cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A155755 Triangle T(n, k) = A143491(n+2, k+2) + A143491(n+2, n-k+2), read by rows.

Original entry on oeis.org

2, 3, 3, 7, 10, 7, 25, 35, 35, 25, 121, 168, 142, 168, 121, 721, 1064, 735, 735, 1064, 721, 5041, 8055, 5399, 3330, 5399, 8055, 5041, 40321, 69299, 49371, 22449, 22449, 49371, 69299, 40321, 362881, 663740, 509830, 223300, 109298, 223300, 509830, 663740, 362881
Offset: 0

Views

Author

Roger L. Bagula, Jan 26 2009

Keywords

Comments

This symmetric summation of the triangle A143491 is equivalent to the coefficient [x^m] (p_n(x) + x^n*p_n(1/x)) of the polynomials defined in A143491 plus their reverses.

Examples

			Triangle begins as:
       2;
       3,      3;,
       7,     10,      7;
      25,     35,     35,     25;
     121,    168,    142,    168,    121;
     721,   1064,    735,    735,   1064,    721;
    5041,   8055,   5399,   3330,   5399,   8055,   5041;
   40321,  69299,  49371,  22449,  22449,  49371,  69299,  40321;
  362881, 663740, 509830, 223300, 109298, 223300, 509830, 663740, 362881;
		

Crossrefs

Cf. A143491.

Programs

  • Mathematica
    (* First program *)
    q[x_, n_]:= Product[x +n-i+1, {i,0,n-1}];
    p[x_, n_]:= q[x, n] + x^n*q[1/x, n];
    Table[CoefficientList[p[x, n], x], {n,0,12}]//Flatten (* modified by G. C. Greubel, Jun 06 2021 *)
    (* Second program *)
    A143491[n_, k_]:= (n-2)!*Sum[(n-k-j+1)*Abs[StirlingS1[j+k-2, k-2]]/(j+k-2)!, {j,0,n-k}];
    A155755[n_, k_]:= A143491[n+2, k+2] + A143491[n+2, n-k+2];
    Table[A155755[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 06 2021 *)
  • Sage
    def A143491(n,k): return factorial(n-2)*sum( (n-k-j+1)*stirling_number1(j+k-2, k-2)/factorial(j+k-2) for j in (0..n-k) )
    def A155755(n,k): return A143491(n+2, k+2) + A143491(n+2, n-k+2)
    flatten([[A155755(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 06 2021

Formula

T(n, k) = A143491(n+2, k+2) + A143491(n+2, n-k+2).
Sum_{k=0..n} T(n, k) = (n+2)!.

A094638 Triangle read by rows: T(n,k) = |s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind A008276 (1 <= k <= n; in other words, the unsigned Stirling numbers of the first kind in reverse order).

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24, 1, 15, 85, 225, 274, 120, 1, 21, 175, 735, 1624, 1764, 720, 1, 28, 322, 1960, 6769, 13132, 13068, 5040, 1, 36, 546, 4536, 22449, 67284, 118124, 109584, 40320, 1, 45, 870, 9450, 63273, 269325, 723680, 1172700, 1026576, 362880
Offset: 1

