A027555 -id:A027555 - cp's OEIS frontend

A002294 a(n) = binomial(5n, n)/(4n + 1).

1, 1, 5, 35, 285, 2530, 23751, 231880, 2330445, 23950355, 250543370, 2658968130, 28558343775, 309831575760, 3390416787880, 37377257159280, 414741863546285, 4628362722856425, 51912988256282175, 584909606696793885, 6617078646960613370

Offset: 0

N. J. A. Sloane

From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates quintic trees (rooted, ordered, incomplete) with n vertices (including the root).
This is the Pfaff-Fuss-Catalan sequence C^{m}_n for m = 5. See the Graham et al. reference, p. 347. eq. 7.66. See also the Pólya-Szegő reference.
Also 5-Raney sequence. See the Graham et al. reference, pp. 346-347. (End)
a(n) = A258708(3*n, 2*n) for n > 0. - Reinhard Zumkeller, Jun 23 2015
Conjecturally, a(n) is the number of 4-uniform words on the alphabet [n] that avoid the patterns 231 and 221 (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
From Stillwell (1995), p. 62: "Eisenstein's Theorem. If y^5 + y = x, then y has a power series expansion y = x - x^5 + 10*x^9/2^1 - 15 * 14 * x^13/3! + 20 * 19 * 18*x^17/4! - ...." - Michael Somos, Sep 19 2019
a(n) is the total number of down steps before the first up step in all 4_1-Dyck paths of length 5*n. A 4_1-Dyck path is a lattice path with steps (1, 4), (1, -1) that starts and ends at y = 0 and stays above the line y = -1. - Sarah Selkirk, May 10 2020
Dropping the first 1 (starting from 1, 5, 35, ... with offset 1), the series reversion gives 1, -5, 15, -35, 70, ... (again offset 1), essentially A000332 and row 5 of A027555. - R. J. Mathar, Aug 17 2023
Number of rooted polyominoes composed of n hexagonal cells of the hyperbolic regular tiling with Schläfli symbol {6,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {6,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 5 of the generalized Catalan family {C(k, n)}_{n>=0} given in a comment of A130564. - Wolfdieter Lang, Feb 05 2024

			There are a(2) = 5 quintic trees (vertex degree <= 5 and 5 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these five trees yields 5*5 + binomial(5,2) = 35 = a(3) such trees.
G.f. = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + 23751*x^6 + 231880*x^7 + ...
G.f. = t + t^5 + 5*t^9 + 35*t^13 + 285*t^17 + 2530*t^21 + 23751*t^25 + 231880*t^29 + ...

Archiv der Mathematik u. Physik, Editor's note: "Über die Bestimmung der Anzahl der verschiedenen Arten, auf welche sich ein n-Eck durch Diagonalen in lauter m-Ecke zerlegen laesst, mit Bezug auf einige Abhandlungen der Herren Lame, Rodrigues, Binet, Catalan und Duhamel in dem Journal de Mathematiques pures et appliquees, publie par Joseph Liouville. T. III. IV.", Archiv der Mathematik u. Physik, 1 (1841), pp. 193ff; see especially p. 198.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 23.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347.
G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, Heidelberg, New York, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nürnberg, Jul 27 1994.
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).

