A000579 -id:A000579 - cp's OEIS frontend

A040977 a(n) = binomial(n+5,5)*(n+3)/3.

1, 8, 35, 112, 294, 672, 1386, 2640, 4719, 8008, 13013, 20384, 30940, 45696, 65892, 93024, 128877, 175560, 235543, 311696, 407330, 526240, 672750, 851760, 1068795, 1330056, 1642473, 2013760, 2452472, 2968064, 3570952, 4272576, 5085465

Offset: 0

Barry E. Williams, Dec 14 1999

Sequence is n^2*(n^2-1)*(n^2-4)/360 if offset 3.
If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-7) is the number of 7-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
6-dimensional square numbers, fifth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+5,i+5)*b(i), where b(i) = [1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Sequence of the absolute values of the z^2 coefficients divided by 5 of the polynomials in the GF2 denominators of A156925. See A157703 for background information. - Johannes W. Meijer, Mar 07 2009
2*a(n) is number of ways to place 5 queens on an (n+5) X (n+5) chessboard so that they diagonally attack each other exactly 10 times. The maximal possible attack number, p=binomial(k,2)=10 for k=5 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015
Ehrhart polynomial for the Chan-Robbins-Yuen polytope CRY_4. [De Loera et al.] - N. J. A. Sloane, Apr 16 2016
Coefficients in the terminating series identity 1 - 8*n/(n + 7) + 35*n*(n - 1)/((n + 7)*(n + 8)) - 112*n*(n - 1)*(n - 2)/((n + 7)*(n + 8)*(n + 9)) + ... = 0 for n = 1,2,3,.... Cf. A005585 and A050486. - Peter Bala, Feb 18 2019

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jesús A. De Loera, Fu Liu and Ruriko Yoshida, A generating function for all semi-magic squares and the volume of the Birkhoff polytope, J. Algebraic Combin., Vol. 30, No. 1 (2009), pp. 113-139. See page 138, n=4 entry in table.
Milan Janjic, Two Enumerative Functions.
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. 15.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Partial sums of A005585.
Cf. A000292, A000579, A050486, A053120, A133111, A133112.
Cf. A156925, A157703. - Johannes W. Meijer, Mar 07 2009

[Binomial(n+5, 5) + 2*Binomial(n+5, 6): n in [0..35]]; // Vincenzo Librandi, Jun 09 2013

with(combinat); A040977 := n->binomial(n+5,5)*(n+3)/3;
a:=n->(sum((numbcomp(n,6)), j=4..n))/3:seq(a(n), n=6..38); # Zerinvary Lajos, Aug 26 2008
nmax:=34; for n from 0 to nmax do fz(n):=product((1-m*z)^(n+1-m),m=1..n); c(n):= abs(coeff(fz(n),z,2))/5; end do: a:=n-> c(n): seq(a(n), n=2..nmax); # Johannes W. Meijer, Mar 07 2009

CoefficientList[Series[(1 + x) / (1 - x)^7, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1},{1,8,35,112,294,672,1386},40] (* Harvey P. Dale, Feb 20 2016 *)