Views

Author

André F. Labossière, May 17 2004

Keywords

Comments

Triangle of coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in decreasing powers of x. - T. D. Noe, Feb 22 2008
Row n also gives the number of permutation of 1..n with complexity 0,1,...,n-1. See the comments in A008275. - N. J. A. Sloane, Feb 08 2019
T(n,k) is the number of deco polyominoes of height n and having k columns. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: T(2,1)=1 and T(2,2)=1 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, having, respectively, 1 and 2 columns. - Emeric Deutsch, Aug 14 2006
Sum_{k=1..n} k*T(n,k) = A121586. - Emeric Deutsch, Aug 14 2006
Let the triangle U(n,k), 0 <= k <= n, read by rows, be given by [1,0,1,0,1,0,1,0,1,0,1,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,...] where DELTA is the operator defined in A084938; then T(n,k) = U(n-1,k-1). - Philippe Deléham, Jan 06 2007
From Tom Copeland, Dec 15 2007: (Start)
Consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5, ...).
Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...). And T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n > 0. Call this sequence the right factorial (n+)!.
Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1, 0, 0, 0, ...), the left factorial (n-)!.
Sampling every other integer to the right, we obtain (1,3,5,7,9,...). T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n > 0, the right double factorial, (n+)!!.
Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7,...). T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left double factorial, (n-)!!.
Sampling every 3 steps to the right, we obtain (1,4,7,10,...). T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...), giving A007559 for n > 0, the right triple factorial, (n+)!!!.
Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...) = signed A008544, the left triple factorial, (n-)!!!.
The list partition transform A133314 of [1,T * c(t)] gives [1,T * c(-t)] with all odd terms negated; e.g., LPT[1,T*c(2)] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147). And e.g.f. for [1,T * c(t)] = (1-xt)^(-1/t).
The above results hold for t any real or complex number. (End)
Let R_n(x) be the real and I_n(x) the imaginary part of Product_{k=0..n} (x + I*k). Then, for n=1,2,..., we have R_n(x) = Sum_{k=0..floor((n+1)/2)}(-1)^k*Stirling1(n+1,n+1-2*k)*x^(n+1-2*k), I_n(x) = Sum_{k=0..floor(n/2)}(-1)^(k+1)*Stirling1(n+1,n-2*k)*x^(n-2*k). - Milan Janjic, May 11 2008
T(n,k) is also the number of permutations of n with "reflection length" k (i.e., obtained from 12..n by k not necessarily adjacent transpositions). For example, when n=3, 132, 213, 321 are obtained by one transposition, while 231 and 312 require two transpositions. - Kyle Petersen, Oct 15 2008
From Tom Copeland, Nov 02 2010: (Start)
[x^(y+1) D]^n = x^(n*y) [T(n,1)(xD)^n + T(n,2)y (xD)^(n-1) + ... + T(n,n)y^(n-1)(xD)], with D the derivative w.r.t. x.
E.g., [x^(y+1) D]^4 = x^(4*y) [(xD)^4 + 6 y(xD)^3 + 11 y^2(xD)^2 + 6 y^3(xD)].
(xD)^m can be further expanded in terms of the Stirling numbers of the second kind and operators of the form x^j D^j. (End)
With offset 0, 0 <= k <= n: T(n,k) is the sum of products of each size k subset of {1,2,...,n}. For example, T(3,2) = 11 because there are three subsets of size two: {1,2},{1,3},{2,3}. 1*2 + 1*3 + 2*3 = 11. - Geoffrey Critzer, Feb 04 2011
The Kn11, Fi1 and Fi2 triangle sums link this triangle with two sequences, see the crossrefs. For the definitions of these triangle sums see A180662. The mirror image of this triangle is A130534. - Johannes W. Meijer, Apr 20 2011
T(n+1,k+1) is the elementary symmetric function a_k(1,2,...,n), n >= 0, k >= 0, (a_0(0):=1). See the T. D. Noe and Geoffrey Critzer comments given above. For a proof see the Stanley reference, p. 19, Second Proof. - Wolfdieter Lang, Oct 24 2011
Let g(t) = 1/d(log(P(j+1,-t)))/dt (see Tom Copeland's 2007 formulas). The Mellin transform (t to s) of t*Dirac[g(t)] gives Sum_{n=1..j} n^(-s), which as j tends to infinity gives the Riemann zeta function for Re(s) > 1. Dirac(x) is the Dirac delta function. The complex contour integral along a circle of radius 1 centered at z=1 of z^s/g(z) gives the same result. - Tom Copeland, Dec 02 2011
Rows are coefficients of the polynomial expansions of the Pochhammer symbol, or rising factorial, Pch(n,x) = (x+n-1)!/(x-1)!. Expansion of Pch(n,xD) = Pch(n,Bell(.,:xD:)) in a polynomial with terms :xD:^k=x^k*D^k gives the Lah numbers A008297. Bell(n,x) are the unsigned Bell polynomials or Stirling polynomials of the second kind A008277. - Tom Copeland, Mar 01 2014
From Tom Copeland, Dec 09 2016: (Start)
The Betti numbers, or dimension, of the pure braid group cohomology. See pp. 12 and 13 of the Hyde and Lagarias link.
Row polynomials and their products appear in presentation of the Jack symmetric functions of R. Stanley. See Copeland link on the Witt differential generator.
(End)
From Tom Copeland, Dec 16 2019: (Start)
The e.g.f. given by Copeland in the formula section appears in a combinatorial Dyson-Schwinger equation of quantum field theory in Yeats in Thm. 2 on p. 62 related to a Hopf algebra of rooted trees. See also the Green function on p. 70.
Per comments above, this array contains the coefficients in the expansion in polynomials of the Euler, or state number, operator xD of the rising factorials Pch(n,xD) = (xD+n-1)!/(xD-1)! = x [:Dx:^n/n!]x^{-1} = L_n^{-1}(-:xD:), where :Dx:^n = D^n x^n and :xD:^n = x^n D^n. The polynomials L_n^{-1} are the Laguerre polynomials of order -1, i.e., normalized Lah polynomials.
The Witt differential operators L_n = x^(n+1) D and the row e.g.f.s appear in Hopf and dual Hopf algebra relations presented by Foissy. The Witt operators satisfy L_n L_k - L_k L_n = (k-n) L_(n+k), as for the dual Hopf algebra. (End)

Examples

			Triangle starts:
  1;
  1,  1;
  1,  3,  2;
  1,  6, 11,  6;
  1, 10, 35, 50, 24;
...
		