T. D. Noe, Table of n, a(n) for n = 0..100
V. E. Adler and A. B. Shabat, Volterra chain and Catalan numbers, arXiv:1810.13198 [nlin.SI], 2018.
Joerg Arndt, Matters Computational (The Fxtbook), pp. 337-338.
Joerg Arndt, Subset-lex: did we miss an order?, arXiv:1405.6503 [math.CO], 2014-2015.
A. Asinowski, B. Hackl, and S. Selkirk, Down step statistics in generalized Dyck paths, arXiv:2007.15562 [math.CO], 2020.
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
L. W. Beineke and R. E. Pippert, On the enumeration of planar trees of hexagons, Glasgow Math. J., 15 (1974), 131-147.
L. W. Beineke and R. E. Pippert, On the enumeration of planar trees of hexagons, Glasgow Math. J., 15 (1974), 131-147. [Annotated scanned copy]
Frits Beukers, Hypergeometric functions, how special are they?, Notices Amer. Math. Soc. 61(1) (2014), 48-56. MR3137256
Wun-Seng Chou, Tian-Xiao He, and Peter J.-S. Shiue, On the Primality of the Generalized Fuss-Catalan Numbers, J. Int. Seqs., Vol. 21 (2018), #18.2.1.
Malin Christensson, Make hyperbolic tilings of images, web page, 2019.
Olivier Danvy, Summa Summarum: Moessner's Theorem without Dynamic Programming, arXiv:2412.03127 [cs.DM], 2024. See p. 25.
C. Defant and N. Kravitz, Stack-sorting for words, arXiv:1809.09158 [math.CO], 2018.
R. W. Gosper, Rope around the earth
F. Harary, E. M. Palmer and R. C. Read, On the cell-growth problem for arbitrary polygons, Discr. Math. 11 (1975), 371-389.
F. Harary, E. M. Palmer, and R. C. Read, On the cell-growth problem for arbitrary polygons, computer printout, circa 1974.
Clemens Heuberger, Sarah J. Selkirk, and Stephan Wagner, Enumeration of Generalized Dyck Paths Based on the Height of Down-Steps Modulo k, arXiv:2204.14023 [math.CO], 2022.
Forrest M. Hilton, Finite Dynamical Laminations, arXiv:2408.01353 [math.DS], 2024. See p. 7.
V. E. Hoggatt, Jr., 7-page typed letter to N. J. A. Sloane with suggestions for new sequences, circa 1977.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 287.
R. P. Loh, A. G. Shannon, and A. F. Horadam, Divisibility Criteria and Sequence Generators Associated with Fermat Coefficients, preprint, 1980.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv:1403.5962 [math.CO], 2014.
Mitchell Paukner, Lucy Pepin, Manda Riehl, and Jarred Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015-2016.
Karol A. Penson and Karol Zyczkowski, Product of Ginibre matrices : Fuss-Catalan and Raney distribution, arXiv version, arXiv:1103.3453 [math-ph], 2011.
Yuan (Friedrich) Qiu, Joe Sawada, and Aaron Williams, Maximize the Rightmost Digit: Gray Codes for Restricted Growth Strings, WALCOM 2025. See p. 5.
John Stillwell, Eisenstein's footnote, Math. Intelligencer 17(2) (1995), 58-62. MR1336074 (96d:01024)
B. Sury, Generalized Catalan numbers: linear recursion and divisibility, JIS 12 (2009), Article 09.7.5.
L. Takacs, Enumeration of rooted trees and forests, Math. Scientist 18 (1993), 1-10, esp. Eq. (5).
Wikipedia, Fuss-Catalan number.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.
S. Yakoubov, Pattern Avoidance in Extensions of Comb-Like Posets, arXiv:1310.2979 [math.CO], 2013-2014.
Jun Yan, Lattice paths enumerations weighted by ascent lengths, arXiv:2501.01152 [math.CO], 2025. See p. 7.

Cf. A001764, A002296, A258708, A346647 (binomial transform), A346665 (inverse binomial transform).
Fourth column of triangle A062993.
Polyominoes: A221184{n-1} (oriented), A004127 (unoriented), A369473 (chiral), A143546 (achiral), A002293 {5,oo}, A002295 {7,oo}.
Cf. A130564.

List([0..22],n->Binomial(5*n,n)/(4*n+1)); # Muniru A Asiru, Nov 01 2018

a002294 n = a002294_list !! n
a002294_list = [a258708 (3 * n) (2 * n) | n <- [1..]]
-- Reinhard Zumkeller, Jun 23 2015

[ Binomial(5*n,n)/(4*n+1): n in [0..100]]; // Vincenzo Librandi, Mar 24 2011

seq(binomial(5*k+1,k)/(5*k+1),k=0..30); # Robert FERREOL, Apr 03 2015
n:=30:G:=series(RootOf(g = 1+x*g^5, g),x=0,n+1):seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 03 2015

CoefficientList[InverseSeries[ Series[ y - y^5, {y, 0, 100}], x], x][[Range[2, 100, 4]]]
Table[Binomial[5n,n]/(4n+1),{n,0,20}] (* Harvey P. Dale, Dec 30 2011 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1, 2, 3, 4}/5, {2, 3, 5}/4, x 5^5/4^4], {x, 0, n}]; (* Michael Somos, May 06 2015 *)
a[ n_] := With[{m = 4 n + 1}, SeriesCoefficient[ InverseSeries @ Series[ x - x^5, {x, 0, m}], {x, 0, m}]]; (* Michael Somos, May 06 2015 *)

{a(n) = binomial( 5 * n, n) / (4*n + 1)}; /* Michael Somos, Mar 17 2011 */

{a(n) = if( n<0, 0, n = 4*n + 1; polcoeff( serreverse( x - x^5 + x * O(x^n) ), n))}; /* Michael Somos, Mar 17 2011 */