vector(20,n,n--;2*binomial(n+6,6)-binomial(n+5,5)) \\ Derek Orr, May 05 2015

Vec((1+x)/(1-x)^7 + O(x^100)) \\ Altug Alkan, Nov 29 2015

a(n) = (-1)^n*A053120(2*n+6, 6)/32, (1/32 of seventh unsigned column of Chebyshev T-triangle, zeros omitted).
G.f.: (1+x)/(1-x)^7.
a(n-3) = Sum_{i+j+k=n} i*j*k^2. - Benoit Cloitre, Nov 01 2002
a(n) = 2*binomial(n+6, 6) - binomial(n+5, 5). - Paul Barry, Mar 04 2003
a(n-3) = 1/(1!*2!*3!)*Sum_{1 <= x_1, x_2, x_3 <= n} |det V(x_1,x_2,x_3)| = 1/12*Sum_{1 <= i,j,k <= n} |(i-j)(i-k)(j-k)|, where V(x_1,x_2,x_3) is the Vandermonde matrix of order 3. - Peter Bala, Sep 13 2007
a(n) = binomial(n+5,5) + 2*binomial(n+5,6). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = (n+1)*(n+2)*(n+3)^2*(n+4)*(n+5)/360. - Wesley Ivan Hurt, May 05 2015
a(n) = A000579(n+5) + A000579(n+6). - R. J. Mathar, Nov 29 2015
Sum_{n>=0} 1/a(n) = 15*Pi^2 - 1175/8. - Jaume Oliver Lafont, Jul 11 2017
Sum_{n>=0} (-1)^n/a(n) = 15*Pi^2/2 - 585/8. - Amiram Eldar, Jan 24 2022

A006542 a(n) = binomial(n,3)*binomial(n-1,3)/4.

Original entry on oeis.org

1, 10, 50, 175, 490, 1176, 2520, 4950, 9075, 15730, 26026, 41405, 63700, 95200, 138720, 197676, 276165, 379050, 512050, 681835, 896126, 1163800, 1495000, 1901250, 2395575, 2992626, 3708810, 4562425, 5573800, 6765440, 8162176, 9791320, 11682825, 13869450

Offset: 4

N. J. A. Sloane

Number of permutations of n+4 that avoid the pattern 132 and have exactly 3 descents. - Mike Zabrocki, Aug 26 2004
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 20 2005
a(n) = number of Dyck n-paths with exactly 4 peaks. - David Callan, Jul 03 2006
Six-dimensional figurate numbers for a hyperpyramid with pentagonal base. This corresponds to the sum(sum(sum(sum(1+sum(5*n))))) interpretation, see the Munafo webpage. - Robert Munafo, Jun 18 2009

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 166, no. 1).
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 238.
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 = 4..200
Isaac Ahern and Sam Cook, Affine Symmetry Tensors in Minkowski Space, American Journal of Undergraduate Research, Volume 13, Issue 3, August 2016.
P. Aluffi, Degrees of projections of rank loci, arXiv:1408.1702 [math.AG], 2014. ["After compiling the results of many explicit computations, we noticed that many of the numbers d_{n,r,S} appear in the existing literature in contexts far removed from the enumerative geometry of rank conditions; we owe this surprising (to us) observation to perusal of [Slo14]."]
Brandy Amanda Barnette, Counting Convex Sets on Products of Totally Ordered Sets, Masters Theses & Specialist Projects, Paper 1484, 2015.
V. E. Hoggatt, Jr., Letter to N. J. A. Sloane, Apr 1977
G. Kreweras, Traitement simultané du "Problème de Young" et du "Problème de Simon Newcomb", Cahiers du Bureau Universitaire de Recherche Opérationnelle. Institut de Statistique, Université de Paris, 10 (1967), 23-31.
G. Kreweras, Traitement simultané du "Problème de Young" et du "Problème de Simon Newcomb", Cahiers du Bureau Universitaire de Recherche Opérationnelle. Institut de Statistique, Université de Paris, 10 (1967), 23-31. [Annotated scanned copy]
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. 6, 25.
Robert Munafo, C(n,3)C(n-1,3)/4
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
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

The expression binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1) for the values m = 2 through 14 produces the sequences A000012, A000217, A002415, A006542, A006857, A108679, A134288, A134289, A134290, A134291, A140925, A140935, A169937.
Cf. A000332, A000579, A001263, A002378, A004068, A005585, A005891, A006322, A006414, A047819, A107891, A114242.
Fourth column of the table of Narayana numbers A001263.
Apart from a scale factor, a column of A124428.