References

  • M. Miyata and J. W. Son, On the complexity of permutations and the metric space of bijections, Tensor, 60 (1998), No. 1, 109-116 (MR1768839).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 18, equations 18:4:2 - 18:4:8 at page 151.
  • R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.

Crossrefs

A008276 gives the (signed) Stirling numbers of the first kind.
Cf. A000108, A014137, A001246, A033536, A000984, A094639, A006134, A082894, A002897, A079727, A000217 (2nd column), A000914 (3rd column), A001303 (4th column), A000915 (5th column), A053567 (6th column), A000142 (row sums).
Triangle sums (see the comments): A124380 (Kn11), A001710 (Fi1, Fi2). - Johannes W. Meijer, Apr 20 2011

Programs

  • GAP
    Flat(List([1..10], n-> List([1..n], k-> Stirling1(n,n-k+1) ))); # G. C. Greubel, Dec 29 2019
  • Haskell
    a094638 n k = a094638_tabl !! (n-1) !! (k-1)
    a094638_row n = a094638_tabl !! (n-1)
    a094638_tabl = map reverse a130534_tabl
    -- Reinhard Zumkeller, Aug 01 2014
    
  • Magma
    [(-1)^(k+1)*StirlingFirst(n,n-k+1): k in [1..n], n in [1..10]]; // G. C. Greubel, Dec 29 2019
    
  • Maple
    T:=(n,k)->abs(Stirling1(n,n+1-k)): for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form. # Emeric Deutsch, Aug 14 2006
  • Mathematica
    Table[CoefficientList[Series[Product[1 + i x, {i,n}], {x,0,20}], x], {n,0,6}] (* Geoffrey Critzer, Feb 04 2011 *)
    Table[Abs@StirlingS1[n, n-k+1], {n, 10}, {k, n}]//Flatten (* Michael De Vlieger, Aug 29 2015 *)
  • Maxima
    create_list(abs(stirling1(n+1,n-k+1)),n,0,10,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
    
  • PARI
    {T(n,k)=if(n<1 || k>n,0,(n-1)!*polcoeff(polcoeff(x*y/(1 - x*y+x*O(x^n))^(1 + 1/y),n,x),k,y))} /* Paul D. Hanna, Jul 21 2011 */
    
  • Sage
    [[stirling_number1(n, n-k+1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, Dec 29 2019
    

Formula

With P(n,t) = Sum_{k=0..n-1} T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n-1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1-tx)^(-1/t). T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] evaluated at t=x=0. (1-tx)^(-1/t) - 1 is the e.g.f. for a plane m-ary tree when t=m-1. See Bergeron et al. in "Varieties of Increasing Trees". - Tom Copeland, Dec 09 2007
First comment and formula above rephrased as o.g.f. for row n: Product_{i=0...n} (1+i*x). - Geoffrey Critzer, Feb 04 2011
n-th row polynomials with alternate signs are the characteristic polynomials of the (n-1)x(n-1) matrices with 1's in the superdiagonal, (1,2,3,...) in the main diagonal, and the rest zeros. For example, the characteristic polynomial of [1,1,0; 0,2,1; 0,0,3] is x^3 - 6*x^2 + 11*x - 6. - Gary W. Adamson, Jun 28 2011
E.g.f.: A(x,y) = x*y/(1 - x*y)^(1 + 1/y) = Sum_{n>=1, k=1..n} T(n,k)*x^n*y^k/(n-1)!. - Paul D. Hanna, Jul 21 2011
With F(x,t) = (1-t*x)^(-1/t) - 1 an e.g.f. for the row polynomials P(n,t) of A094638 with P(0,t)=0, G(x,t)= [1-(1+x)^(-t)]/t is the comp. inverse in x. Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+x)^(t+1),
P(n,t) = [(H(x,t)*d/dx)^n] x evaluated at x=0; i.e.,
F(x,t) = exp[x*P(.,t)] = exp[x*H(u,t)*d/du] u, evaluated at u = 0.
Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 20 2011
T(n,k) = |A008276(n,k)|. - R. J. Mathar, May 19 2016
The row polynomials of this entry are the reversed row polynomials of A143491 multiplied by (1+x). E.g., (1+x)(1 + 5x + 6x^2) = (1 + 6x + 11x^2 + 6x^3). - Tom Copeland, Dec 11 2016
Regarding the row e.g.f.s in Copeland's 2007 formulas, e.g.f.s for A001710, A001715, and A001720 give the compositional inverses of the e.g.f. here for t = 2, 3, and 4 respectively. - Tom Copeland, Dec 28 2019

Extensions

Edited by Emeric Deutsch, Aug 14 2006

A143494 Triangle read by rows: 2-Stirling numbers of the second kind.