For the connection with the solution of the quintic, hypergeometric series, and Lagrange inversion, see Beukers (2014). - N. J. A. Sloane, Mar 12 2014
G.f.: hypergeometric([1, 2, 3, 4] / 5, [2, 3, 5] / 4, x * 5^5 / 4^4). - Michael Somos, Mar 17 2011
O.g.f. A(x) satisfies A(x) = 1 + x * A(x)^5 = 1 / (1 - x * A(x)^4).
Given g.f. A(x) then z = t * A(t^4) satisfies 0 = z^5 - z + t. - Michael Somos, Mar 17 2011
a(n) = binomial(5*n, n - 1)/n, n >= 1, a(0) = 1. From the Lagrange series of the o.g.f. A(x) with its above given implicit equation.
a(n) = upper left term in M^n, M = the production matrix:
   1,  1;
   4,  4,  1;
  10, 10,  4,  1;
  20, 20, 10,  4,  1;
  ...
where (1, 4, 10, 20, ...) is the tetrahedral sequence, A000292. - Gary W. Adamson, Jul 08 2011
D-finite with recurrence: 8*n*(4*n+1)*(2*n-1)*(4*n-1)*a(n) - 5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1) = 0. - R. J. Mathar, Dec 02 2014
a(n) = binomial(5*n + 1, n)/(5*n + 1) = A062993(n+3,3). - Robert FERREOL, Apr 03 2015
a(0) = 1; a(n) = Sum_{i1 + i2 + ... + i5 = n - 1} a(i1) * a(i2) * ... *a(i5) for n >= 1. - Robert FERREOL, Apr 03 2015
From Ilya Gutkovskiy, Jan 15 2017: (Start)
O.g.f.: 5F4([1/5, 2/5, 3/5, 4/5, 1]; [1/2, 3/4, 1, 5/4]; 3125*x/256).[Cancellation of the 1s, see G.f. the above. - Wolfdieter Lang, Feb 05 2024]
E.g.f.: 4F4([1/5, 2/5, 3/5, 4/5]; [1/2, 3/4, 1, 5/4]; 3125*x/256).
a(n) ~ 5^(5*n + 1/2)/(sqrt(Pi) * 2^(8*n + 7/2) * n^(3/2)). (End)
x*A'(x)/A(x) = (A(x) - 1)/(- 4*A(x) + 5) = x + 9*x^2 + 91*x^3 + 969*x^4 + ... is the o.g.f. of A163456. Cf. A001764 and A002293 - A002296. - Peter Bala, Feb 04 2022
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^9). - Seiichi Manyama, Jun 16 2025

More terms from Olivier Gérard, Jul 05 2001

A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

Original entry on oeis.org

0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422

Offset: 0

Clark Kimberling

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

			4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - _Bruno Berselli_, Feb 05 2014

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

T. D. Noe, Table of n, a(n) for n = 0..1000
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.
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 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. 16 (2013) #13.5.7.
Alexander R. Povolotsky, Problem 1147, Pi Mu Epsilon Fall 2006 Problems.
Alexander R. Povolotsky, Problem, Pi Mu Epsilon Spring 2007 Problems.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Cf. A000292, A000332, A000389, A000579, A000580, A024166, A027555, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086025, A086026, A086027, A086028, A086029, A086030, A087127.

Cf. A000330, A000537 (first diffs), A001286, A003215, A100536, A101094 (partial sums), A101097, A101102, A222716.

a024166 n = sum $ zipWith (*) [n+1,n..0] a000578_list
-- Reinhard Zumkeller, Oct 14 2001

[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017

A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017

Nest[Accumulate,Range[0,40]^3,2] (* Harvey P. Dale, Jan 10 2016 *)
Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n,0,30}] (* G. C. Greubel, Nov 21 2017 *)

a(n)=sum(j=1,n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008

for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022

A087127 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of triangular numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 2p-1, where a(i,p) satisfies Sum_{i=1..n} C(i+1,2)^p = 3 C(n+2,3) * Sum_{i=1..2p-1} a(i,p) C(n-1,i-1)/(i+2).

Original entry on oeis.org

1, 1, 2, 1, 1, 8, 19, 18, 6, 1, 26, 163, 432, 564, 360, 90, 1, 80, 1135, 6354, 18078, 28800, 26100, 12600, 2520, 1, 242, 7291, 77400, 405060, 1210680, 2211570, 2520000, 1751400, 680400, 113400, 1, 728, 45199, 862218, 7667646, 38350080, 118848420

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 08 2018: (Start)
The table entries T(n,k) are the coefficients when expressing the polynomial C(x+2,2)^p of degree 2*p in terms of falling factorials: C(x+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 0..2*p} T(p,k)*C(n,k+1).
The sum of the p-th powers of the triangular numbers is also given by Sum_{i = 0..n-1} C(i+2,2)^p = Sum_{k = 2..2*p} A122193(p,k)*C(n+2,k+1) for p >= 1. (End)

			Row 3 contains 1,8,19,18,6, so Sum_{i=1..n} C(i+1,2)^3 = (n+2) * C(n+1,2) * [ a(1,3)/3 + a(2,3)*C(n-1,1)/4 + a(3,3)*C(n-1,2)/5 + a(4,3)*C(n-1,3)/6 + a(5,3)*C(n-1,4)/7 ] = [ (n+2)*(n+1)*n/2 ] * [ 1/3 + (8/4)*C(n-1,1) + (19/5)*C(n-1,2) + (18/6)*C(n-1,3) + (6/7)*C(n-1,4). Cf. A085438 for more details.
From _Peter Bala_, Mar 08 2018: (Start)
Table begins
n=0 |1
n=1 |1   2     1
n=2 |1   8    19    18      6
n=3 |1  26   163   432    564    360     90
n=4 |1  80  1135  6354  18078  28800  26100  12600  2520
...
Row 2: C(i+2,2)^2 = C(i,0) + 8*C(i,1) + 19*C(i,2) + 18*C(i,3) + 6*C(i,4). Hence, Sum_{i = 0..n-1} C(i+2,2)^2 =  C(n,1) + 8*C(n,2) + 19*C(n,3) + 18*C(n,4) + 6*C(n,5). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
M. Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A000217, A122193.

