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
Examples
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)
Links
- 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.
Crossrefs
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).
Classes of patterns:
- A000142 = strict
- A019536 = necklace
- A032011 = distinct multiplicities
- A060223 = Lyndon
- A296975 = aperiodic
- A349058 = weakly alternating
- A351200 = distinct runs
- A351292 = distinct run-lengths
Programs
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Julia
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
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Maple
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
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Mathematica
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 *)
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PARI
{T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(k + i) * binomial(k, i) * i^n))}; /* Michael Somos, Jul 08 2018 */
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SageMath
@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
Formula
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
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