Original entry on oeis.org

1, 2, 1, 4, 5, 1, 8, 19, 9, 1, 16, 65, 55, 14, 1, 32, 211, 285, 125, 20, 1, 64, 665, 1351, 910, 245, 27, 1, 128, 2059, 6069, 5901, 2380, 434, 35, 1, 256, 6305, 26335, 35574, 20181, 5418, 714, 44, 1, 512, 19171, 111645, 204205, 156660, 58107, 11130, 1110, 54, 1
Offset: 2

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 2 of the r-Stirling numbers of the second kind. The 2-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1 and 2 belong to distinct subsets.
More generally, the r-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the numbers 1, 2, ..., r belong to distinct subsets. The case r = 1 gives the usual Stirling numbers of the second kind A008277; for other cases see A143495 (r = 3) and A143496 (r = 4).
The lower unitriangular array of r-Stirling numbers of the second kind equals the matrix product P^(r-1) * S (with suitable offsets in the row and column indexing), where P is Pascal's triangle, A007318 and S is the array of Stirling numbers of the second kind, A008277.
For the definition of and entries relating to the corresponding r-Stirling numbers of the first kind see A143491. For entries on r-Lah numbers refer to A143497. The theory of r-Stirling numbers of both kinds is developed in [Broder].
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-2)*E^n*x^2 = Sum_{k = 0..n} T(n+2,k+2)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 2..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_2(x) = x^2. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 2-Eulerian numbers E_2(n,j) := A144696(n,j): T(n,k) = 2!/k!*Sum_ {j = n-k..n-2} E_2(n,j)*binomial(j,n-k) for n >= k >= 2. (End)
From Wolfdieter Lang, Sep 29 2011: (Start)
T(n,k) = S(n,k,2), n>=k>=2, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column no. k from (A20) with k->2, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(2*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(2*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. (End)

Examples

			Triangle begins
  n\k|...2....3....4....5....6....7
  =================================
  2..|...1
  3..|...2....1
  4..|...4....5....1
  5..|...8...19....9....1
  6..|..16...65...55...14....1
  7..|..32..211..285..125...20....1
  ...
T(4,3) = 5. The set {1,2,3,4} can be partitioned into three subsets such that 1 and 2 belong to different subsets in 5 ways: {{1}{2}{3,4}}, {{1}{3}{2,4}}, {{1}{4}{2,3}}, {{2}{3}{1,4}} and {{2}{4}{1,3}}; the remaining possibility {{1,2}{3}{4}} is not allowed.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 2    1           |     | 2   1          ||0  1           ||0  1          |
| 4    5   1       |  =  | 4   3   1      ||0  2   1       ||0  0  1       | ...
| 8   19   9   1   |     | 8   7   4  1   ||0  4   3  1    ||0  0  2  1    |
|16   65  55  14  1|     |16  15  11  6  1||0  8   7  4  1 ||0  0  4  3  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - x)). See A055248. (End)
		

Crossrefs

A001047 (column 3), A005493 (row sums), A008277, A016269 (column 4), A025211 (column 5), A049444 (matrix inverse), A074051 (alt. row sums).