List([4..40], n-> n*(n-1)^2*(n-2)^2*(n-3)/144); # G. C. Greubel, Feb 24 2019

[ n*((n-1)*(n-2))^2*(n-3)/144 : n in [4..40] ]; // Wesley Ivan Hurt, Jun 17 2014

A006542:=-(1+3*z+z**2)/(z-1)**7; # conjectured by Simon Plouffe in his 1992 dissertation
A006542:=n->n*((n-1)*(n-2))^2*(n-3)/144; seq(A006542(n), n=4..40); # Wesley Ivan Hurt, Jun 17 2014

Table[Binomial[n, 3]*Binomial[n-1, 3]/4, {n, 4, 40}]

PARI
```
a(n)=n*((n-1)*(n-2))^2*(n-3)/144
```

[n*(n-1)^2*(n-2)^2*(n-3)/144 for n in (4..40)] # G. C. Greubel, Feb 24 2019

a(n) = C(n, 3)*C(n-1, 3)/4 = n*(n-1)^2*(n-2)^2*(n-3)/144.
a(n) = A000292(n-3)*A000292(n-2)/4.
E.g.f.: x^4*(6 + 6*x + x^2)*exp(x)/144. - Vladeta Jovovic, Jan 29 2003
a(n) = Sum(Sum(Sum(Sum(1 + Sum(5*n))))) = Sum (A006414). - Xavier Acloque, Oct 08 2003
a(n) = C(n, 6) + 3*C(n+1, 6) + C(n+2, 6). - Mike Zabrocki, Aug 26 2004
G.f.: x^4*(1 + 3*x + x^2)/(1-x)^7. - Emeric Deutsch, Jun 20 2005
a(n) = C(n-2, n-4)*C(n-1, n-3)*C(n, n-2)/18. - Zerinvary Lajos, Jul 29 2005
a(n) = C(n,4)*C(n,3)/n. - Mitch Harris, Jul 06 2006
a(n+2) = (1/4)*Sum_{1 <= x_1, x_2 <= n} x_1*x_2*(det V(x_1,x_2))^2 = (1/4)*Sum_{1 <= i,j <= n} i*j*(i-j)^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
a(n) = C(n-1,3)^2 - C(n-1,2)*C(n-1,4). - Gary Detlefs, Dec 05 2011
a(n) = A000292(A000217(n-1)) - A000217(A000292(n-1)). - Ivan N. Ianakiev, Jun 17 2014
a(n) = Product_{i=1..3} A002378(n-4+i)/A002378(i). - Bruno Berselli, Nov 12 2014 (Rewritten, Sep 01 2016.)
Sum_{n>=4} 1/a(n) = 238 - 24*Pi^2. - Jaume Oliver Lafont, Jul 10 2017
Sum_{n>=4} (-1)^n/a(n) = 134 - 192*log(2). - Amiram Eldar, Oct 19 2020
a(n) = A000332(n) + 5*A000579(n+1). - Yasser Arath Chavez Reyes, Aug 18 2024

Zabroki and Lajos formulas offset corrected by Gary Detlefs, Dec 05 2011

A092392 Triangle read by rows: T(n,k) = C(2*n - k,n), 0 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 20, 10, 4, 1, 70, 35, 15, 5, 1, 252, 126, 56, 21, 6, 1, 924, 462, 210, 84, 28, 7, 1, 3432, 1716, 792, 330, 120, 36, 8, 1, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 48620, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1, 184756, 92378, 43758, 19448, 8008, 3003, 1001, 286, 66, 11, 1

Offset: 0

Ralf Stephan, Mar 21 2004

First column is C(2*n,n) or A000984. Central coefficients are C(3*n,n) or A005809. - Paul Barry, Oct 14 2009
T(n,k) = A046899(n,n-k), k = 0..n-1. - Reinhard Zumkeller, Jul 27 2012
From  Peter Bala, Nov 03 2015: (Start)
Viewed as the square array [binomial (2*n + k, n + k)]n,k>=0  this is the generalized Riordan array ( 1/sqrt(1 - 4*x),c(x) ) in the sense of the Bala link, where c(x) is the o.g.f. for A000108.
The square array factorizes as ( 1/(2 - c(x)),x*c(x) ) * ( 1/(1 - x),1/(1 - x) ), which equals the matrix product of A100100 with the square Pascal matrix [binomial (n + k,k)]n,k>=0. See the example below. (End)

