dec 12 2021

A007588 Stella octangula numbers: a(n) = n(2n^2 - 1).

Original entry on oeis.org

0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, 2651, 3444, 4381, 5474, 6735, 8176, 9809, 11646, 13699, 15980, 18501, 21274, 24311, 27624, 31225, 35126, 39339, 43876, 48749, 53970, 59551, 65504, 71841, 78574, 85715, 93276, 101269, 109706, 118599, 127960

Offset: 0

N. J. A. Sloane

Also as a(n)=(1/6)*(12*n^3-6*n), n>0: structured hexagonal anti-diamond numbers (vertex structure 13) (Cf. A005915 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The only known square stella octangula number for n>1 is a(169) = 169*(2*169^2 - 1) = 9653449 = 3107^2. - Alexander Adamchuk, Jun 02 2008
Ljunggren proved that 9653449 = (13*239)^2 is the only square stella octangula number for n>1. See A229384 and the Wikipedia link. - Jonathan Sondow, Sep 30 2013
4*A007588 = A144138(ChebyshevU[3,n]). - Vladimir Joseph Stephan Orlovsky, Jun 30 2011
If A016813 is regarded as a regular triangle (with leading terms listed in A001844), a(n) provides the row sums of this triangle: 1, 5+9=14, 13+17+21=51 and so on. - J. M. Bergot, Jul 05 2013
Shares its digital root, A267017, with n*(n^2 + 1)/2 ("sum of the next n natural numbers" see A006003). - Peter M. Chema, Aug 28 2016

J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 51.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
W. Ljunggren, Zur Theorie der Gleichung x^2 + 1 = Dy^4, Avh. Norske Vid. Akad. Oslo. I. 1942 (5): 27.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alexander Adamchuk and Vincenzo Librandi, Table of n, a(n) for n = 0..10000 [Alexander Adamchuk computed terms 0 - 169, Jun 02, 2008; Vincenzo Librandi computed the first 10000 terms, Aug 18 2011]
A. Bremner, R. Høibakk and D. Lukkassen, Crossed ladders and Euler’s quartic, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.
John Elias, Nesting Cubes of the Surface Points of a Hexagonal Prism
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Amelia Carolina Sparavigna, Generalized Sum of Stella Octangula Numbers, Politecnico di Torino (Italy, 2021).
Amelia Carolina Sparavigna, Cardano Formula and Some Figurate Numbers, Politecnico di Torino (Italy, 2021).
Eric Weisstein's World of Mathematics, Stella Octangula Number
Wikipedia, Stella octangula number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Backwards differences give star numbers A003154: A003154(n)=a(n)-a(n-1).
1/12*t*(n^3-n)+ n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A001653 = Numbers n such that 2*n^2 - 1 is a square.
a(169) = (A229384(3)*A229384(4))^2.
Cf. A267017, A006003, A135503.
Cf. A002378, A005843, A061317, A062392.

[n*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Aug 18 2011

A007588:=n->n*(2*n^2 - 1); seq(A007588(n), n=0..40); # Wesley Ivan Hurt, Mar 10 2014

Table[ n(2n^2-1), {n,0,169} ] (* Alexander Adamchuk, Jun 02 2008 *)
LinearRecurrence[{4,-6,4,-1},{0,1,14,51},50] (* Harvey P. Dale, Sep 16 2011 *)

PARI
```
a(n)=n*(2*n^2-1)
```

def A007588(n): return n*(2*n**2-1) # Chai Wah Wu, Feb 18 2022

G.f.: x*(1+10*x+x^2)/(1-x)^4.
a(n) = n*A056220(n).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Harvey P. Dale, Sep 16 2011
From Ilya Gutkovskiy, Jul 02 2016: (Start)
E.g.f.: x*(1 + 6*x + 2*x^2)*exp(x).
Dirichlet g.f.: 2*zeta(s-3) - zeta(s-1). (End)
a(n) = A004188(n) + A135503(n). - Miquel Cerda, Dec 25 2016
a(n) = A061317(n) - A005843(n) = A062392(n) - A062392(n-1). - J.S. Seneschal, Jul 01 2025

In the formula given in the 1995 Encyclopedia of Integer Sequences, the second 2 should be an exponent.