Programs

  • Maple
    with combinat: T := (n, k) -> (1/(k-2)!)*add ((-1)^(k-i)*binomial(k-2,i)*(i+2)^(n-2),i = 0..k-2): for n from 2 to 11 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := StirlingS2[n, k] - StirlingS2[n-1, k]; Flatten[ Table[ t[n, k], {n, 2, 11}, {k, 2, n}]] (* Jean-François Alcover, Dec 02 2011 *)
  • Sage
    @CachedFunction
    def stirling2r(n, k, r) :
        if n < r: return 0
        if n == r: return 1 if k == r else 0
        return stirling2r(n-1,k-1,r) + k*stirling2r(n-1,k,r)
    A143494 = lambda n,k: stirling2r(n, k, 2)
    for n in (2..6):
        [A143494(n, k) for k in (2..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+2,k+2) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*C(k,i)*(i+2)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - Stirling2(n-1,k) for n, k >= 2.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*T(n-1,k) for n > 2, with boundary conditions T(n,1) = T(1,n) = 0 for all n, T(2,2) = 1 and T(2,k) = 0 for k > 2. Special cases: T(n,2) = 2^(n-2); T(n,3) = 3^(n-2) - 2^(n-2).
As a sum of monomial functions of degree m: T(n+m,n) = Sum_{2 <= i_1 <= ... <= i_m <= n} (i_1*i_2*...*i_m). For example, T(6,4) = Sum_{2 <= i <= j <= 4} (i*j) = 2*2 + 2*3 + 2*4 + 3*3 + 3*4 + 4*4 = 55.
E.g.f. column k+2 (with offset 2): 1/k!*exp(2*x)*(exp(x) - 1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-2*x)*(1-3*x)*...*(1-k*x)).
E.g.f.: exp(2*t + x*(exp(t) - 1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+2,k+2) *x^k*t^n/n! = Sum_{n >= 0} B_n(2;x)*t^n/n! = 1 + (2 + x)*t/1! + (4 + 5*x + x^2)*t^2/2! + ..., where the row polynomial B_n(2;x) := Sum_{k = 0..n} T(n+2,k+2)*x^k denotes the 2-Bell polynomial.
Dobinski-type identities: Row polynomial B_n(2;x) = exp(-x)*Sum_{i >= 0} (i + 2)^n*x^i/i!. Sum_{k = 0..n} k!*T(n+2,k+2)*x^k = Sum_{i >= 0} (i + 2)^n*x^i/(1 + x)^(i+1).
The T(n,k) are the connection coefficients between falling factorials and the shifted monomials (x + 2)^(n-2). For example, from row 4 we have 4 + 5*x + x*(x - 1) = (x + 2)^2, while from row 5 we have 8 + 19*x + 9*x*(x - 1) + x*(x - 1)*(x - 2) = (x + 2)^3.
The row sums of the array are the 2-Bell numbers, B_n(2;1), equal to A005493(n-2). The alternating row sums are the complementary 2-Bell numbers, B_n(2;-1), equal to (-1)^n*A074051(n-2).
This array is the matrix product P * S, where P denotes the Pascal triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
Also, this array equals the transpose of the upper triangular array A126351. The inverse array is A049444, the signed 2-Stirling numbers of the first kind. See A143491 for the unsigned version of the inverse.
Let f(x) = exp(exp(x)). Then for n >= 1, the row polynomials R(n,x) are given by R(n+2,exp(x)) = 1/f(x)*(d/dx)^n(exp(2*x)*f(x)). Similar formulas hold for A008277, A039755, A105794, A111577 and A154537. - Peter Bala, Mar 01 2012

A049444 Generalized Stirling number triangle of first kind.

Original entry on oeis.org

1, -2, 1, 6, -5, 1, -24, 26, -9, 1, 120, -154, 71, -14, 1, -720, 1044, -580, 155, -20, 1, 5040, -8028, 5104, -1665, 295, -27, 1, -40320, 69264, -48860, 18424, -4025, 511, -35, 1, 362880, -663696, 509004, -214676, 54649, -8624, 826, -44, 1, -3628800, 6999840, -5753736
Offset: 0

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Author

Keywords

Comments

T(n, k) = ^2P_n^k in the notation of the given reference with T(0, 0) := 1. The monic row polynomials s(n,x) := Sum_{m=0..n} T(n, k)*x^k which are s(n, x) = Product_{j=0..n-1} (x-(2+j)), n >= 1 and s(0, x)=1 satisfy s(n, x+y) = Sum_{k=0..n} binomial(n, k)*s(k,x)*S1(n-k, y), with the Stirling1 polynomials S1(n, x) = Sum_{m=1..n} (A008275(n, m)*x^m) and S1(0, x)=1.
In the umbral calculus (see the S. Roman reference given in A048854) the s(n, x) polynomials are called Sheffer polynomials for (exp(2*t), exp(t)-1). This translates to the usual exponential Riordan (Sheffer) notation (1/(1+x)^2, log(1+x)).
See A143491 for the unsigned version of this array and A143494 for the inverse. - Peter Bala, Aug 25 2008
Corresponding to the generalized Stirling number triangle of second kind A137650. - Peter Luschny, Sep 18 2011
Unsigned, reversed rows (cf. A145324, A136124) are the dimensions of the cohomology of a complex manifold with a symmetric group (S_n) action. See p. 17 of the Hyde and Lagarias link. See also the Murri link for an interpretation as the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces. - Tom Copeland, Dec 09 2016
The row polynomials s(n, x) = (-1)^n*risingfactorial(2 - x, n) are related to the column sequences of the unsigned Abel triangle A137452(n, k), for k >= 2. See the formula there. - Wolfdieter Lang, Nov 21 2022