			From _Paul Barry_, Oct 14 2009: (Start)
Triangle begins
  1,
  2, 1,
  6, 3, 1,
  20, 10, 4, 1,
  70, 35, 15, 5, 1,
  252, 126, 56, 21, 6, 1,
  924, 462, 210, 84, 28, 7, 1,
  3432, 1716, 792, 330, 120, 36, 8, 1
Production array is
  2, 1,
  2, 1, 1,
  2, 1, 1, 1,
  2, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
As a square array = A100100 * square Pascal matrix:
  /1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
  |2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
  |6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
  |20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
  |70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n=0..149 of triangle, flattened
P. Bala, Notes on generalized Riordan arrays
P. Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), #13.5.1.
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Ik-Pyo Kim, Michael J. Tsatsomeros, Inverse Relations in Shapiro's Open Questions, arXiv:1707.06590 [math.CO], 2017. See p. 7.

Rows 0-14 are A000984, A001700, A001791, A002054, A002694, A003516, A002696, A030053, A004310, A030054, A004311, A030055, A004312, A030056, A004313.
Columns are A000217, A000292, A000332, A000389, A000579.
Diagonals are A005809, A045721, A025174, A004319, A013698, A003408.
Cf. A100100.

a092392 n k = a092392_tabl !! (n-1) !! (k-1)
a092392_row n = a092392_tabl !! (n-1)
a092392_tabl = map reverse a046899_tabl
-- Reinhard Zumkeller, Jul 27 2012

/* As a triangle */ [[Binomial(2*n-k, n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 22 2017

A092392 := proc(n,k)
    binomial(2*n-k,n-k) ;
end proc:
seq(seq(A092392(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Table[Binomial[2 n - k, n], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Mar 19 2016 *)

C(x):=(1-sqrt(1-4*x))/2;
A(x,y):=(1/sqrt(1-4*x))/(1-y*C(x));
taylor(A(x,y),y,0,10,x,0,10); /* Vladimir Kruchinin, Mar 19 2016 */

for(n=0,10, for(k=0,n, print1(binomial(2*n - k,n), ", "))) \\ G. C. Greubel, Nov 22 2017

As a number triangle, this is T(n, k) = if(k <= n, C(2*n - k, n), 0). Its row sums are C(2*n + 1, n + 1) = A001700. Its diagonal sums are A176287. - Paul Barry, Apr 23 2005
G.f. of column k: 2^k/[sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))^k].
As a number triangle, this is the Riordan array (1/sqrt(1 - 4*x), x*c(x)), c(x) the g.f. of A000108. - Paul Barry, Jun 24 2005
G.f.: A(x,y)=1/sqrt(1 - 4*x)/(1-y*x*C(x)), where C(x) is g.f. of Catalan numbers. - Vladimir Kruchinin, Mar 19 2016

Diagonal sums comment corrected by Paul Barry, Apr 14 2010

Offset corrected by R. J. Mathar, Feb 08 2013

A060540 Square array read by antidiagonals downwards: T(n,k) = (nk)!/(k!^nn!), (n>=1, k>=1), the number of ways of dividing nk labeled items into n unlabeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 15, 1, 1, 35, 280, 105, 1, 1, 126, 5775, 15400, 945, 1, 1, 462, 126126, 2627625, 1401400, 10395, 1, 1, 1716, 2858856, 488864376, 2546168625, 190590400, 135135, 1, 1, 6435, 66512160, 96197645544, 5194672859376, 4509264634875, 36212176000, 2027025, 1

