A053134 -id:A053134 - cp's OEIS frontend

A000332 Binomial coefficient binomial(n,4) = n(n-1)(n-2)*(n-3)/24.

0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410

Offset: 0

N. J. A. Sloane

Number of intersection points of diagonals of convex n-gon where no more than two diagonals intersect at any point in the interior.
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation. - Ignacio Larrosa Cañestro, Apr 09 2002. [See Les Reid link for proof. - N. J. A. Sloane, Apr 02 2016] [See Peter Kagey link for alternate proof. - Sameer Gauria, Jul 29 2025]
Start from cubane and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink on chemistry. - Robert G. Wilson v, Aug 02 2002
For n>0, a(n) = (-1/8)*(coefficient of x in Zagier's polynomial P_(2n,n)). (Zagier's polynomials are used by PARI/GP for acceleration of alternating or positive series.)
Figurate numbers based on the 4-dimensional regular convex polytope called the regular 4-simplex, pentachoron, 5-cell, pentatope or 4-hypertetrahedron with Schlaefli symbol {3,3,3}. a(n)=((n*(n-1)*(n-2)*(n-3))/4!). - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 07 2009
Maximal number of crossings that can be created by connecting n vertices with straight lines. - Cameron Redsell-Montgomerie (credsell(AT)uoguelph.ca), Jan 30 2007
If X is an n-set and Y a fixed (n-1)-subset of X then a(n) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Product of four consecutive numbers divided by 24. - Artur Jasinski, Dec 02 2007
The only prime in this sequence is 5. - Artur Jasinski, Dec 02 2007
For strings consisting entirely of 0's and 1's, the number of distinct arrangements of four 1's such that 1's are not adjacent. The shortest possible string is 7 characters, of which there is only one solution: 1010101, corresponding to a(5). An eight-character string has 5 solutions, nine has 15, ten has 35 and so on, congruent to A000332. - Gil Broussard, Mar 19 2008
For a(n)>0, a(n) is pentagonal if and only if 3 does not divide n. All terms belong to the generalized pentagonal sequence (A001318). Cf. A000326, A145919, A145920. - Matthew Vandermast, Oct 28 2008
Nonzero terms = row sums of triangle A158824. - Gary W. Adamson, Mar 28 2009
Except for the 4 initial 0's, is equivalent to the partial sums of the tetrahedral numbers A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009
If the first 3 zeros are disregarded, that is, if one looks at binomial(n+3, 4) with n>=0, then it becomes a 'Matryoshka doll' sequence with alpha=0: seq(add(add(add(i,i=alpha..k),k=alpha..n),n=alpha..m),m=alpha..50). - Peter Luschny, Jul 14 2009
For n>=1, a(n) is the number of n-digit numbers the binary expansion of which contains two runs of 0's. - Vladimir Shevelev, Jul 30 2010
For n>0, a(n) is the number of crossing set partitions of {1,2,..,n} into n-2 blocks. - Peter Luschny, Apr 29 2011
The Kn3, Ca3 and Gi3 triangle sums of A139600 are related to the sequence given above, e.g., Gi3(n) = 2*A000332(n+3) - A000332(n+2) + 7*A000332(n+1). For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 29 2011
For n > 3, a(n) is the hyper-Wiener index of the path graph on n-2 vertices. - Emeric Deutsch, Feb 15 2012
Except for the four initial zeros, number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts. - V.J. Pohjola, Aug 31 2012
a(n+3) is the number of different ways to color the faces (or the vertices) of a regular tetrahedron with n colors if we count mirror images as the same.
a(n) = fallfac(n,4)/4! is also the number of independent components of an antisymmetric tensor of rank 4 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Does not satisfy Benford's law [Ross, 2012] - N. J. A. Sloane, Feb 12 2017
Number of chiral pairs of colorings of the vertices (or faces) of a regular tetrahedron with n available colors. Chiral colorings come in pairs, each the reflection of the other. - Robert A. Russell, Jan 22 2020
From Mircea Dan Rus, Aug 26 2020: (Start)
a(n+3) is the number of lattice rectangles (squares included) in a staircase of order n; this is obtained by stacking n rows of consecutive unit lattice squares, aligned either to the left or to the right, which consist of 1, 2, 3, ..., n squares and which are stacked either in the increasing or in the decreasing order of their lengths. Below, there is a staircase or order 4 which contains a(7) = 35 rectangles. [See the Teofil Bogdan and Mircea Dan Rus link, problem 3, under A004320]
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(End)
a(n+4) is the number of strings of length n on an ordered alphabet of 5 letters where the characters in the word are in nondecreasing order. E.g., number of length-2 words is 15: aa,ab,ac,ad,ae,bb,bc,bd,be,cc,cd,ce,dd,de,ee. - Jim Nastos, Jan 18 2021
From Tom Copeland, Jun 07 2021: (Start)
Aside from the zeros, this is the fifth diagonal of the Pascal matrix A007318, the only nonvanishing diagonal (fifth) of the matrix representation IM = (A132440)^4/4! of the differential operator D^4/4!, when acting on the row vector of coefficients of an o.g.f., or power series.
M = e^{IM} is the matrix of coefficients of the Appell sequence p_n(x) = e^{D^4/4!} x^n = e^{b. D} x^n = (b. + x)^n = Sum_{k=0..n} binomial(n,k) b_n x^{n-k}, where the (b.)^n = b_n have the e.g.f. e^{b.t} = e^{t^4/4!}, which is that for A025036 aerated with triple zeros, the first column of M.
See A099174 and A000292 for analogous relationships for the third and fourth diagonals of the Pascal matrix. (End)
For integer m and positive integer r >= 3, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (3 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024