Flat(List([0..6],n->List([0..2*n],k->Sum([0..k],i->(-1)^(k-i)*Binomial(k,i)*Binomial(i+2,2)^n)))); # Muniru A Asiru, Mar 22 2018

seq(seq(add( (-1)^(k-i)*binomial(k,i)*binomial(i+2,2)^n, i = 0..k), k = 0..2*n), n = 0..8); # Peter Bala, Mar 08 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 3, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 2, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 2*p - 1}]//Flatten (* G. C. Greubel, Nov 23 2017 *)
a[i_,p_]:=(-1)^i HypergeometricPFQ[ConstantArray[3,p]~Join~{2-i},ConstantArray[1,p],1];Table[a[i,p],{p,0,10},{i,2,2 p+2}]//Flatten (* Jonathan Burns, Mar 20 2018 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 3, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 2, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 2*p-1, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(1, p) = 1, a(2, p) = 3^(p-1)-1, a(3, p) = 3^(p-1)*[2^(p-1)-2]+1, ..., a(2*p-3, p) = [ (6*p^4-20*p^3+21*p^2-7*p)*(2*p-4)! ]/[3*2^(p-1)], a(2*p-2, p) = [ (p^2-p)*(2*p-3)! ]/2^(p-2), a(2*p-1, p) = [ (p-1)*(2*p-3)! ]/2^(p-2).
a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+3, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+2, i-2*k)^(p-1) ]
From Peter Bala, Mar 08 2018: (Start)
The following remarks assume row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*C(i+2,2)^n. Equivalently, let v_n denote the sequence (1, 3^n, 6^n, 10^n, ...) regarded as an infinite column vector, where 1, 3, 6, 10, ... is the sequence of triangular numbers A000217. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318. Cf. A122193.
T(n+1,k) = C(k+2,2)*T(n,k) + 2*C(k+1,2)*T(n,k-1) + C(k,2)*T(n,k-2), with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 2*n.
Let R(n,x) denote the n-th row polynomial.
R(n+1,x) = 1/2!*(1 + x)^2*(d/dx)^2 (x^2*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+2,2)^n*x^i/(1 + x)^(i+1).
R(n,x) = (1 + x)^2 o (1 + x)^2 o ... o (1 + x)^2 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^2 o ... o x^2 (n factors) is the n-th row polynomial of A122193.
x^2*R(n,x) = (1 + x)^2 * the n-th row polynomial of A122193 (End)

Edited by Dean Hickerson, Aug 16 2003

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378

Offset: 0

Fabian Rothelius, Feb 04 2001

T(n,k) is the number of ways to distribute k identical objects in n distinct containers; containers may be left empty.
T(n,k) is the number of nondecreasing functions f from {1,...,k} to {1,...,n}. - Dennis P. Walsh, Apr 07 2011
Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003
Consider k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].
An element of CP(n,k) is a n-tuple T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,...,n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j > i_(j+1), T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j > i_(j+1)).
The indicator function which tests on i_j = i_(j+1) generates A158497, which contains further examples of this type of counting.
Triangle of the numbers of combinations of k elements with repetitions from n elements {1,2,...,n} (when every element i, i=1,...,n, appears in a k-combination either 0, or 1, or 2, ..., or k times). - Vladimir Shevelev, Jun 19 2012
G.f. for Faber polynomials is -log(-t*x-(1-sqrt(1-4*t))/2+1)=sum(n>0, T(n,k)*t^k/n). - Vladimir Kruchinin, Jul 04 2013
Values of complete homogeneous symmetric polynomials with all arguments equal to 1, or, equivalently, the number of monomials of degree k in n variables. - Tom Copeland, Apr 07 2014
Row k >= 0 of the infinite square array A[k,n] = C(n,k), n >= 0, would start with k zeros in front of the first nonzero element C(k,k) = 1; this here is the triangle obtained by taking the first k+1 nonzero terms C(k .. 2k, k) of rows k = 0, 1, 2, ... of that array. - M. F. Hasler, Mar 05 2017