Offset: 1

Henry Bottomley, Apr 02 2001

The Copeland link gives the associations of this entry with the operator calculus of Appell Sheffer polynomials, the combinatorics of simple set partitions encoded in the Faa di Bruno formula for composition of analytic functions (formal Taylor series), the Pascal matrix, and the geometry of the n-dimensional simplices (hypertriangles, or hypertetrahedra). These, in turn, are related to simple instances of the application of the exponential formula / principle / schema giving the number of not-necessarily-connected objects composed from an ensemble of connected objects. - Tom Copeland, Jun 09 2021

			Array begins:
  1,   1,       1,          1,             1,                 1, ...
  1,   3,      10,         35,           126,               462, ...
  1,  15,     280,       5775,        126126,           2858856, ...
  1, 105,   15400,    2627625,     488864376,       96197645544, ...
  1, 945, 1401400, 2546168625, 5194672859376, 11423951396577720, ...
  ...

Seiichi Manyama, Antidiagonals n = 1..50, flattened (first 20 antidiagonals from Harry J. Smith)
Tom Copeland, Calculus, Combinatorics, and Geometry Underlying OEIS A060540, and the Exponential Formula, 2021.
Nattawut Phetmak and Jittat Fakcharoenphol, Uniformly Generating Derangements with Fixed Number of Cycles in Polynomial Time, Thai J. Math. (2023) Vol. 21, No. 4, 899-915. See pp. 901, 914.
Elena L. Wang and Guoce Xin, On Ward Numbers and Increasing Schröder Trees, arXiv:2507.15654 [math.CO], 2025. See p. 13.

Rows include A000012, A001700, A060542, A082368, A322252.
Columns k=1..10 give A000012, A001147, A025035, A025036, A025037, A025038, A025040, A025041, A025042.
Main diagonal is A057599.
Related to A057599, see also A096126 and A246048.
Cf. A060358, A361948 (includes row/col 0).
Cf. A000217, A000292, A000332, A000389, A000579, A000580, A007318, A036040, A099174, A133314, A132440, A135278 (associations in Copeland link).

T[n_, k_] := (n*k)!/(k!^n*n!);
Table[T[n-k+1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 29 2018 *)

{ i=0; for (m=1, 20, for (n=1, m, k=m - n + 1; write("b060540.txt", i++, " ", (n*k)!/(k!^n*n!))); ) } \\ Harry J. Smith, Jul 06 2009

T(n,k) = (n*k)!/(k!^n*n!) = T(n-1,k)*A060543(n,k) = A060538(n,k)/k!.

T(n,k) = Product_{j=2..n} binomial(j*k-1,k-1). - M. F. Hasler, Aug 22 2014

Definition reworded by M. F. Hasler, Aug 23 2014

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

A000435 Normalized total height of all nodes in all rooted trees with n labeled nodes.

Original entry on oeis.org

0, 1, 8, 78, 944, 13800, 237432, 4708144, 105822432, 2660215680, 73983185000, 2255828154624, 74841555118992, 2684366717713408, 103512489775594200, 4270718991667353600, 187728592242564421568, 8759085548690928992256, 432357188322752488126152, 22510748754252398927872000

Offset: 1

N. J. A. Sloane

This is the sequence that started it all: the first sequence in the database!
The height h(V) of a node V in a rooted tree is its distance from the root. a(n) = Sum_{all nodes V in all n^(n-1) rooted trees on n nodes} h(V)/n.
In the trees which have [0, n-1] = (0, 1, ..., n-1) as their ordered set of nodes, the number of nodes at distance i from node 0 is f(n,i) = (n-1)...(n-i)(i+1)n^(j-1), 0 <= i < n-1, i+j = n-1 (and f(n,n-1) = (n-1)!): (n-1)...(n-i) counts the words coding the paths of length i from any node to 0, n^(j-1) counts the Pruefer codes of the rest, words build by iterated deletion of the greater node of degree 1 ... except the last one, (i+1), necessary pointing at the path. If g(n,i) = (n-1)...(n-i)n^j, i+j = n-1, f(n,i) = g(n,i) - g(n,i+1), g(n,i) = Sum_{k>=i} f(n,k), the sequence is Sum_{i=1..n-1} g(n,i). - Claude Lenormand (claude.lenormand(AT)free.fr), Jan 26 2001
If one randomly selects one ball from an urn containing n different balls, with replacement, until exactly one ball has been selected twice, the probability that this ball was also the second ball to be selected once is a(n)/n^n. See also A001865. - Matthew Vandermast, Jun 15 2004
a(n) is the number of connected endofunctions with no fixed points. - Geoffrey Critzer, Dec 13 2011
a(n) is the number of weakly connected simple digraphs on n labeled nodes where every node has out-degree 1. A digraph where all out-degrees are 1 can be called a functional digraph due to the correspondence with endofunctions. - Andrew Howroyd, Feb 06 2024