			a(5) = 5 from the five independent components of an antisymmetric tensor A of rank 4 and dimension 5, namely A(1,2,3,4), A(1,2,3,5), A(1,2,4,5), A(1,3,4,5) and A(2,3,4,5). See the Dec 10 2015 comment. - _Wolfdieter Lang_, Dec 10 2015

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 74, Problem 8.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 70.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 7.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 294.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Charles W. Trigg, Mathematical Quickies, New York: Dover Publications, Inc., 1985, p. 53, #191.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 127.

Franklin T. Adams-Watters, Table of n, a(n) for n = 0..1002
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Brandy Amanda Barnette, Counting Convex Sets on Products of Totally Ordered Sets, Masters Theses & Specialist Projects, Paper 1484, 2015.
Gaston A. Brouwer, Jonathan Joe, Abby A. Noble, and Matt Noble, Problems on the Triangular Lattice, arXiv:2405.12321 [math.CO], 2024. Mentions this sequence.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
Paul Erdős, Norbert Kaufman, R. H. Koch, and Arthur Rosenthal, E750 (Interior diagonal points), Amer. Math. Monthly, 54 (Jun, 1947), p. 344.
Th. Grüner, A. Kerber, R. Laue, and M. Meringer, Mathematics for Combinatorial Chemistry.
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See p. 4.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 254.
Milan Janjic, Two Enumerative Functions.
Peter Kagey, A Proof Without Words: Triangles in the Triangular Grid, arXiv:2211.00186 [math.HO], 2022.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Iva Kodrnja and Helena Koncul, Polynomials vanishing on a basis of S_m(Gamma_0(N)), Glasnik Matematički (2024) Vol. 59, No. 79, 313-325. See p. 324.
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 pp. 13, 15.
Tim McDevitt and Kathryn Sutcliffe, A New Look at an Old Triangle Counting Problem. The Mathematics Teacher. Vol. 110, No. 6 (February 2017), pp. 470-474.
Rajesh Kumar Mohapatra and Tzung-Pei Hong, On the Number of Finite Fuzzy Subsets with Analysis of Integer Sequences, Mathematics (2022) Vol. 10, No. 7, 1161.
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
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.
Les Reid, Counting Triangles in an Array.
Les Reid, Counting Triangles in an Array. [Cached copy]
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
Kenneth A. Ross, First Digits of Squares and Cubes, Math. Mag. 85 (2012) 36-42.
Kirill S. Shardakov and Vladimir P. Bubnov, Stochastic Model of a High-Loaded Monitoring System of Data Transmission Network, Selected Papers of the Models and Methods of Information Systems Research Workshop, CEUR Workshop Proceedings, (St. Petersburg, Russia, 2019), 29-34.
Eric Weisstein's World of Mathematics, Composition.
Eric Weisstein's World of Mathematics, Pentatope Number.
Eric Weisstein's World of Mathematics, Pentatope.
A. F. Y. Zhao, Pattern Popularity in Multiply Restricted Permutations, Journal of Integer Sequences, 17 (2014), #14.10.3.
Index to sequences related to pyramidal numbers.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Index entries for sequences related to Benford's law.

