A299041 -id:A299041 - cp's OEIS frontend

A131689 Triangle of numbers T(n,k) = k!Stirling2(n,k) = A000142(k)A048993(n,k) read by rows, T(n, k) for 0 <= k <= n.

1, 0, 1, 0, 1, 2, 0, 1, 6, 6, 0, 1, 14, 36, 24, 0, 1, 30, 150, 240, 120, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 126, 1806, 8400, 16800, 15120, 5040, 0, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 0, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880

Offset: 0

Philippe Deléham, Sep 14 2007

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,6,...] where DELTA is the operator defined in A084938; another version of A019538.
See also A019538: version with n > 0 and k > 0. - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 21 2014: (Start)
T(n,k) gives the number of (k-1)-dimensional faces in the interior of the first barycentric subdivision of the standard (n-1)-dimensional simplex. For example, the barycentric subdivision of the 1-simplex is o--o--o, with 1 interior vertex and 2 interior edges, giving T(2,1) = 1 and T(2,2) = 2.
This triangle is used when calculating the face vectors of the barycentric subdivision of a simplicial complex. Let S be an n-dimensional simplicial complex and write f_k for the number of k-dimensional faces of S, with the usual convention that f_(-1) = 1, so that F := (f_(-1), f_0, f_1,...,f_n) is the f-vector of S. If M(n) denotes the square matrix formed from the first n+1 rows and n+1 columns of the present triangle, then the vector F*M(n) is the f-vector of the first barycentric subdivision of the simplicial complex S (Brenti and Welker, Lemma 2.1). For example, the rows of Pascal's triangle A007318 (but with row and column indexing starting at -1) are the f-vectors for the standard n-simplexes. It follows that A007318*A131689, which equals A028246, is the array of f-vectors of the first barycentric subdivision of standard n-simplexes. (End)
This triangle T(n, k) appears in the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} (x^k/(1 - x)^(k+2))*T(n, k). See also the Eulerian triangle A008292 with a Mar 31 2017 comment for a rewritten form. For the e.g.f. see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
T(n,k) = the number of alignments of length k of n strings each of length 1. See Slowinski. An example is given below. Cf. A122193 (alignments of strings of length 2) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018
The row polynomials R(n,x) are the Fubini polynomials. - Emanuele Munarini, Dec 05 2020
From Gus Wiseman, Feb 18 2022: (Start)
Also the number of patterns of length n with k distinct parts (or with maximum part k), where we define a pattern to be a finite sequence covering an initial interval of positive integers. For example, row n = 3 counts the following patterns:
  (1,1,1)  (1,2,2)  (1,2,3)
           (2,1,2)  (1,3,2)
           (2,2,1)  (2,1,3)
           (1,1,2)  (2,3,1)
           (1,2,1)  (3,1,2)
           (2,1,1)  (3,2,1)
(End)
Regard A048994 as a lower-triangular matrix and divide each term A048994(n,k) by n!, then this is the matrix inverse. Because Sum_{k=0..n} (A048994(n,k) * x^n / n!) = A007318(x,n), Sum_{k=0..n} (A131689(n,k) * A007318(x,k)) = x^n. - Natalia L. Skirrow, Mar 23 2023
T(n,k) is the number of ordered partitions of [n] into k blocks. - Alois P. Heinz, Feb 21 2025

			The triangle T(n,k) begins:
  n\k 0 1    2     3      4       5        6        7        8        9      10 ...
  0:  1
  1:  0 1
  2:  0 1    2
  3:  0 1    6     6
  4:  0 1   14    36     24
  5:  0 1   30   150    240     120
  6:  0 1   62   540   1560    1800      720
  7:  0 1  126  1806   8400   16800    15120     5040
  8:  0 1  254  5796  40824  126000   191520   141120    40320
  9:  0 1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 0 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... reformatted and extended. - _Wolfdieter Lang_, Mar 31 2017
From _Peter Bala_, Feb 04 2018: (Start)
T(4,2) = 14 alignments of length 2 of 4 strings of length 1. Examples include
  (i) A -    (ii) A -    (iii) A -
      B -         B -          - B
      C -         - C          - C
      - D         - D          - D
There are C(4,1) = 4 alignments of type (i) with a single gap character - in column 1, C(4,2) = 6 alignments of type (ii) with two gap characters in column 1 and C(4,3) = 4 alignments of type (iii) with three gap characters in column 1, giving a total of 4 + 6 + 4 = 14 alignments. (End)