			For n = 3 there are 3^2 = 9 rooted labeled trees on 3 nodes, namely (with o denoting a node, O the root node):
   o
   |
   o     o   o
   |      \ /
   O       O
The first can be labeled in 6 ways and contains nodes at heights 1 and 2 above the root, so contributes 6*(1+2) = 18 to the total; the second can be labeled in 3 ways and contains 2 nodes at height 1 above the root, so contributes 3*2=6 to the total, giving 24 in all. Dividing by 3 we get a(3) = 24/3 = 8.
For n = 4 there are 4^3 = 64 rooted labeled trees on 4 nodes, namely (with o denoting a node, O the root node):
   o
   |
   o     o        o   o
   |     |         \ /
   o     o   o      o     o o o
   |      \ /       |      \|/
   O       O        O       O
  (1)     (2)      (3)     (4)
Tree (1) can be labeled in 24 ways and contains nodes at heights 1, 2, 3 above the root, so contributes 24*(1+2+3) = 144 to the total;
tree (2) can be labeled in 24 ways and contains nodes at heights 1, 1, 2 above the root, so contributes 24*(1+1+2) = 96 to the total;
tree (3) can be labeled in 12 ways and contains nodes at heights 1, 2, 2 above the root, so contributes 12*(1+2+2) = 60 to the total;
tree (4) can be labeled in 4 ways and contains nodes at heights 1, 1, 1 above the root, so contributes 4*(1+1+1) = 12 to the total;
giving 312 in all. Dividing by 4 we get a(4) = 312/4 = 78.

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).