Examples

			The Triangle  begins:
n\k       0       1        2       3       4      5      6    7   8 9 ...
0:        1
1:       -2       1
2:        6      -5        1
3:      -24      26       -9       1
4:      120    -154       71     -14       1
5      -720    1044     -580     155     -20      1
6:     5040   -8028     5104   -1665     295    -27      1
7:   -40320   69264   -48860   18424   -4025    511    -35    1
8:   362880 -663696   509004 -214676   54649  -8624    826  -44
9: -3628800 6999840 -5753736 2655764 -761166 140889 -16884 1266 -54 1
...  [reformatted by _Wolfdieter Lang_, Nov 21 2022]
		

References

  • Y. Manin, Frobenius Manifolds, Quantum Cohomology and Moduli Spaces, American Math. Soc. Colloquium Publications Vol. 47, 1999. [From Tom Copeland, Jun 29 2008]
  • S. Roman, The Umbral Calculus, Academic Press, 1984 (also Dover Publications, 2005).

Crossrefs

Unsigned column sequences are A000142(n+1), A001705-A001709. Row sums (signed triangle): n!*(-1)^n, row sums (unsigned triangle): A001710(n-2). Cf. A008275 (Stirling1 triangle).

Programs

  • Haskell
    a049444 n k = a049444_tabl !! n !! k
    a049444_row n = a049444_tabl !! n
    a049444_tabl = map fst $ iterate (\(row, i) ->
       (zipWith (-) ([0] ++ row) $ map (* i) (row ++ [0]), i + 1)) ([1], 2)
    -- Reinhard Zumkeller, Mar 11 2014
  • Maple
    A049444_row := proc(n) local k,i;
    add(add(Stirling1(n, n-i), i=0..k)*x^(n-k-1),k=0..n-1);
    seq(coeff(%,x,k),k=1..n-1) end:
    seq(print(A049444_row(n)),n=1..7); # Peter Luschny, Sep 18 2011
    A049444:= (n, k)-> add((-1)^(n-j)*(n-j+1)!*binomial(n, j)*Stirling1(j, k), j=0..n):
    seq(print(seq(A049444(n, k), k=0..n)), n=0..11);  # Mélika Tebni, May 02 2022
  • Mathematica
    t[n_, i_] = Sum[(-1)^k*Binomial[n, k]*(k+1)!*StirlingS1[n-k, i], {k, 0, n-i}]; Flatten[Table[t[n, i], {n, 0, 9}, {i, 0, n}]] [[1 ;; 48]]
    (* Jean-François Alcover, Apr 29 2011, after Milan Janjic *)

Formula

T(n, k) = T(n-1, k-1) - (n+1)*T(n-1, k), n >= k >= 0; T(n, k) = 0, n < k; T(n, -1) = 0, T(0, 0) = 1.
E.g.f. for k-th column of signed triangle: ((log(1+x))^k)/(k!*(1+x)^2).
Triangle (signed) = [-2, -1, -3, -2, -4, -3, -5, -4, -6, -5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, ...]; triangle (unsigned) = [2, 1, 3, 2, 4, 3, 5, 4, 6, 5, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...], where DELTA is Deléham's operator defined in A084938 (unsigned version in A143491).
E.g.f.: (1 + x)^(y-2). - Vladeta Jovovic, May 17 2004 [For row polynomials s(n, y)]
With P(n, t) = Sum_{j=0..n-2} T(n-2,j) * t^j and P(1, t) = -1 and P(0, t) = 1, then G(x, t) = -1 + exp[P(.,t)*x] = [(1+x)^t - 1 - t^2 * x] / [t(t-1)], whose compositional inverse in x about 0 is given in A074060. G(x, 0) = -log(1+x) and G(x, 1) = (1+x) log(1+x) - 2x. G(x, q^2) occurs in formulas on pages 194-196 of the Manin reference. - Tom Copeland, Feb 17 2008
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then T(n,i) = f(n,i,2), for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
T(n, k) = Sum_{j=0..n} (-1)^(n-j)*(n-j+1)!*binomial(n, j)*Stirling1(j, k). - Mélika Tebni, May 02 2022
From Wolfdieter Lang, Nov 24 2022: (Start)
Recurrence for row polynomials {s(n, x)}_{n>=0}: s(0, x) = 1, s(n, x) = (x - 2)*exp(-(d/dx)) s(n-1, x), for n >= 1. This is adapted from the general Sheffer result given by S. Roman, Corollary 3.7.2., p. 50.
Recurrence for column sequence {T(n, k)}{n>=k}: T(n, n) = 1, T(n, k) = (n!/(n-k))*Sum{j=k..n-1} (1/j!)*(a(n-1-j) + k*beta(n-1-j))*T(n-1, k), for k >= 0, where alpha = repeat(-2, 2) and beta(n) = [x^n] (d/dx)log(log(x)/x) = (-1)^(n+1)*A002208(n+1)/A002209(n+1), for n >= 0. This is the adapted Boas-Buck recurrence, also given in Rainville, Theorem 50., p. 141, For the references and a comment see A046521. (End)