Vincenzo Librandi, Rows n = 0..100, flattened
Peter Bala, Deformations of the Hadamard product of power series
F. Brenti and V. Welker, f-vectors of barycentric subdivisions, arXiv:math/0606356 [math.CO], Math. Z., 259(4), 849-865, 2008.
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
Germain Kreweras, Une dualité élémentaire souvent utile dans les problèmes combinatoires, Mathématiques et Sciences Humaines 3 (1963): 31-41.
Jerry Metzger and Thomas Richards, A Prisoner Problem Variation, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.7.
Massimo Nocentini, An algebraic and combinatorial study of some infinite sequences of numbers supported by symbolic and logic computation, PhD Thesis, University of Florence, 2019. See Ex. 36.
Mircea Dan Rus, Yet another note on notation, arXiv:2501.08762 [math.HO], 2025. See p. 6.
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522
M. Z. Spivey, On Solutions to a General Combinatorial Recurrence, J. Int. Seq. 14 (2011) # 11.9.7.
Wikipedia, Barycentric subdivision
Wikipedia, Simplicial complex
Wikipedia, Simplex
Gus Wiseman, Sequences counting and ranking compositions by the patterns they match or avoid.

Columns k=0..10 are A000007, A000012, A000918, A001117, A000919, A001118, A000920, A135456, A133068, A133360, A133132,
Case m=1 of the polynomials defined in A278073.
Cf. A000142 (diagonal), A000670 (row sums), A000012 (alternating row sums), A210029 (central terms).
Cf. A001286, A037960, A037961, A037962, A037963.
Cf. A008292, A028246 (o.g.f. and e.g.f. of sums of powers).
Cf. A019538, A122193, A299041.
A version for partitions is A116608, or by maximum A008284.
A version for compositions is A235998, or by maximum A048004.
Classes of patterns:
- A000142 = strict
- A005649 = anti-run, complement A069321
- A019536 = necklace
- A032011 = distinct multiplicities
- A060223 = Lyndon
- A226316 = (1,2,3)-avoiding, weakly A052709, complement A335515
- A296975 = aperiodic
- A345194 = alternating, up/down A350354, complement A350252
- A349058 = weakly alternating
- A351200 = distinct runs
- A351292 = distinct run-lengths
Cf. A007318, A097805, A335456, A335457, A344605, A349050.

function T(n, k)
    if k < 0 || k > n return 0 end
    if n == 0 && k == 0 return 1 end
    k*(T(n-1, k-1) + T(n-1, k))
end
for n in 0:7
    println([T(n, k) for k in 0:n])
end
# Peter Luschny, Mar 26 2020

A131689 := (n,k) -> Stirling2(n,k)*k!: # Peter Luschny, Sep 17 2011
# Alternatively:
A131689_row := proc(n) 1/(1-t*(exp(x)-1)); expand(series(%,x,n+1)); n!*coeff(%,x,n); PolynomialTools:-CoefficientList(%,t) end:
for n from 0 to 9 do A131689_row(n) od; # Peter Luschny, Jan 23 2017

t[n_, k_] := k!*StirlingS2[n, k]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
T[n_, k_] := If[n <= 0 || k <= 0, Boole[n == 0 && k == 0], Sum[(-1)^(i + k) Binomial[k, i] i^(n + k), {i, 0, k}]]; (* Michael Somos, Jul 08 2018 *)