A006566 Dodecahedral numbers: a(n) = n(3n - 1)(3n - 2)/2.

Original entry on oeis.org

0, 1, 20, 84, 220, 455, 816, 1330, 2024, 2925, 4060, 5456, 7140, 9139, 11480, 14190, 17296, 20825, 24804, 29260, 34220, 39711, 45760, 52394, 59640, 67525, 76076, 85320, 95284, 105995, 117480, 129766, 142880, 156849, 171700, 187460, 204156

Offset: 0

N. J. A. Sloane

Schlaefli symbol for this polyhedron: {5,3}.
A093485 = first differences; A124388 = second differences; third differences = 27. - Reinhard Zumkeller, Oct 30 2006
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
From Peter Bala, Sep 09 2013: (Start)
a(n) = binomial(3*n,3). Two related sequences are binomial(3*n+1,3) (A228887) and binomial(3*n+2,3) (A228888). The o.g.f.'s for these three sequences are rational functions whose numerator polynomials are obtained from the fourth row [1, 4, 10, 16, 19, 16, 10, 4, 1] of the triangle of trinomial coefficients A027907 by taking every third term:
Sum_{n >= 1} binomial(3*n,3)*x^n = (x + 16*x^2 + 10*x^3)/(1-x)^4;
Sum_{n >= 1} binomial(3*n+1,3)*x^n = (4*x + 19*x^2 + 4*x^3)/(1-x)^4;
Sum_{n >= 1} binomial(3*n+2,3)*x^n = (10*x + 16*x^2 + x^3)/(1-x)^4. (End)

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

T. D. Noe, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Victor Meally, Letter to N. J. A. Sloane, no date.
Ed Pegg Jr, Dodecahedral 2024.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A027907, A035006, A060544, A093485, A124388, A228887, A228888, A295421.

Cf. A000292 (tetrahedral numbers), A000578 (cubes), A005900 (octahedral numbers), A006564 (icosahedral numbers).

a006566 n = n * (3 * n - 1) * (3 * n - 2) `div` 2
a006566_list = scanl (+) 0 a093485_list  -- Reinhard Zumkeller, Jun 16 2013

[n*(3*n-1)*(3*n-2)/2: n in [0..40]]; // Vincenzo Librandi, Dec 11 2015

A006566:=(1+16*z+10*z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation

Table[n(3n-1)(3n-2)/2,{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,20,84},40] (* Harvey P. Dale, Jul 24 2013 *)
CoefficientList[Series[x (1 + 16 x + 10 x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 11 2015 *)

PARI
```
a(n)=n*(3*n-1)*(3*n-2)/2
```

G.f.: x(1 + 16x + 10x^2)/(1 - x)^4.
a(n) = A000292(3n-3) = A054776(n)/6 = n*A060544(n).
a(n) = C(n+2,3) + 16 C(n+1,3) + 10 C(n,3).
a(0)=0, a(1)=1, a(2)=20, a(3)=84, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Jul 24 2013
a(n) = binomial(3*n,3). a(-n) = - A228888(n). Sum_{n>=1} 1/a(n) = 1/2*( sqrt(3)*Pi - 3*log(3) ). Sum_{n>=1} (-1)^n/a(n) = 1/3*sqrt(3)*Pi - 4*log(2). - Peter Bala, Sep 09 2013
a(n) = A006564(n) + A035006(n). - Peter M. Chema, May 04 2016
E.g.f.: x*(2 + 18*x + 9*x^2)*exp(x)/2. - Ilya Gutkovskiy, May 04 2016
From Amiram Eldar, Jan 09 2024: (Start)
Sum_{n>=1} 1/a(n) = (sqrt(3)*Pi - 3*log(3))/2 (A295421).
Sum_{n>=1} (-1)^(n+1)/a(n) = (12*log(2) - sqrt(3)*Pi)/3. (End)

More terms from Henry Bottomley, Nov 23 2001

A058187 Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.