Extensions

Second formula corrected by Philippe Deléham, Nov 09 2008

A145324 Triangle read by rows: coefficients of 1; 1(X+2); 1(X+2)(X+3); 1(X+2)(X+3)(X+4); ....

Original entry on oeis.org

1, 1, 2, 1, 5, 6, 1, 9, 26, 24, 1, 14, 71, 154, 120, 1, 20, 155, 580, 1044, 720, 1, 27, 295, 1665, 5104, 8028, 5040, 1, 35, 511, 4025, 18424, 48860, 69264, 40320, 1, 44, 826, 8624, 54649, 214676, 509004, 663696, 362880, 1, 54, 1266, 16884, 140889, 761166
Offset: 1

Views

Author

Jose Ramon Real, Oct 07 2008

Keywords

Comments

The last number of row n is n!.
Essentially the triangle given by [1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [2, 1, 3, 2, 4, 3, 5, 4, 6, 5, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 09 2008
T(n+1,k+1) = a_k(2,3,...,n+1), n >= 0, k = 0..n, with the elementary symmetric function a_k(x[1],x[2],...,x[n]), with a_0(0):=1. E.g., a_2(2,3,4) = 2*3 + 2*4 + 3*4 = 26 = T(4,3). - Wolfdieter Lang, Oct 24 2011

Examples

			From _Wolfdieter Lang_, Oct 24 2011: (Start)
n\k 1  2   3    4     5    6     7 ...
1:  1
2:  1  2
3:  1  5   6
4:  1  9  26   24
5:  1 14  71  154   120
6:  1 20 155  580  1044  720
7:  1 27 295 1665  5104 8028  5040
...
T(4,3)= 26 = |s(5,3)| - |s(5,4)| + |s(5,5)| = 35 - 10 + 1.
(End)
		

Programs

  • Maple
    A145324 := proc(n,k) coeftayl( 1*mul(x+i,i=2..n),x=0,n-k) ; end: for n from 1 to 11 do for k from 1 to n do printf("%d,",A145324(n,k)) ; od: od: # R. J. Mathar, Oct 10 2008
  • Mathematica
    Table[Reverse[CoefficientList[Product[x+j, {j, 2, k}], x]], {k, 1, 15}] // Flatten (* Robert A. Russell, Sep 29 2018 *)

Formula

T(n,k) = A143491(n+1,n+2-k). - R. J. Mathar, Oct 10 2008
T(n,k) = Sum_{m=0..k-1} (-1)^m*|s(n+1, n+2-k+m)|, n >= 1, k = 1..n, with the Stirling numbers of the first kind s(n,k) = A048994(n,k). - Wolfdieter Lang, Oct 24 2011
T(n,k) = T(n-1,k)+n*T(n-1,k-1). - Mikhail Kurkov, Jun 26 2018

Extensions

More terms from R. J. Mathar, Oct 10 2008

A143497 Triangle of unsigned 2-Lah numbers.

Original entry on oeis.org

1, 4, 1, 20, 10, 1, 120, 90, 18, 1, 840, 840, 252, 28, 1, 6720, 8400, 3360, 560, 40, 1, 60480, 90720, 45360, 10080, 1080, 54, 1, 604800, 1058400, 635040, 176400, 25200, 1890, 70, 1, 6652800, 13305600, 9313920, 3104640, 554400, 55440, 3080, 88, 1
Offset: 2