{T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(k + i) * binomial(k, i) * i^n))};
/* Michael Somos, Jul 08 2018 */

@cached_function
def F(n): # Fubini polynomial
    R. = PolynomialRing(ZZ)
    if n == 0: return R(1)
    return R(sum(binomial(n, k)*F(n - k)*x for k in (1..n)))
for n in (0..9): print(F(n).list()) # Peter Luschny, May 21 2021

T(n,k) = k*(T(n-1,k-1) + T(n-1,k)) with T(0,0)=1. Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A000629(n), A033999(n), A000007(n), A000670(n), A004123(n+1), A032033(n), A094417(n), A094418(n), A094419(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. [corrected by Philippe Deléham, Feb 11 2013]
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A000670(n), A122704(n) for x=-1, 0, 1, 2 respectively. - Philippe Deléham, Oct 09 2007
Sum_{k=0..n} (-1)^k*T(n,k)/(k+1) = Bernoulli numbers A027641(n)/A027642(n). - Peter Luschny, Sep 17 2011
G.f.: F(x,t) = 1 + x*t + (x+x^2)*t^2/2! + (x+6*x^2+6*x^3)*t^3/3! + ... = Sum_{n>=0} R(n,x)*t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x. - Philippe Deléham, Feb 11 2013
T(n,k) = [t^k] (n! [x^n] (1/(1-t*(exp(x)-1)))). - Peter Luschny, Jan 23 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See also Bala, Example E8. - Peter Bala, Jan 08 2018

A087107 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of tetrahedral numbers. The p-th row (p>=1) contains a(i,p) for i=1 to 3p-2, where a(i,p) satisfies Sum_{i=1..n} C(i+2,3)^p = 4 C(n+3,4) * Sum_{i=1..3p-2} a(i,p) C(n-1,i-1)/(i+3).

Original entry on oeis.org

1, 1, 3, 3, 1, 1, 15, 69, 147, 162, 90, 20, 1, 63, 873, 5191, 16620, 31560, 36750, 25830, 10080, 1680, 1, 255, 9489, 130767, 919602, 3832650, 10238000, 18244380, 21990360, 17745000, 9198000, 2772000, 369600, 1, 1023, 97953, 2903071, 40317780

Offset: 1

André F. Labossière, Aug 11 2003

Let s_n denote the sequence (1, 4^n, 10^n, 20^n, ...) regarded as an infinite column vector, where 1, 4, 10, 20, ... is the sequence of tetrahedral numbers A000292. It appears that the n-th row of this table is determined by the matrix product P^(-1)s_n, where P denotes Pascal's triangle A007318. - Peter Bala, Nov 26 2017
From Peter Bala, Mar 11 2018: (Start)
The observation above is correct.
The table entries T(n,k) are the coefficients when expressing the polynomial C(x+3,3)^p of degree 3*p in terms of falling factorials: C(x+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 0..3*p} T(p,k)*C(n,k+1).
The sum of the p-th powers of the tetrahedral numbers is also given by Sum_{i = 0..n-1} C(i+3,3)^p = Sum_{k = 3..3*p} A299041(p,k)*C(n+3,k+1) for p >= 1. (End)

			Row 3 contains 1,15,69,147,162,90,20, so Sum_{i=1..n} C(i+2,3)^3 = 4 * C(n+3,4) * [ a(1,3)/4 + a(2,3)*C(n-1,1)/5 + a(3,3)*C(n-1,2)/6 + ... + a(7,3)*C(n-1,6)/10 ] = 4 * C(n+3,4) * [ 1/4 + 15*C(n-1,1)/5 + 69*C(n-1,2)/6 + 147*C(n-1,3)/7 + 162*C(n-1,4)/8 + 90*C(n-1,5)/9 + 20*C(n-1,6)/10 ]. Cf. A086021 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
n=0 | 1
n=1 | 1  3   3    1
n=2 | 1 15  69  147   162    90    20
n=3 | 1 63 873 5191 16620 31560 36750 25830 10080 1680
...
Row 2: C(i+3,3)^2 = C(i,0) + 15*C(i,1) + 69*C(i,2) + 147*C(i,3) + 162*C(i,4) + 90*C(i,5) + 20*C(i,6). Hence, Sum_{i = 0..n-1} C(i+3,3)^2 =  C(n,1) + 15*C(n,2) + 69*C(n,3) + 147*C(n,4) + 162*C(n,5) + 90*C(n,6) + 20*C(n,7). (End)