			The triangle T(n,k), n >= 0, 0 <= k <= n, begins
  1      A000217
  1 1   /     A000292
  1 2  3    /    A000332
  1 3  6  10    /    A000389
  1 4 10  20  35    /     A000579
  1 5 15  35  70 126     /
  1 6 21  56 126 252  462
  1 7 28  84 210 462  924 1716
  1 8 36 120 330 792 1716 3432 6435
.
T(3,2)=6 considers the CP with the 3^2=9 elements (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3), and does not count the 3 of them which are (2,1),(3,1) and (3,2).
T(3,3) = 10 because the ways to distribute the 3 objects into the three containers are: (3,0,0) (0,3,0) (0,0,3) (2,1,0) (1,2,0) (2,0,1) (1,0,2) (0,1,2) (0,2,1) (1,1,1), for a total of 10 possibilities.
T(3,3)=10 since (x^2/(x-1))^3 = (x+1+1/x+O(1/x^2))^3 = x^3+3x^2+6x+10+O(x).
T(4,2)=10 since there are 10 nondecreasing functions f from {1,2} to {1,2,3,4}. Using <f(1),f(2)> to denote such a function, the ten functions are <1,1>, <1,2>, <1,3>, <1,4>, <2,2>, <2,3>, <2,4>, <3,3>, <3,4>, and <4,4>. - _Dennis P. Walsh_, Apr 07 2011
T(4,0) + T(4,1) + T(4,2) + T(4,3) = 1 + 4 + 10 + 20 = 35 = T(4,4). - _Jonathan Sondow_, Jun 28 2014
From _Paul Curtz_, Jun 18 2018: (Start)
Consider the array
2,    1,    1,    1,    1,    1,     ... = A054977(n)
1,    1/2,  1/3,  1/4,  1/5,  1/6,   ... = 1/(n+1) = 1/A000027(n)
1/3,  1/6,  1/10, 1/15, 1/21, 1/28,  ... = 2/((n+2)*(n+3)) = 1/A000217(n+2)
1/10, 1/20, 1/35, 1/56, 1/84, 1/120, ... = 6/((n+3)*(n+4)*(n+5)) =1/A000292(n+2) (see the triangle T(n,k)).
Every row is an autosequence of the second kind. (See OEIS Wiki, Autosequence.)
By decreasing antidiagonals the denominator of the array is a(n).
Successive vertical denominators: A088218(n), A000984(n), A001700(n), A001791(n+1), A002054(n), A002694(n).
Successive diagonal denominators: A165817(n), A005809(n), A045721(n), A025174(n+1), A004319(n). (End)
Without the first row (2, 1, 1, 1, ... ), the array leads to A165257(n) instead of a(n). - _Paul Curtz_, Jun 19 2018

R. Grimaldi, Discrete and Combinatorial Mathematics, Addison-Wesley, 4th edition, chapter 1.4.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6, (referring to A158498 before recycling)
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, and C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014 and J. Int. Seq. 18 (2015) 15.10.6.
C. M. Ringel, The Catalan combinatorics of the hereditary artin algebras, arXiv:1502.06553 [math.RT], 2015.
Dennis Walsh, Notes on finite monotonic and non-monotonic functions

Columns: T(n,1) = A000027(n), n >= 1. T(n,2) = A000217(n) = A161680(n+1), n >= 2. T(n,3) = A000292(n), n >= 3. T(n,4) = A000332(n+3), n >= 4. T(n,5) = A000389(n+4), n >= 5. T(n,6) = A000579(n+5), n >=6. T(n,k) = A001405(n+k-1) for k <= n <= k+2. [Corrected and extended by M. F. Hasler, Mar 05 2017]
Rows: T(5,k) = A000332(k+4). T(6,k) = A000389(k+5). T(7,k) = A000579(k+6).
Diagonals: T(n,n) = A001700(n-1). T(n,n-1) = A000984(n-1).
T(n,k) = A046899(n-1,k). - R. J. Mathar, Mar 26 2009
Take Pascal's triangle A007318, delete entries to the right of a vertical line just right of center, then scan the diagonals.
For a signed version of this triangle see A027555.
Row sums give A000984.
Cf. A007318, A158497, A100100 (mirrored), A009766.
A000984, A001700, A001791, A002054, A002694, A004319, A005809, A025174, A045721, A088218, A165817, A054977, A165257, A000027, A000217, A006134 (trace of the symmetric Pascal matrix).