Robert G. Wilson v, Table of n, a(n) for n = 1..1000 (first 100 terms from T. D. Noe)
Vijayakumar Ambat, Article in the Malayalam newspaper Ayala Manorama - Padhippura, 12 June 2015, that mentions the OEIS, and in particular this sequence.
V. I. Arnold, Topological classification of complex trigonometric polynomials and the combinatorics of graphs with the same number of edges and vertices, Functional Anal. Appl., 30 (1996), 1-17.
Shalosh B. Ekhad and Doron Zeilberger, Going Back to Neil Sloane's FIRST LOVE (OEIS Sequence A435): On the Total Heights in Rooted Labeled Trees, arXiv:1607.05776 [math.CO], 2016.
Shalosh B. Ekhad and Doron Zeilberger, Going Back to Neil Sloane's FIRST LOVE (OEIS Sequence A435): On the Total Heights in Rooted Labeled Trees, Version on DZ's home page with more links; Local copy, pdf file only, no active links
I. P. Goulden and D. M. Jackson, A proof of a conjecture for the number of ramified coverings of the sphere by the torus, arXiv:math/9902009 [math.AG], 1999.
I. P. Goulden, D. M. Jackson, and A. Vainshtein, The number of ramified coverings of the sphere by the torus and surfaces of higher genera, arXiv:math/9902125 [math.AG], 1999.
I. P. Goulden, D. M. Jackson, and A. Vainshtein, The number of ramified coverings of the sphere by the torus and surfaces of higher genera Ann. Comb. 4 (2000), no. 1, 27-46. (See Theorem 1.1.)
Brady Haran, The Number Collector (with Neil Sloane), Numberphile Podcast (2019)
Andrew Lohr and Doron Zeilberger, On the limiting distributions of the total height on families of trees, Integers (2018) 18, Article #A32.
T. Kyle Petersen, Exponential generating functions and Bell numbers, Inquiry-Based Enumerative Combinatorics (2019) Chapter 7, Undergraduate Texts in Mathematics, Springer, Cham, 98-99.
A. Rényi and G. Szekeres, On the height of trees, Journal of the Australian Mathematical Society 7.04 (1967): 497-507. See (4.7).
Marko Riedel et al., Connected endofunctions with no fixed points, Mathematics Stack Exchange, Dec 2014.
J. Riordan, Letter to N. J. A. Sloane, Aug. 1970
J. Riordan and N. J. A. Sloane, Enumeration of rooted trees by total height, J. Austral. Math. Soc., vol. 10 pp. 278-282, 1969.
N. J. A. Sloane, Page from 1964 notebook showing start of OEIS [includes A000027, A000217, A000292, A000332, A000389, A000579, A000110, A007318, A000058, A000215, A000289, A000324, A234953 (= A001854(n)/n), A000435, A000169, A000142, A000272, A000312, A000111]
N. J. A. Sloane, Cover of same notebook
N. J. A. Sloane, Lengths of Cycle Times in Random Neural Networks, Ph. D. Dissertation, Cornell University, February 1967; also Report No. 10, Cognitive Systems Research Program, Cornell University, 1967. This sequence appears on page 119.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Illustration of a(3) and a(4)
Yukun Yao and Doron Zeilberger, An Experimental Mathematics Approach to the Area Statistic of Parking Functions, arXiv:1806.02680 [math.CO], 2018.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 3.
Index entries for sequences related to rooted trees
Index entries for sequences related to trees

Cf. A001863, A001864, A001854, A002862 (unlabeled version), A234953, A259334.

Column k=1 of A350452.

A000435 := n-> (n-1)!*add (n^k/k!, k=0..n-2);
seq(simplify((n-1)*GAMMA(n-1,n)*exp(n)),n=1..20); # Vladeta Jovovic, Jul 21 2005

f[n_] := (n - 1)! Sum [n^k/k!, {k, 0, n - 2}]; Array[f, 18] (* Robert G. Wilson v, Aug 10 2010 *)
nx = 18; Rest[ Range[0, nx]! CoefficientList[ Series[ LambertW[-x] - Log[1 + LambertW[-x]], {x, 0, nx}], x]] (* Robert G. Wilson v, Apr 13 2013 *)

x='x+O('x^30); concat(0, Vec(serlaplace(lambertw(-x)-log(1+lambertw(-x))))) \\ Altug Alkan, Sep 05 2018

A000435(n)=(n-1)*A001863(n) \\ M. F. Hasler, Dec 10 2018

from math import comb
def A000435(n): return ((sum(comb(n,k)*(n-k)**(n-k)*k**k for k in range(1,(n+1>>1)))<<1) + (0 if n&1 else comb(n,m:=n>>1)*m**n))//n # Chai Wah Wu, Apr 25-26 2023

a(n) = (n-1)! * Sum_{k=0..n-2} n^k/k!.
a(n) = A001864(n)/n.
E.g.f.: LambertW(-x) - log(1+LambertW(-x)). - Vladeta Jovovic, Apr 10 2001
a(n) = A001865(n) - n^(n-1).
a(n) = A001865(n) - A000169(n). - Geoffrey Critzer, Dec 13 2011
a(n) ~ sqrt(Pi/2)*n^(n-1/2). - Vaclav Kotesovec, Aug 07 2013
a(n)/A001854(n) ~ 1/2 [See Renyi-Szekeres, (4.7)]. Also a(n) = Sum_{k=0..n-1} k*A259334(n,k). - David desJardins, Jan 20 2017
a(n) = (n-1)*A001863(n). - M. F. Hasler, Dec 10 2018