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

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A087110, A174266, A299041.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+3, 3)^n, i= 0..k), k = 0..3*n), n = 0..8); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 4, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 3, 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, 3*p - 2}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

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

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+4, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+3, i-2*k)^(p-1) ].
From Peter Bala, Nov 26 2017: (Start)
Conjectural formula for table entries: T(n,k) = Sum_{j = 0..k} (-1)^(k+j)*binomial(k,j)*binomial(j+3,3)^n.
Conjecturally, the n-th row polynomial R(n,x) = 1/(1 + x)*Sum_{i >= 0} binomial(i+3,3)^n *(x/(1 + x))^n. (End)
From Peter Bala, Mar 11 2018: (Start)
The conjectures above are correct.
The following remarks assume the row and column indices start at 0.
T(n+1,k) = C(k+3,3)*T(n,k) + 3*C(k+2,3)*T(n,k-1) + 3*C(k+1,3)*T(n,k-2) + C(k,3)*T(n,k-3) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 3*n.
Sum_{k = 0..3*n} T(n,k)*binomial(x,k) = (binomial(x+3,3))^n.
x^3*R(n,x) = (1 + x)^3 * the n-th row polynomial of A299041.
R(n+1,x) = 1/3!*(1 + x)^3*(d/dx)^3 (x^3*R(n,x)).
(1 - x)^(3*n)*R(n,x/(1 - x)) gives the n-th row polynomial of A174266.
R(n,x) = (1 + x)^3 o (1 + x)^3 o ... o (1 + x)^3 (n factors), where o denotes the black diamond product of power series defined in Dukes and White. Note the polynomial x^3 o ... o x^3 (n factors) is the n-th row polynomial of A299041. (End)

Edited by Dean Hickerson, Aug 16 2003

A122193 Triangle T(n,k) of number of loopless multigraphs with n labeled edges and k labeled vertices and without isolated vertices, n >= 1; 2 <= k <= 2*n.

Original entry on oeis.org

1, 1, 6, 6, 1, 24, 114, 180, 90, 1, 78, 978, 4320, 8460, 7560, 2520, 1, 240, 6810, 63540, 271170, 604800, 730800, 453600, 113400, 1, 726, 43746, 774000, 6075900, 25424280, 61923960, 90720000, 78813000, 37422000, 7484400

Offset: 1

Vladeta Jovovic, Aug 24 2006

T(n,k) equals the number of arrangements on a line of n (nondegenerate) finite closed intervals having k distinct endpoints. See the 'IBM Ponder This' link. An example is given below. - Peter Bala, Jan 28 2018

T(n,k) equals the number of alignments of length k of n strings each of length 2. See Slowinski. Cf. A131689 (alignments of strings of length 1) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018