binomial(n, k): A161680 (k = 2), A000389 (k = 5), A000579 (k = 6), A000580 (k = 7), A000581 (k = 8), A000582 (k = 9).
Cf. A053134, A053126, A075733, A006322, A006325.
Cf. also A000583, A014820, A092181, A092182, A092183.
Cf. A000217, A000292, A007318 (column k = 4).
Cf. A158824.
Cf. A006008 (Number of ways to color the faces (or vertices) of a regular tetrahedron with n colors when mirror images are counted as two).
Cf. A104712 (third column, k=4).
See A269747 for a 3-D analog.
Cf. A006008 (oriented), A006003 (achiral) tetrahedron colorings.
Row 3 of A325000, col. 4 of A007318.
Cf. A000292, A025036, A099174.

A000332 := List([1..10^2], n -> Binomial(n, 4)); # Muniru A Asiru, Oct 16 2017

[Binomial(n,4): n in [0..50]]; // Vincenzo Librandi, Nov 23 2014

A000332 := n->binomial(n,4); [seq(binomial(n,4), n=0..100)];

Table[ Binomial[n, 4], {n, 0, 45} ] (* corrected by Harvey P. Dale, Aug 22 2011 *)
Table[(n-4)(n-3)(n-2)(n-1)/24, {n, 100}] (* Artur Jasinski, Dec 02 2007 *)
LinearRecurrence[{5,-10,10,-5,1}, {0,0,0,0,1}, 45] (* Harvey P. Dale, Aug 22 2011 *)
CoefficientList[Series[x^4 / (1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 23 2014 *)

PARI
```
a(n)=binomial(n,4);
```

# Starts at a(3), i.e. computes n*(n+1)*(n+2)*(n+3)/24
# which is more in line with A000217 and A000292.
def A000332():
    x, y, z, u = 1, 1, 1, 1
    yield 0
    while True:
        yield x
        x, y, z, u = x + y + z + u + 1, y + z + u + 1, z + u + 1, u + 1
a = A000332(); print([next(a) for i in range(41)]) # Peter Luschny, Aug 03 2019

print([n*(n-1)*(n-2)*(n-3)//24 for n in range(50)])
# Gennady Eremin, Feb 06 2022