Flat(List([0..10], n->List([0..n], k->Binomial(n+k-1, k)))); # Stefano Spezia, Oct 30 2018

a059481 n k = a059481_tabl !! n !! n
a059481_row n = a059481_tabl !! n
a059481_tabl = map reverse a100100_tabl
-- Reinhard Zumkeller, Jan 15 2014

&cat [[&*[ Binomial(n+k-1,k)]: k in [0..n]]: n in [0..30] ]; // Vincenzo Librandi, Apr 08 2011

for n from 0 to 10 do for k from 0 to n do print(binomial(n+k-1,k)) ; od: od: # R. J. Mathar, Mar 31 2009

t[n_, k_] := Binomial[n+k-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 09 2013 *)
(* The combinatorial objects defined in the first comment can, for n >= 1, be generated by: *) r[n_, k_] := FrobeniusSolve[ConstantArray[1,n],k]; (* Peter Luschny, Jan 24 2019 *)

sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 0 thru n do disp(sjoin(makelist(binomial(i+j-1, j), j, 0, i), " ")); display_triangle(10); /* triangle output */ /* Stefano Spezia, Oct 30 2018 */

{T(n, k) = binomial( n+k-1, k)}; \\ Michael Somos, Sep 09 2003, edited by M. F. Hasler, Mar 05 2017

{T(n, k) = if( n<0, 0, polcoeff( Pol(((1 / (x - x^2) + x * O(x^n))^n + O(x)) * x^n), k))}; /* Michael Somos, Sep 09 2003 */

[[binomial(n+k-1,k) for k in range(n+1)] for n in range(11)] # G. C. Greubel, Nov 21 2018

T(n,0) + T(n,1) + . . . + T(n,n-1) = T(n,n). - Jonathan Sondow, Jun 28 2014
From Peter Bala, Jul 21 2015: (Start)
T(n, k) = Sum_{j = k..n} (-1)^(k+j)*binomial(2*n,n+j)*binomial(n+j-1,j)* binomial(j,k) (gives the correct value T(n,k) = 0 for k > n).
O.g.f.: 1/2*( x*(2*x - 1)/(sqrt(1 - 4*t*x)*(1 - x - t)) + (1 + 2*x)/sqrt(1 - 4*t*x) + (1 - t)/(1 - x - t) ) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ....
n-th row polynomial R(n,t) = [x^n] ( (1 + x)^2/(1 + x(1 - t)) )^n.
exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + t)*x + (1 + 2*t + 2*t^2)*x^2 + (1 + 3*t + 5*t^2 + 5*t^3)*x^3 + ... is the o.g.f for A009766. (End)
a(n) = abs(A027555(n)). - M. F. Hasler, Mar 05 2017
For n >= k > 0, T(n, k) = Sum_{j=1..n} binomial(k + j - 2, k - 1) = Sum_{j=1..n} A007318(k + j - 2, k - 1). - Stefano Spezia, Oct 30 2018
T(n, k) = RisingFactorial(n, k) / k!. - Peter Luschny, Nov 24 2023

Offset changed from 1 to 0 by R. J. Mathar, Jan 15 2013

Edited by M. F. Hasler, Mar 05 2017

A086020 a(n) = Sum_(i=1..n) binomial(i+2,3)^2 [ Sequential sums of the tetragonal numbers or "tetras" (pyramidal, square) raised to power 2 (drawn from the 4th diagonal - left or right - of Pascal's Triangle) ].

Original entry on oeis.org

1, 17, 117, 517, 1742, 4878, 11934, 26334, 53559, 101959, 183755, 316251, 523276, 836876, 1299276, 1965132, 2904093, 4203693, 5972593, 8344193, 11480634, 15577210, 20867210, 27627210, 36182835, 46915011, 60266727, 76750327

Offset: 1

André F. Labossière, Jul 17 2003

Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference, p. 243; expression in (13.26) yields same sequence with offset 0). - Emeric Deutsch, Aug 02 2005

Partial sums of A001249. - R. J. Mathar, Aug 19 2008

			a(8) = Sum_{i=1..8} binomial(i+2,3)^2 = (20*(8^7) + 210*(8^6) + 854*(8^5) + 1680*(8^4) + 1610*(8^3) + 630*(8^2) + 36*8)/7! = 26334.

T. D. Noe, Table of n, a(n) for n = 1..1000
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988.
John Engbers and Christopher Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
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.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000292, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A000332, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520: n in [1..30]]; // G. C. Greubel, Nov 22 2017

a:=n->n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520: seq(a(n),n=1..31); # Emeric Deutsch

Accumulate[Binomial[Range[30]+2,3]^2]  (* Harvey P. Dale, Mar 24 2011 *)
LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{1,17,117,517,1742,4878, 11934, 26334},30] (* Harvey P. Dale, Aug 17 2014 *)

a(n)=n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520 \\ Charles R Greathouse IV, May 18 2015

a(n) = Sum_(i=1..n) binomial(i+2, 3)^2.
a(n) = ( C(n+3, 4)/35 )*( 35 + 84*C(n-1, 1) + 70*C(n-1, 2) + 20*C(n-1, 3) ).
a(n) = n*(n+1)*(n+2)*(n+3)*(2*n+3)(5*n^2 + 15*n + 1)/2520. - Emeric Deutsch, Aug 02 2005
O.g.f: x*(1+x)*(1 + 8*x + x^2)/(1-x)^8. - R. J. Mathar, Aug 19 2008

A085438 a(n) = Sum_{i=1..n} binomial(i+1,2)^3.