			Triangle begins:
  1;
  1,  6,   6;
  1, 24, 114,  180,   90;
  1, 78, 978, 4320, 8460, 7560, 2520;
  ...
From _Francisco Santos_, Nov 17 2017: (Start)
For n=3 edges and k=4 vertices there are three loopless multigraphs without isolated vertices: a path, a Y-graph, and the multigraph {12, 34, 34}. The number of labelings in each is 3!4!/a, where a is the number of automorphisms. This gives respectively 3!4!/2 = 72, 3!4!/6 = 24 and 3!4!/8 = 18, adding up to 72 + 24 + 18 = 114. (End)
From _Peter Bala_, Jan 28 2018: (Start)
T(2,3) = 6: Consider 2 (nondegenerate) finite closed intervals [a, b] and [c, d]. There are 6 arrangements of these two intervals with 3 distinct endpoints:
  ...a--b--d....  a < b = c < d
  ...a...c--b...  a < c < b = d
  ...a--d...b...  a = c < d < b
  ...a--b...d...  a = c < b < d
  ...c...a--d...  c < a < b = d
  ...c--a--b....  c < a = d < b
T(2,4) = 6: There are 6 arrangements of the two intervals with 4 distinct endpoints:
  ...a--b...c--d.....  no intersection a < b < c < d
  ...a...c...b...d...  a < c < b < d
  ...a...c--d...b....  [c,d] is a proper subset of [a,b]
  ...c...a...d...b...  c < a < d < b
  ...c...a--b...d... [a,b] is a proper subset of [c,d]
  ...c--d...a--b.....  no intersection c < d < a < b.
Sums of powers of triangular numbers:
Row 2: Sum_{i = 2..n-1} C(i,2)^2 = C(n,3) + 6*C(n,4) + 6*C(n,5);
Row 3: Sum_{i = 2..n-1} C(i,2)^3 = C(n,3) + 24*C(n,4) + 114*C(n,5) + 180*C(n,6) + 90*C(n,7). See A024166 and A085438.
exp( Sum_{n >= 1} R(n,2)*x^n/n ) = (1 + x + 19*x^2 + 1147*x^3 + 145606*x^4 + 31784062*x^5 + ... )^4
exp( Sum_{n >= 1} R(n,3)*x^n/n ) = (1 + x + 37*x^2 + 4453*x^3 + 1126375*x^4 + 489185863*x^5 + ... )^9
exp( Sum_{n >= 1} R(n,4)*x^n/n ) = (1 + x + 61*x^2 + 12221*x^3 + 5144411*x^4 + 3715840571*x^5 + ... )^16 (End)
From _Peter Bala_, Feb 04 2018: (Start)
T(3,3) = 24 alignments of length 3 of 3 strings each of length 2. Examples include
  (i) A B -    (ii) A - B
      - C D         - C D
      - E F         E F -
There are 18 alignments of type (i) with two gap characters in one of the columns (3 ways of putting 2 gap characters in a column x 2 ways to place the other letter in the row which doesn't yet have a gap character x 3 columns: there are 6 alignments of type (ii) with a single gap character in each column (3 ways to put a single gap character in the first column x 2 ways to then place a single gap character in the second column). (End)

Michael De Vlieger, Table of n, a(n) for n = 1..10000 (rows 1 <= n <= 100).
Peter Bala, Deformations of the Hadamard product of power series
Peter Bala, Notes on A122193
S. Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
J. Engbers and C. Stocker, Two Combinatorial Proofs of Identities Involving Sums of Powers of Binomial Coefficients, Integers 16 (2016), #A58.
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.
IBM Ponder This, Jan 01 2001
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522

Row sums give A055203.
Cf. A078739, A005799, A019538, A131689, A154283, A299041, A087127.
For Sum_{i = 2..n} C(i,2)^k see A024166 (k = 2), A085438 - A085442 ( k = 3 thru 7).

# Note that the function implements the full triangle because it can be
# much better reused and referenced in this form.
A122193 := (n,k) -> A078739(n,k)*k!/2^n:
# Displays the truncated triangle from the definition:
seq(print(seq(A122193(n,k),k=2..2*n)),n=1..6); # Peter Luschny, Mar 25 2011

t[n_, k_] := Sum[(-1)^(n - r) Binomial[n, r] StirlingS2[n + r, k], {r, 0, n}]; Table[t[n, k] k!/2^n, {n, 6}, {k, 2, 2 n}] // Flatten (* Michael De Vlieger, Nov 18 2017, after Jean-François Alcover at A078739 *)