a(n) = n*(n-1)*(n-2)*(n-3)/24.
G.f.: x^4/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = n*a(n-1)/(n-4). - Benoit Cloitre, Apr 26 2003, R. J. Mathar, Jul 07 2009
a(n) = Sum_{k=1..n-3} Sum_{i=1..k} i*(i+1)/2. - Benoit Cloitre, Jun 15 2003
Convolution of natural numbers {1, 2, 3, 4, ...} and A000217, the triangular numbers {1, 3, 6, 10, ...}. - Jon Perry, Jun 25 2003
a(n) = A110555(n+1,4). - Reinhard Zumkeller, Jul 27 2005
a(n+1) = ((n^5-(n-1)^5) - (n^3-(n-1)^3))/24 - (n^5-(n-1)^5-1)/30; a(n) = A006322(n-2)-A006325(n-1). - Xavier Acloque, Oct 20 2003; R. J. Mathar, Jul 07 2009
a(4*n+2) = Pyr(n+4, 4*n+2) where the polygonal pyramidal numbers are defined for integers A>2 and B>=0 by Pyr(A, B) = B-th A-gonal pyramid number = ((A-2)*B^3 + 3*B^2 - (A-5)*B)/6; For all positive integers i and the pentagonal number function P(x) = x*(3*x-1)/2: a(3*i-2) = P(P(i)) and a(3*i-1) = P(P(i) + i); 1 + 24*a(n) = (n^2 + 3*n + 1)^2. - Jonathan Vos Post, Nov 15 2004
First differences of A000389(n). - Alexander Adamchuk, Dec 19 2004
For n > 3, the sum of the first n-2 tetrahedral numbers (A000292). - Martin Steven McCormick (mathseq(AT)wazer.net), Apr 06 2005 [Corrected by Doug Bell, Jun 25 2017]
Starting (1, 5, 15, 35, ...), = binomial transform of [1, 4, 6, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
Sum_{n>=4} 1/a(n) = 4/3, from the Taylor expansion of (1-x)^3*log(1-x) in the limit x->1. - R. J. Mathar, Jan 27 2009
A034263(n) = (n+1)*a(n+4) - Sum_{i=0..n+3} a(i). Also A132458(n) = a(n)^2 - a(n-1)^2 for n>0. - Bruno Berselli, Dec 29 2010
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=1. - Harvey P. Dale, Aug 22 2011
a(n) = (binomial(n-1,2)^2 - binomial(n-1,2))/6. - Gary Detlefs, Nov 20 2011
a(n) = Sum_{k=1..n-2} Sum_{i=1..k} i*(n-k-2). - Wesley Ivan Hurt, Sep 25 2013
a(n) = (A000217(A000217(n-2) - 1))/3 = ((((n-2)^2 + (n-2))/2)^2 - (((n-2)^2 + (n-2))/2))/(2*3). - Raphie Frank, Jan 16 2014
Sum_{n>=0} a(n)/n! = e/24. Sum_{n>=3} a(n)/(n-3)! = 73*e/24. See A067764 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
Sum_{n>=4} (-1)^(n+1)/a(n) = 32*log(2) - 64/3 = A242023 = 0.847376444589... . - Richard R. Forberg, Aug 11 2014
4/(Sum_{n>=m} 1/a(n)) = A027480(m-3), for m>=4. - Richard R. Forberg, Aug 12 2014
E.g.f.: x^4*exp(x)/24. - Robert Israel, Nov 23 2014
a(n+3) = C(n,1) + 3*C(n,2) + 3*C(n,3) + C(n,4). Each term indicates the number of ways to use n colors to color a tetrahedron with exactly 1, 2, 3, or 4 colors.
a(n) = A080852(1,n-4). - R. J. Mathar, Jul 28 2016
From Gary W. Adamson, Feb 06 2017: (Start)
G.f.: Starting (1, 5, 14, ...), x/(1-x)^5 can be written
as (x * r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = (1+x)^5;
as (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...) where r(x) = (1+x+x^2)^5;
as (x * r(x) * r(x^4) * r(x^16) * r(x^64) * ...) where r(x) = (1+x+x^2+x^3)^5;
... (as a conjectured infinite set). (End)
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = A006008(n) - a(n+3) = (A006008(n) - A006003(n)) / 2 = a(n+3) - A006003(n).
a(n+3) = A006008(n) - a(n) = (A006008(n) + A006003(n)) / 2 = a(n) + A006003(n).
a(n) = A007318(n,4).
a(n+3) = A325000(3,n). (End)
Product_{n>=5} (1 - 1/a(n)) = cosh(sqrt(15)*Pi/2)/(100*Pi). - Amiram Eldar, Jan 21 2021

Some formulas that referred to another offset corrected by R. J. Mathar, Jul 07 2009

A048854 Triangle read by rows. A generalization of unsigned Lah numbers, called L[4,1].

Original entry on oeis.org

1, 2, 1, 12, 12, 1, 120, 180, 30, 1, 1680, 3360, 840, 56, 1, 30240, 75600, 25200, 2520, 90, 1, 665280, 1995840, 831600, 110880, 5940, 132, 1, 17297280, 60540480, 30270240, 5045040, 360360, 12012, 182, 1, 518918400, 2075673600, 1210809600, 242161920, 21621600, 960960, 21840, 240, 1, 17643225600, 79394515200, 52929676800, 12350257920, 1323241920, 73513440, 2227680, 36720, 306, 1