Additional references from Valery A. Liskovets
Editorial changes by N. J. A. Sloane, Feb 03 2012
Edited by M. F. Hasler, Dec 10 2018

A110555 Triangle of partial sums of alternating binomial coefficients: T(n, k) = Sum_{j=0..k} binomial(n, j)*(-1)^j, for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, -1, 0, 1, -2, 1, 0, 1, -3, 3, -1, 0, 1, -4, 6, -4, 1, 0, 1, -5, 10, -10, 5, -1, 0, 1, -6, 15, -20, 15, -6, 1, 0, 1, -7, 21, -35, 35, -21, 7, -1, 0, 1, -8, 28, -56, 70, -56, 28, -8, 1, 0, 1, -9, 36, -84, 126, -126, 84, -36, 9, -1, 0, 1, -10, 45, -120, 210, -252, 210, -120

Offset: 0

Reinhard Zumkeller, Jul 27 2005

			Triangle T(n, k) starts:
  [0] 1;
  [1] 1,  0;
  [2] 1, -1,  0;
  [3] 1, -2,  1,   0;
  [4] 1, -3,  3,  -1,  0;
  [5] 1, -4,  6,  -4,  1,   0;
  [6] 1, -5, 10, -10,  5,  -1,  0;
  [7] 1, -6, 15, -20, 15,  -6,  1,  0;
  [8] 1, -7, 21, -35, 35, -21,  7, -1,  0.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Ângela Mestre and José Agapito, A Family of Riordan Group Automorphisms, J. Int. Seq., Vol. 22 (2019), Article 19.8.5.
Index entries for triangles and arrays related to Pascal's triangle

T(n,1)  = -n + 1 for n>0;
T(n,2)  =  A000217(n-2) for n > 1;
T(n,3)  = -A000292(n-4) for n > 2;
T(n,4)  =  A000332(n-1) for n > 3;
T(n,5)  = -A000389(n-1) for n > 5;
T(n,6)  =  A000579(n-1) for n > 6;
T(n,7)  = -A000580(n-1) for n > 7;
T(n,8)  =  A000581(n-1) for n > 8;
T(n,9)  = -A000582(n-1) for n > 9;
T(n,10) =  A001287(n-1) for n > 10;
T(n,11) = -A001288(n-1) for n > 11;
T(n,12) =  A010965(n-1) for n > 12;
T(n,13) = -A010966(n-1) for n > 13;
T(n,14) =  A010967(n-1) for n > 14;
T(n,15) = -A010968(n-1) for n > 15;
T(n,16) =  A010969(n-1) for n > 16.
Cf. A071919 (variant), A000007 (row sums), A110556 (central terms).
Cf. A008949, A007318.

T := (n, k) -> (-1)^k * binomial(n-1, k):
seq(print(seq(T(n, k), k = 0..n)), n = 0..7); # Peter Luschny, Apr 13 2023

T[0, 0] := 1;  T[n_, n_] := 0; T[n_, k_] := (-1)^k*Binomial[n - 1, k]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 31 2017 *)

concat(1, for(n=1,10, for(k=0,n, print1(if(k != n, (-1)^k*binomial(n-1,k), 0), ", ")))) \\ G. C. Greubel, Aug 31 2017

T(n, 0) = 1, T(n, n) = 0^n, T(n, k) = -T(n-1, k-1) + T(n-1, k), for 0 < k < n.
T(n, k) = binomial(n-1, k)*(-1)^k, 0 <= k < n, T(n, n) = 0^n.
T(n, n-k-1) = -T(n, k), for 0 < k < n.
T(n, k) = A071919(n, k)*(-1)^k and A071919(n, k) = abs(T(n, k)).
Triangle T(n,k), 0 <= k <= n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [0, -1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 05 2005
G.f.: (1 + x*y) / (1 + x*y - x). - R. J. Mathar, Aug 11 2015

Typo in name corrected by Andrey Zabolotskiy, Feb 22 2022

Offset corrected by Peter Luschny, Apr 13 2023

cp's OEIS Frontend

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