Double e.g.f.: exp(-x)*Sum_{n>=0} exp(binomial(n,2)*y)*x^n/n!.
T(n,k) = S_{2,2}(n,k)*k!/2^n; S_{2,2} the generalized Stirling numbers A078739. - Peter Luschny, Mar 25 2011
From Peter Bala, Jan 28 2018: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i*(i-1)/2)^n.
T(n,k) = k*(k-1)/2*( T(n-1,k) + 2*T(n-1,k-1) + T(n-1,k-2) ) for 2 < k <= 2*n with boundary conditions T(n,2) = 1 for n >= 1 and T(n,k) = 0 if (k < 2) or (k > 2*n).
n-th row polynomial R(n,x) = Sum_{i >= 2} (i*(i-1)/2)^n * x^i/(1+x)^(i+1) for n >= 1.
1/(1-x)*R(n,x/(1-x)) = Sum_{i >= 2} (i*(i-1)/2)^n*x^i for n >= 1.
R(n,x) = 1/2^n*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*F(n+k,x), where F(n,x) = Sum_{k = 0..n} k!*Stirling2(n,k)*x^k is the n-th Fubini polynomial, the n-th row polynomial of A131689.
R(n,x) = x/(1+x)*1/2^n*Sum_{k = 0..n} binomial(n,k)*F(n+k,x) for n >= 1.
The polynomials Sum_{k = 2..2*n} T(n,k)*x^(k-2)*(1-x)^(2*n-k) are the row polynomials of A154283.
A154283 * A007318 equals the row reverse of this array.
Sum_{k = 2..2*n} T(n,k)*binomial(x,k) = ( binomial(x,2) )^n. Equivalently, Sum_{k = 2..2*n} (-1)^k*T(n,k)*binomial(x+k,k) = ( binomial(x+2,2) )^n. Cf. the Worpitzky-type identity Sum_{k = 1..n} A019538(n,k)*binomial(x,k) = x^n.
Sum_{i = 2..n-1} (i*(i-1)/2)^m = Sum_{k = 2..2*m} T(m,k) * binomial(n,k+1) for m >= 1. See Examples below.
R(n,x) = x^2 o x^2 o ... o x^2 (n factors), where o is the black diamond product of power series defined in Dukes and White. Note the polynomial x o x o ... o x (n factors) is the n-th row polynomial of A019538.
x^2*R(n,-1-x) = (1+x)^2*R(n,x) for n >= 1.
R(n+1,x) = 1/2*x^2*(d/dx)^2 ((1+x)^2*R(n,x)).
The zeros of R(n,x) belong to the interval [-1, 0].
Alternating row sums equal 1, that is R(n,-1) = 1.
R(n,-2) = 4*R(n,1) = 4*A055203(n).
4^n*Sum_{k = 2..2*n} T(n,k)*(-1/2)^k appears to equal (-1)^(n+1)*A005799(n) for n >= 1.
For k a nonzero integer, the power series A(k,x) := exp( Sum_{n >= 1} 1/k^2*R(n,k)*x^n/n ) appear to have integer coefficients. See the Example section.
Sum_{k = 2..2*n} T(n,k)*binomial(x,k-2) = binomial(x,2)^n - 2*binomial(x+1,2)^n + binomial(x+2,2)^n. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane (the corresponding property also holds for the row polynomials of A019538 with a factor of x removed). (End)
From Peter Bala, Mar 08 2018: (Start)
n-th row polynomial R(n,x) = coefficient of (z_1)^2 * ... * (z_n)^2 in the expansion of the rational function 1/(1 + x - x*(1 + z_1)*...*(1 + z_n)).
The n-th row of the table is given by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318 and v_n is the sequence (0, 0, 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. Cf. A087127. (End)

Definition corrected by Francisco Santos, Nov 17 2017

A062208 a(n) = Sum_{m>=0} binomial(m,3)^n*2^(-m-1).