Offset: 0

Wolfdieter Lang

s(n,x) := Sum_{m=0..n} T(n,m)*x^m are monic polynomials satisfying s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,x)*p(n-k,y), with polynomials p(n,x) = Sum_{m=1..n} A048786(n,m)*x^m (row polynomials of triangle A048786) and p(0,x)=1.
In the umbral calculus (see the Roman reference, p. 21) the s(n,x) are called Sheffer polynomials for(1/sqrt(1+4*t),t/(1+4*t)). Here the Sheffer notation differs. See the W. Lang link under A006232.
For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[4,1], the Sheffer triangle ((1 - 4*t)^(-1/2), t/(1 - 4*t)). It is defined as transition matrix
risefac[4,1](x, n) = Sum_{m=0..n} L[4,1](n, m)*fallfac[4,1](x, m), where risefac[4,1](x, n) := Product_{0..n-1} (x  + (1 + 4*j)) for n >= 1 and risefac[4,1](x, 0) := 1, and  fallfac[4,1](x, n) := Product_{0..n-1} (x - (1 + 4*j)) for n >= 1 and fallfac[4,1](x, 0) := 1.
In matrix notation: L[4,1] = S1phat[4,1]*S2hat[4,1] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A290319 and A111578 (but here with offsets 0), respectively.
The a- and z-sequences for this Sheffer matrix have e.g.f.s Ea(t) = 1 + 4*t and Ez(t) = (1 + 4*t)*(1 - (1 + 4*t)^(-1/2))/t, respectively. That is, a = {1, 4, repeat(0)} and z(n) = 2*A292220(n). See the W. Lang link on a- and z-sequences there.
The inverse matrix T^(-1) = L^(-1)[4,1] is Sheffer ((1 + 4*t)^(-1/2), t/(1 + 4*t)). This means that T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[4,1](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[4,1](x, m), n >= 0.
Diagonal sequences have o.g.f. G(d, x) = A001813(d)*Sum_{m=0..d} A091042(d, m)*x^m/(1 - x)^{2*d + 1}, for d >= 0 (d=0 main diagonal). G(d, x) generates {A001813(d)*binomial(2*(n + d),2*d)}{n >= 0}. See the second W. Lang link on how to compute o.g.f.s of diagonal sequences of general Sheffer triangles. - _Wolfdieter Lang, Oct 12 2017

			The triangle T(n, m) begins:
n\m         0          1          2         3        4      5     6   7 8  ...
0:          1
1:          2          1
2:         12         12          1
3:        120        180         30         1
4:       1680       3360        840        56        1
5:      30240      75600      25200      2520       90      1
6:     665280    1995840     831600    110880     5940    132     1
7:   17297280   60540480   30270240   5045040   360360  12012   182   1
8:  518918400 2075673600 1210809600 242161920 21621600 960960 21840 240 1
...
n = 9: 17643225600 79394515200 52929676800 12350257920 1323241920 73513440 2227680 36720 306 1,
n = 10: 670442572800 3352212864000 2514159648000 670442572800 83805321600 5587021440 211629600 4651200 58140 380 1.
...
Recurrence from a-sequence: T(4, 2) = 2*T(3, 1) + 4*4*T(3, 2) = 2*180 + 16*30 = 840.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2)+ z(3)*T(3, 3)) = 4*(2*120 + 2*180 - 8*30 + 60*1) = 1680.
Four term recurrence: T(4, 2) = T(3, 1) + 2*13*T(3, 2) - 8*3*5*T(2, 2) =  180 + 26*30 - 120*1 = 840.
Meixner type identity for n = 2: (D_x - 4*(D_x)^2)*(12 + 12*x + 1*x^2) = (12 + 2*x) - 4*2 = 2*(2 + x).
Sheffer recurrence for R(3, x): [(2 + x) + 8*(1 + x)*D_x + 16*x*(D_x)^2] (12 + 12*x + 1*x^2) = (2 + x)*(12 + 12*x + x^2) + 8*(1 + x)*(12 + 2*x) + 16*2*x = 120 + 180*x + 30*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (4!*10/2)*(1*30/3! + 4*1/2!) = 840.
Diagonal sequence d = 2: {12, 180, 840 ...} has o.g.f. 12*(1 + 10*x + 5*x^2)/(1 - x)^5  (see A001813(2) and row n=2 of A091042) generating
{12*binomial(2*(n + 2), 4)}_{n >= 0}. - _Wolfdieter Lang_, Oct 12 2017