Original entry on oeis.org

1, 28, 244, 1244, 4619, 13880, 35832, 82488, 173613, 339988, 627484, 1102036, 1855607, 3013232, 4741232, 7256688, 10838265, 15838476, 22697476, 31958476, 44284867, 60479144, 81503720, 108503720, 142831845, 186075396, 240085548, 307008964, 389321839

Offset: 1

André F. Labossière, Jun 30 2003

			a(10) = (90*(10^7)+630*(10^6)+1638*(10^5)+1890*(10^4)+840*(10^3)-48*(10))/5040 = 339988.

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

G. C. Greubel, Table of n, a(n) for n = 1..5000
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.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Column k=3 of A334781.

Cf. A000292, A087127, A024166, A024166, A085439, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(90*n^7 +630*n^6 +1638*n^5 +1890*n^4+ 840*n^3 -48*n)/ Factorial(7): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n)/7!, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)

Vec(x*(x^4+20*x^3+48*x^2+20*x+1)/(x-1)^8 + O(x^100)) \\ Colin Barker, May 02 2014

a(n) = sum(i=1, n, binomial(i+1, 2)^3); \\ Michel Marcus, Nov 22 2017

a(n) = (90*n^7 +630*n^6 +1638*n^5 +1890*n^4+ 840*n^3 -48*n)/7!.
a(n) = (C(n+2, 3)/35)*(35 +210*C(n-1, 1) +399*C(n-1, 2) +315*C(n-1, 3) +90*C(n-1, 4)).
G.f.: x*(x^4+20*x^3+48*x^2+20*x+1) / (x-1)^8. - Colin Barker, May 02 2014

More terms from Colin Barker, May 02 2014

Formula and example edited by Colin Barker, May 02 2014

A085442 a(n) = Sum_{i=1..n} binomial(i+1,2)^7.

Original entry on oeis.org

1, 2188, 282124, 10282124, 181141499, 1982230040, 15475158552, 93839322648, 467508775773, 1989944010148, 7445104711204, 25010673566116, 76686775501847, 217396817767472, 575714897767472, 1436257466526768, 3398894618986905, 7674255436599996, 16612972826599996

Offset: 1

André F. Labossière, Jul 07 2003

T. D. Noe, Table of n, a(n) for n = 1..1000
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.
Index entries for linear recurrences with constant coefficients, signature (16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1).

Column k=7 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085440, A085441, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(1/823680) *n*(n+1)*(n+2)*(429*n^12 +5148*n^11 +24123*n^10 +52470*n^9 +43047*n^8 -8856*n^7 +4109*n^6 +50430*n^5 -18796*n^4 -44472*n^3 +26864*n^2 +8352*n -5568): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[Sum[Binomial[k+1,2]^7, {k,1,n}], {n,1,30}] (* G. C. Greubel, Nov 22 2017 *)
LinearRecurrence[{16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1},{1,2188,282124,10282124,181141499,1982230040,15475158552,93839322648,467508775773,1989944010148,7445104711204,25010673566116,76686775501847,217396817767472,575714897767472,1436257466526768},20] (* Harvey P. Dale, May 11 2022 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^7), ", ")) \\ G. C. Greubel, Nov 22 2017

a(n) = (1/823680) *n*(n+1)*(n+2)*(429*n^12 +5148*n^11 +24123*n^10 +52470*n^9 +43047*n^8 -8856*n^7 +4109*n^6 +50430*n^5 -18796*n^4 -44472*n^3 +26864*n^2 +8352*n -5568). - Vladeta Jovovic, Jul 07 2003

G.f.: x*(x^12 +2172*x^11 +247236*x^10 +6030140*x^9 +49258935*x^8 +163809288*x^7 +242384856*x^6 +163809288*x^5 +49258935*x^4 +6030140*x^3 +247236*x^2 +2172*x+ 1) / (x -1)^16. - Colin Barker, May 02 2014

A085439 a(n) = Sum_{i=1..n} binomial(i+1,2)^4.

Original entry on oeis.org

1, 82, 1378, 11378, 62003, 256484, 871140, 2550756, 6651381, 15802006, 34776742, 71791798, 140366759, 261917384, 469277384, 811379400, 1359360681, 2214396762, 3517606762, 5462416762, 8309813083, 12406965164, 18209748140, 26309748140, 37466388765, 52644875166

Offset: 1

André F. Labossière, Jul 03 2003

			a(15) = (2520*(15^9) +22680*(15^8) +79920*(15^7) +136080*(15^6) +107352*(15^5) +22680*(15^4) -10080*(15^3) +1728*15)/9! = 469277384.