Original entry on oeis.org

1, 1, 4, 4, 10, 10, 20, 20, 35, 35, 56, 56, 84, 84, 120, 120, 165, 165, 220, 220, 286, 286, 364, 364, 455, 455, 560, 560, 680, 680, 816, 816, 969, 969, 1140, 1140, 1330, 1330, 1540, 1540, 1771, 1771, 2024, 2024, 2300, 2300, 2600, 2600, 2925, 2925, 3276, 3276

Offset: 0

Henry Bottomley, Nov 20 2000

For n >= i, i = 6,7, a(n - i) is the number of incongruent two-color bracelets of n beads, i of which are black (cf. A005513, A032280), having a diameter of symmetry. The latter means the following: if we imagine (0,1)-beads as points (with the corresponding labels) dividing a circumference of a bracelet into n identical parts, then a diameter of symmetry is a diameter (connecting two beads or not) such that a 180-degree turn of one of two sets of points around it (obtained by splitting the circumference by this diameter) leads to the coincidence of the two sets (including their labels). - Vladimir Shevelev, May 03 2011
From Johannes W. Meijer, May 20 2011: (Start)
The Kn11, Kn12, Kn13, Fi1 and Ze1 triangle sums, see A180662 for their definitions, of the Connell-Pol triangle A159797 are linear sums of shifted versions of the duplicated tetrahedral numbers, e.g., Fi1(n) = a(n-1) + 5*a(n-2) + a(n-3) + 5*a(n-4).
The Kn11, Kn12, Kn13, Kn21, Kn22, Kn23, Fi1, Fi2, Ze1 and Ze2 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above. (End)
The number of quadruples of integers [x, u, v, w] that satisfy x > u > v > w >= 0, n + 5 = x + u. - Michael Somos, Feb 09 2015
Also, this sequence is the fourth column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - Mohammad K. Azarian, Jul 18 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..880
Hansraj Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., Vol. 10, No. 8 (1979), 964-999.
Vladimir Shevelev, A problem of enumeration of two-color bracelets with several variations, arXiv:0710.1370 [math.CO], 2007-2011.
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A057884. Sum of 2 consecutive terms gives A006918, whose sum of 2 consecutive terms gives A002623, whose sum of 2 consecutive terms gives A000292, which is this sequence without the duplication. Continuing to sum 2 consecutive terms gives A000330, A005900, A001845, A008412 successively.
Cf. A096338, A005513, A032280
Cf. A000292, A190717, A190718. - Johannes W. Meijer, May 20 2011

a058187 n = a058187_list !! n
a058187_list = 1 : f 1 1 [1] where
   f x y zs = z : f (x + y) (1 - y) (z:zs) where
     z = sum $ zipWith (*) [1..x] [x,x-1..1]
-- Reinhard Zumkeller, Dec 21 2011

A058187:= proc(n) option remember; A058187(n):= binomial(floor(n/2)+3, 3) end: seq(A058187(n), n=0..51); # Johannes W. Meijer, May 20 2011

a[n_]:= Length @ FindInstance[{x>u, u>v, v>w, w>=0, x+u==n+5}, {x, u, v, w}, Integers, 10^9]; (* Michael Somos, Feb 09 2015 *)
With[{tetra=Binomial[Range[30]+2,3]},Riffle[tetra,tetra]] (* Harvey P. Dale, Mar 22 2015 *)

{a(n) = binomial(n\2+3, 3)}; /* Michael Somos, Jun 07 2005 */

[binomial((n//2)+3, 3) for n in (0..60)] # G. C. Greubel, Feb 18 2022

a(n) = A006918(n+1) - a(n-1).
a(2*n) = a(2*n+1) = A000292(n) = (n+1)*(n+2)*(n+3)/6.
a(n) = (2*n^3 + 21*n^2 + 67*n + 63)/96 + (n^2 + 7*n + 11)(-1)^n/32. - Paul Barry, Aug 19 2003
a(n) = A108299(n-3,n)*(-1)^floor(n/2) for n > 2. - Reinhard Zumkeller, Jun 01 2005
Euler transform of finite sequence [1, 3]. - Michael Somos, Jun 07 2005
G.f.: 1 / ((1 - x) * (1 - x^2)^3) = 1 / ((1 + x)^3 * (1 - x)^4). a(n) = -a(-7-n) for all n in Z.
a(n) = binomial(floor(n/2) + 3, 3). - Vladimir Shevelev, May 03 2011
a(-n) = -a(n-7); a(n) = A000292(A008619(n)). - Guenther Schrack, Sep 13 2018
Sum_{n>=0} 1/a(n) = 3. - Amiram Eldar, Aug 18 2022

A004068 Number of atoms in a decahedron with n shells.