S. Roman, The Umbral Calculus, Academic Press, New York, 1984.

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
Orli Herscovici, Ross G. Pinsky, An identity involving stirling numbers of both kinds and its connection to right-to-left minima of certain set partitions, (2019).
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017, section C) 4.
Wolfdieter Lang, On Generating functions of Diagonal Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Emanuele Munarini, Combinatorial identities involving the central coefficients of a Sheffer matrix, Applicable Analysis and Discrete Mathematics (2019) Vol. 13, 495-517.

Related to triangle A046521. Cf. A048786. a(n, 0) = A001813.
A111578, A271703 L[1,0], A286724 L[2,1], A290319, A290596 L[3,1], A290597 L[3,2], A292220.
The diagonal sequences are: A000012, 2*A000384(n+1), 12*A053134, 120*A053135, 1680*A053137, ... - Wolfdieter Lang, Oct 12 2017

A290604_row := proc(n) exp(x*t/(1-4*t))/sqrt(1-4*t): series(%, t, n+2): seq(n!*coeff(coeff(%,t,n),x,j), j=0..n) end: seq(A290604_row(n), n=0..9); # Peter Luschny, Sep 23 2017

T[n_, m_] := n!/m! * Binomial[2*n, n] * Binomial[n, m] / Binomial[2*m, m]; Table[a[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013 *)
T[0, 0] = 1; T[-1, ] = T[, -1] = 0; T[n_, m_] /; n < m = 0; T[n_, m_] := T[n, m] = T[n-1, m-1] + 2*(4*n-3)*T[n-1, m] - 8*(n-1)*(2*n-3)*T[n-2, m]; Table[T[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Sep 23 2017 *)

T(n, m) = (n!/m!)*A046521(n, m) = (n!/m!)* binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0, a(n, m) := 0, n < m.
Sum_{n>=0, k>=0} T(n, k)*x^n*y^k/(2*n)! = exp(x)*cosh(sqrt(x*y)). - Vladeta Jovovic, Feb 21 2003
T(n, m) = L[4,1](n,m) = Sum_{k=m..n} A290319(n, k)*A111578(k+1, m+1), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 4*t)^(-1/2)*exp(x*t/(1 - 4*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 4*t)^(-1/2)*(t/(1 - 4*t))^m/m!, m >= 0.
Three term recurrence for column entries m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 4*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0..n-1} z(j)*T(n-1, j), n >= 1, T(0, 0) = 0, from the a-sequence {1, 4 repeat(0)} and the z(j) = 2*A292220(j) (see above).
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(4*n - 3)*T(n-1, m) - 8*(n-1)*(2*n - 3)*T(n-2, m), n >= m >= 0, with T(0, 0) =1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 4*D_x)) * R(n, x) = n * R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-4)^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(2 + x)*1 + 8*(1 + x)*D_x + 16*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (n!/(n-m))*(2 + 4*m)*Sum_{p=0..n-1-m} 4^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.
Explicit form (from the o.g.f.s of diagonal sequences): ((2*(n-m))!/(n-m)!)*binomial(2*n,2*(n-m)), n >= m >= 0, and vanishing for n < m. - Wolfdieter Lang, Oct 12 2017

Name changed, after merging my newer duplicate, from Wolfdieter Lang, Oct 10 2017

A006324 a(n) = n(n + 1)(2n^2 + 2n - 1)/6.