G. C. Greubel, Table of n, a(n) for n = 1..5000
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.
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Column k=4 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(2520*n^9 +22680*n^8 +79920*n^7 +136080*n^6 +107352*n^5 +22680*n^4 -10080*n^3 +1728*n)/Factorial(9): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(2520*(n^9) + 22680*(n^8) + 79920*(n^7) + 136080*(n^6) + 107352*(n^5) + 22680*(n^4) - 10080*(n^3) + 1728*n)/9!, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)

Vec(x*(x^6+72*x^5+603*x^4+1168*x^3+603*x^2+72*x+1)/(x-1)^10 + O(x^100)) \\ Colin Barker, May 02 2014

a(n) = sum(i=1, n, binomial(i+1, 2)^4); \\ Michel Marcus, Nov 22 2017

a(n) = (2520*n^9 +22680*n^8 +79920*n^7 +136080*n^6 +107352*n^5 +22680*n^4 -10080*n^3 +1728*n)/9!.

G.f.: x*(x^6+72*x^5+603*x^4+1168*x^3+603*x^2+72*x+1) / (x-1)^10. - Colin Barker, May 02 2014

More terms from Colin Barker, May 02 2014

Typo in example fixed by Colin Barker, May 02 2014

A085440 a(n) = Sum_{i=1..n} binomial(i+1,2)^5.

Original entry on oeis.org

1, 244, 8020, 108020, 867395, 4951496, 22161864, 82628040, 267156165, 770440540, 2022773116, 4909947484, 11150268935, 23913084560, 48796284560, 95322158736, 179163294729, 325374464580, 572984364580, 981394464580, 1639143014731, 2675722491224, 4277290592600

Offset: 1

André F. Labossière, Jun 30 2003

Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

T. D. Noe, Table of n, a(n) for n = 1..1000
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.
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Column k=5 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n )/Factorial(11): n in [1..30]]; // G. C. Greubel, Nov 22 2017

Table[(113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!, {n,1,50}] (* G. C. Greubel, Nov 22 2017 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^5), ", ")) \\ G. C. Greubel, Nov 22 2017

a(n) = (113400*n^11 +1247400*n^10 +5544000*n^9 +12474000*n^8 +14196600*n^7 +6237000*n^6 -831600*n^5 +1108800*n^3 -172800*n)/11!.

G.f.: x*(x^8+232*x^7+5158*x^6+27664*x^5+47290*x^4+27664*x^3+5158*x^2+232*x+1) / (x-1)^12. - Colin Barker, May 02 2014

Formula edited by Colin Barker, May 02 2014

A085441 a(n) = Sum_{i=1..n} binomial(i+1,2)^6.

Original entry on oeis.org

1, 730, 47386, 1047386, 12438011, 98204132, 580094436, 2756876772, 11060642397, 38741283022, 121395233038, 346594833742, 914464085783, 2254559726408, 5240543726408, 11568062614344, 24395756421273, 49397866465794, 96443747465794, 182209868465794

Offset: 1

André F. Labossière, Jul 07 2003

			a(5) = C(7,3)*[191*106 + 450*(18*C(14,10) + 3851*C(13,10) + 61839*C(12,10) + 225352*C(11,10) + 225352*C(10,10))]/10010 = 12438011.

T. D. Noe, Table of n, a(n) for n = 1..1000
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.
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Column k=6 of A334781.

Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085440, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030, A234253.

[(n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12): n in [1..30]]; // G. C. Greubel, Nov 22 2017

f:= sum(binomial(1+i,2)^6,i=1..n):
seq(f, n=1..30); # Robert Israel, Nov 22 2017

Table[Sum[Binomial[i+1,2]^6,{i,n}],{n,20}] (* or *) LinearRecurrence[ {14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1},{1,730,47386,1047386,12438011, 98204132,580094436, 2756876772,11060642397, 38741283022,121395233038, 346594833742, 914464085783, 2254559726408},20] (* Harvey P. Dale, Jun 05 2017 *)

for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^6), ", ")) \\ G. C. Greubel, Nov 22 2017

G.f.: x*(x^10 +716*x^9 +37257*x^8 +450048*x^7 +1822014*x^6 +2864328*x^5 +1822014*x^4 +450048*x^3 +37257*x^2 +716*x +1) / (x -1)^14. - Colin Barker, May 02 2014

a(n) = (n/960960)*(6112 - 40040*n^2 + 78078*n^4 + 15015*n^5 + 19305*n^6 + 225225*n^7 + 335335*n^8 + 225225*n^9 + 80535*n^10 + 15015*n^11 + 1155*n^12). - G. C. Greubel, Nov 22 2017

cp's OEIS Frontend

A002294 a(n) = binomial(5*n, n)/(4*n + 1).

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A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

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A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

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A086020 a(n) = Sum_(i=1..n) binomial(i+2,3)^2 [ Sequential sums of the tetragonal numbers or "tetras" (pyramidal, square) raised to power 2 (drawn from the 4th diagonal - left or right - of Pascal's Triangle) ].

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A002294 a(n) = binomial(5n, n)/(4n + 1).