Original entry on oeis.org

0, 1, 7, 23, 54, 105, 181, 287, 428, 609, 835, 1111, 1442, 1833, 2289, 2815, 3416, 4097, 4863, 5719, 6670, 7721, 8877, 10143, 11524, 13025, 14651, 16407, 18298, 20329, 22505, 24831, 27312, 29953, 32759, 35735, 38886, 42217, 45733, 49439

Offset: 0

Albert D. Rich (Albert_Rich(AT)msn.com)

Also as a(n)=(n/6)*(5*n^2+1), n>0: structured pentagonal diamond numbers (vertex structure 6) (cf. A081436 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of atoms in decahedron with n shells, number = 5/6*(n^3) + 1/6*(n) (T. P. Martin, Shells of atoms, eq.(3)). - Brigitte Stepanov, Jul 02 2011
a(n+1) is the number of triples (w,x,y) having all terms in {0,...,n} and x+y >= w. - Clark Kimberling, Jun 14 2012
a(n) = Sum_{k=1..n} A215630(n,k) for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n) - a(n-2) = A010001(n-1), for n>1. - K. G. Stier, Dec 21 2012
a(n) is also a figurate number representing a cube of side n with a vertex cut off by a tetrahedron of side n-1. As such, a(n) = A000578(n) - A000292(n-1), n > 0. - Jean M. Morales, Aug 11 2013
The sequence starting with 1 is the third partial sum of (1, 4, 5, 5, 5, ...) and the binomial transform of (1, 6, 10, 5, 0, 0, 0, ...). - Gary W. Adamson, Sep 27 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (3).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

Cf. A000292, A005891, A006322 (partial sums), A010001, A081436, A100145, A215630.

[5*n^3/6+n/6: n in [0..50]]; // Vincenzo Librandi, May 15 2011

Table[5*n^3/6+n/6,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)

A004068(n):=5*n^3/6+n/6$ makelist(A004068(n),n,0,20); /* Martin Ettl, Jan 07 2013 */

a(n)=5*n^3/6+n/6 \\ Charles R Greathouse IV, Sep 24 2015

def A004068(n): return n*(5*n**2+1)//6 # Chai Wah Wu, Mar 25 2025

a(n) = 5*binomial(n + 1, 3) + binomial(n, 1).
a(n) = 5*n^3/6 + n/6.
a(n) = Sum_{i=0..n-1} A005891(i). - Xavier Acloque, Oct 08 2003
G.f.: x*(1+3*x+x^2) / (1-x)^4. - R. J. Mathar, Jun 05 2011
E.g.f.: (x/6)*(5x^2 + 15x + 6)*exp(x). - G. C. Greubel, Sep 27 2015
Sum_{n>0} 1/a(n) = 3*(2*gamma + polygamma(0, 1-i/sqrt(5)) + polygamma(0, 1+i/sqrt(5))) = 1.233988011257952852492845364799197179252... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

Typo in definition corrected by Jean M. Morales, Aug 11 2013

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

Original entry on oeis.org

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

cp's OEIS Frontend

A004006 a(n) = C(n,1) + C(n,2) + C(n,3), or n*(n^2 + 5)/6.

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A006527 a(n) = (n^3 + 2*n)/3.

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A063496 a(n) = (2*n - 1)*(8*n^2 - 8*n + 3)/3.

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A006331 a(n) = n*(n+1)*(2*n+1)/3.

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A006325 4-dimensional analog of centered polygonal numbers.

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A063496 a(n) = (2n - 1)(8n^2 - 8n + 3)/3.

A006331 a(n) = n(n+1)(2*n+1)/3.

A007588 Stella octangula numbers: a(n) = n(2n^2 - 1).

A006566 Dodecahedral numbers: a(n) = n(3n - 1)(3n - 2)/2.

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