Original entry on oeis.org

1, 11, 46, 130, 295, 581, 1036, 1716, 2685, 4015, 5786, 8086, 11011, 14665, 19160, 24616, 31161, 38931, 48070, 58730, 71071, 85261, 101476, 119900, 140725, 164151, 190386, 219646, 252155, 288145, 327856, 371536, 419441, 471835, 528990, 591186

Offset: 1

Albert Rich (Albert_Rich(AT)msn.com), Jun 14 1998

4-dimensional analog of centered polygonal numbers.
Partial sums of A000447. - Zak Seidov, May 19 2006
From Johannes W. Meijer, Jun 27 2009: (Start)
Equals the absolute values of the coefficients that precede the a(n-1) factors of the recurrence relations RR(n) of A162011.
This sequence enabled the analysis of A162012 and A162013. (End)
Equals the number of integer quadruples (x,y,z,w) such that min(x,y) < min(z,w), max(x,y) < max(z,w), and 0 <= x,y,z,w <= n. - Andrew Woods, Apr 21 2014
For n>3 a(n)=twice the area of an irregular quadrilateral with vertices at the points (C(n,4),C(n+1,4)), (C(n+1,4),C(n+2,4)), (C(n+2,4),C(n+3,4)), and (C(n+3,4),C(n+4,4)). - J. M. Bergot, Jun 14 2014

Delbert L. Johnson, Table of n, a(n) for n = 1..20000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000447, A000539, A000217, A002415, A053134.

Cf. A162011, A162012, a(n-2), and A162013, a(n-3). - Johannes W. Meijer, Jun 27 2009

[ n*(n + 1)*(2*n^2 + 2*n - 1)/6 : n in [1..30] ]; // Wesley Ivan Hurt, Jun 14 2014

A006324:=n->n*(n + 1)*(2*n^2 + 2*n - 1)/6; seq(A006324(n), n=1..30); # Wesley Ivan Hurt, Jun 14 2014

Table[Sum[k^5,{k,n}]/Sum[k,{k,n}], {n,40}] (* Alexander Adamchuk, Apr 12 2006 *)

a(n) = 8*C(n + 2, 4) + C(n + 1, 2).
a(n) = (Sum_{k=1..n} k^5) / (Sum_{k=1..n} k) = A000539(n) / A000217(n). - Alexander Adamchuk, Apr 12 2006
From Johannes W. Meijer, Jun 27 2009: (Start)
Recurrence relation 0 = Sum_{k=0..5} (-1)^k*binomial(5,k)*a(n-k).
G.f.: (1+6*z+z^2)/(1-z)^5. (End)
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Wesley Ivan Hurt, May 02 2021
Sum_{n>=1} 1/a(n) = 6 + 2*sqrt(3)*Pi*tan(sqrt(3)*Pi/2). - Amiram Eldar, Aug 23 2022
a(n) = A053134(n-1) - 4*A002415(n). - Yasser Arath Chavez Reyes, Feb 12 2024

Simpler definition from Alexander Adamchuk, Apr 12 2006

More terms from Zak Seidov

A258993 Triangle read by rows: T(n,k) = binomial(n+k,n-k), k = 0..n-1.

Original entry on oeis.org

1, 1, 3, 1, 6, 5, 1, 10, 15, 7, 1, 15, 35, 28, 9, 1, 21, 70, 84, 45, 11, 1, 28, 126, 210, 165, 66, 13, 1, 36, 210, 462, 495, 286, 91, 15, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1, 66, 715, 3003, 6435, 8008, 6188, 3060, 969, 190, 21

Offset: 1

Reinhard Zumkeller, Jun 22 2015

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

			.  n\k |  0  1    2    3     4     5     6     7    8    9  10 11
. -----+-----------------------------------------------------------
.   1  |  1
.   2  |  1  3
.   3  |  1  6    5
.   4  |  1 10   15    7
.   5  |  1 15   35   28     9
.   6  |  1 21   70   84    45    11
.   7  |  1 28  126  210   165    66    13
.   8  |  1 36  210  462   495   286    91    15
.   9  |  1 45  330  924  1287  1001   455   120   17
.  10  |  1 55  495 1716  3003  3003  1820   680  153   19
.  11  |  1 66  715 3003  6435  8008  6188  3060  969  190  21
.  12  |  1 78 1001 5005 12870 19448 18564 11628 4845 1330 231 23  .