Original entry on oeis.org

1, 1, 63, 16081, 10681263, 14638956721, 35941784497263, 143743469278461361, 874531783382503604463, 7687300579969605991710001, 93777824804632275267836362863, 1537173608464960118370398000894641, 32970915649974341628739088902163732463

Offset: 0

Angelo Dalli, Jun 13 2001

Number of alignments of n strings of length 3.

Conjectures: a(2*n) = 3 (mod 60) and a(2*n+1) = 1 (mod 60); for fixed k, the sequence a(n) (mod k) eventually becomes periodic with exact period a divisor of phi(k), where phi(k) is Euler's totient function A000010. - Peter Bala, Feb 04 2018

Alois P. Heinz, Table of n, a(n) for n = 0..100
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522

Cf. A000670, A055203, A001850, A126086.
See A062204 for further references, formulas and comments.
Cf. A001850, A062204, A062205, A299041.
Row n=3 of A262809.

A000629 := proc(n) local k ; sum( k^n/2^k,k=0..infinity) ; end: A062208 := proc(n) local a,stir,ni,n1,n2,n3,stir2,i,j,tmp ; a := 0 ; if n = 0 then RETURN(1) ; fi ; stir := combinat[partition](n) ; stir2 := {} ; for i in stir do if nops(i) <= 3 then tmp := i ; while nops(tmp) < 3 do tmp := [op(tmp),0] ; od: tmp := combinat[permute](tmp) ; for j in tmp do stir2 := stir2 union { j } ; od: fi ; od: for ni in stir2 do n1 := op(1,ni) ; n2 := op(2,ni) ; n3 := op(3,ni) ; a := a+combinat[multinomial](n,n1,n2,n3)*(A000629(3*n1+2*n2+n3)-1/2-2^(3*n1+2*n2+n3)/4)*(-3)^n2*2^n3 ; od: a/(2*6^n) ; end: seq(A062208(n),n=0..14) ; # R. J. Mathar, Apr 01 2008
a:=proc(n) options operator, arrow: sum(binomial(m, 3)^n*2^(-m-1),m=0.. infinity) end proc: seq(a(n),n=0..12); # Emeric Deutsch, Mar 22 2008

a[n_] = Sum[2^(-1-m)*((m-2)*(m-1)*m)^n, {m, 0, Infinity}]/6^n; a /@ Range[0, 12] (* Jean-François Alcover, Jul 13 2011 *)
With[{r = 3}, Flatten[{1, Table[Sum[Sum[(-1)^i*Binomial[j, i]*Binomial[j - i, r]^k, {i, 0, j}], {j, 0, k*r}], {k, 1, 15}]}]] (* Vaclav Kotesovec, Mar 22 2016 *)

From Vaclav Kotesovec, Mar 22 2016: (Start)
a(n) ~ 3^(2*n + 1/2) * n!^3 / (Pi * n * 2^(n+3) * (log(2))^(3*n+1)).
a(n) ~ sqrt(Pi)*3^(2*n+1/2)*n^(3*n+1/2) / (2^(n+3/2)*exp(3*n)*(log(2))^(3*n+1)).
(End)
a(n) = Sum_{k = 3..3*n} Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)* binomial(i,3)^n. Row sums of A299041. - Peter Bala, Feb 04 2018

New definition from Vladeta Jovovic, Mar 01 2008

Edited by N. J. A. Sloane, Sep 19 2009 at the suggestion of Max Alekseyev

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A122193 Triangle T(n,k) of number of loopless multigraphs with n labeled edges and k labeled vertices and without isolated vertices, n >= 1; 2 <= k <= 2*n.

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A062208 a(n) = Sum_{m>=0} binomial(m,3)^n*2^(-m-1).

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A131689 Triangle of numbers T(n,k) = k!Stirling2(n,k) = A000142(k)A048993(n,k) read by rows, T(n, k) for 0 <= k <= n.