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

If a diagonal of 1's is added on the right, this becomes A085478.
Essentially the same as A143858.
Cf. A007318, A004736, A094727.
Cf. A027941 (row sums), A117671 (central terms), A143858, A000967, A258708.
T(n,k): A000217 (k=1), A000332 (k=2), A000579 (k=3), A000581 (k=4), A001287 (k=5), A010965 (k=6), A010967 (k=7), A010969 (k=8), A010971 (k=9), A010973 (k=10), A010975 (k=11), A010977 (k=12), A010979 (k=13), A010981 (k=14), A010983 (k=15), A010985 (k=16), A010987 (k=17), A010989 (k=18), A010991 (k=19), A010993 (k=20), A010995 (k=21), A010997 (k=22), A010999 (k=23), A011001 (k=24), A017714 (k=25), A017716 (k=26), A017718 (k=27), A017720 (k=28), A017722 (k=29), A017724 (k=30), A017726 (k=31), A017728 (k=32), A017730 (k=33), A017732 (k=34), A017734 (k=35), A017736 (k=36), A017738 (k=37), A017740 (k=38), A017742 (k=39), A017744 (k=40), A017746 (k=41), A017748 (k=42), A017750 (k=43), A017752 (k=44), A017754 (k=45), A017756 (k=46), A017758 (k=47), A017760 (k=48), A017762 (k=49), A017764 (k=50).
T(n+k,n): A005408 (k=1), A000384 (k=2), A000447 (k=3), A053134 (k=4), A002299 (k=5), A053135 (k=6), A053136 (k=7), A053137 (k=8), A053138 (k=9), A196789 (k=10).
Cf. A165253.

Flat(List([1..12], n-> List([0..n-1], k-> Binomial(n+k,n-k) ))); # G. C. Greubel, Aug 01 2019

a258993 n k = a258993_tabl !! (n-1) !! k
a258993_row n = a258993_tabl !! (n-1)
a258993_tabl = zipWith (zipWith a007318) a094727_tabl a004736_tabl

[Binomial(n+k,n-k): k in [0..n-1], n in [1..12]]; // G. C. Greubel, Aug 01 2019

Table[Binomial[n+k,n-k], {n,1,12}, {k,0,n-1}]//Flatten (* G. C. Greubel, Aug 01 2019 *)

T(n,k) = binomial(n+k,n-k);
for(n=1, 12, for(k=0,n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019

[[binomial(n+k,n-k) for k in (0..n-1)] for n in (1..12)] # G. C. Greubel, Aug 01 2019

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

Number of binary pattern classes with 4 black beads in the (2,n)-rectangular grid; two patterns are in the same class if one of them can be obtained by reflection or rotation of the other one. - Yosu Yurramendi, Sep 12 2008

This sequence is the case k=n+3 of b(n,k) = n*(n+1)*((k-2)*n-(k-5))/6, which is the n-th k-gonal pyramidal number. Therefore, apart from 0, this sequence is the 3rd diagonal of the array in A080851. - Luciano Ancora, Apr 10 2015

From Wolfdieter Lang, Sep 19 2012: (Start)
The triangle a(n,k) appears in the formula F(2*l+1)^(2*n) = (sum(a(n,k)*L(2*(n-k)*(2*l+1)),k=0..n-1) + a(n,n))/5^n, n>=0, l>=0, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
The signed triangle as(n,k):=a(n,k)*(-1)^k appears in the formula F(2*l)^(2*n) = (sum(as(n,k)*L(4*(n-k)*l),k=0..n-1) + as(n,n))/5^n, n>=0, l>=0. Proof with the Binet-de Moivre formula for F and L and the binomial formula. (End)

cp's OEIS Frontend

A000332 Binomial coefficient binomial(n,4) = n*(n-1)*(n-2)*(n-3)/24.

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A048854 Triangle read by rows. A generalization of unsigned Lah numbers, called L[4,1].

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A006324 a(n) = n*(n + 1)*(2*n^2 + 2*n - 1)/6.

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A258993 Triangle read by rows: T(n,k) = binomial(n+k,n-k), k = 0..n-1.

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A071239 a(n) = n*(n+1)*(n^2 + 2)/6.

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A053135 Binomial coefficients C(2*n+6,6).

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A000332 Binomial coefficient binomial(n,4) = n(n-1)(n-2)*(n-3)/24.

A006324 a(n) = n(n + 1)(2n^2 + 2n - 1)/6.

A071239 a(n) = n(n+1)(n^2 + 2)/6.