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

For a signed version of this triangle see A062137. The unsigned 2-Lah number L(2; n,k) gives the number of partitions of the set {1, 2, ..., n} into k ordered lists with the restriction that the elements 1 and 2 must belong to different lists. More generally, the unsigned r-Lah number L(r; n, k) gives the number of partitions of the set {1, 2, ..., n} into k ordered lists with the restriction that the elements 1, 2, ..., r belong to different lists. If r = 1 there is no restriction and we obtain the unsigned Lah numbers A105278. For other cases see A143498 (r=3) and A143499 (r=4). We make some remarks on the general case.
The unsigned r-Lah numbers occur as connection constants in the generalized Lah identity (x + 2*r - 1)*(x + 2*r)*...*(x + 2*r + n - r - 2) = Sum_{k=r..n} L(r; n, k)*(x - 1)*(x - 2)*...*(x - k + r) for n >= r and where any empty products are taken equal to 1 (for a bijective proof of the identity, follow the proof of [Petkovsek and Pisanski] but restrict the first r of the Argonauts to different paths).
The unsigned r-Lah numbers satisfy the same recurrence as the unsigned Lah numbers, namely, L(r; n, k) = (n + k - 1)*L(r; n - 1,k) + L(r; n - 1,k - 1), but with the boundary conditions: L(r; n, k) = 0 if n < r or if k < r; L(r; r, r) = 1. The recurrence has the explicit solution L(r; n, k) = ((n - r)!/(k - r)!)*binomial(n + r - 1, k + r - 1) for n, k >= r. It follows that the unsigned r-Lah numbers have 'vertical' generating functions for k >= r of the form Sum_{n>=k} L(r; n, k)*t^n/(n -r)! = 1/(k - r)!*t^k/(1 - t)^(k + r). This yields the e.g.f. for the array of unsigned r-restricted Lah numbers in the form: Sum_{n,k>=r} L(r; n, k)*x^k*t^n/(n-r)! = (x*t)^r * 1/(1 - t)^(2*r) * exp(x*t/(1 - t)) = (x*t)^r (1 + (2*r + x)*t + (2r*(2*r + 1) + 2*(2*r + 1)*x + x^2)*t^2/2! + ...).
The array of unsigned r-Lah numbers begins
1
2r 1
2r*(2r+1) 2*(2r+1) 1
2r*(2r+1)*(2r+2) 3*(2r+1)*(2r+2) 3*(2r+2) 1
...
and equals exp(D(r)), where D(r) is the array with the sequence (2*r, 2*(2*r + 1), 3*(2*r + 2), 4*(2*r + 3), ...) on the main subdiagonal and zeros everywhere else.
The unsigned r-Lah numbers are related to the r-Stirling numbers: the lower triangular array of unsigned r-Lah numbers may be expressed as the matrix product St1(r) * St2(r), where St1(r) and St2(r) denote the arrays of r-Stirling numbers of the first and second kind respectively. The theory of r-Stirling numbers is developed in [Broder]. See A143491 - A143496 for tables of r-Stirling numbers. An alternative factorization for the array is as St1 * P^(2r - 2) * St2, where P denotes Pascal's triangle, A007318, St1 is the triangle of unsigned Stirling numbers of the first kind, abs(A008275) and St2 denotes the triangle of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
The array of unsigned r-Lah numbers is an example of the fundamental matrices sketched in A133314. So redefining the offset as n=0, given matrices A and B with A(n, k) = T(n, k)*a(n - k) and B(n, k) = T(n, k)*b(n - k), then A*B = C where C(n, k) = T(n,k)*[a(.) + b(.)]^(n - k), umbrally. An e.g.f. for the row polynomials of A is exp(x*t) exp{-x*t*[a*t/(a*t - 1)]}/(1 - a*t)^4 = exp(x*t) exp[(.)!*Laguerre(., 3, -x*t)* a(.)*t)], umbrally. - Tom Copeland, Sep 19 2008

Examples

			Triangle begins:
=========================================
n\k |     2     3     4     5     6     7
----+------------------------------------
  2 |     1
  3 |     4     1
  4 |    20    10     1
  5 |   120    90    18     1
  6 |   840   840   252    28     1
  7 |  6720  8400  3360   560    40     1
 ...
T(4,3) = 10. The ten partitions of {1,2,3,4} into 3 ordered lists such that the elements 1 and 2 lie in different lists are: {1}{2}{3,4} and {1}{2}{4,3}, {1}{3}{2,4} and {1}{3}{4,2}, {1}{4}{2,3} and {1}{4}{3,2}, {2}{3}{1,4} and {2}{3}{4,1}, {2}{4}{1,3} and {2}{4}{3,1}. The remaining two partitions {3}{4}{1,2} and {3}{4}{2,1} are not allowed because the elements 1 and 2 belong to the same block.
		

Crossrefs

Cf. A001715 (column 2), A007318, A008275, A008277, A061206 (column 3), A062137, A062141 - A062144 ( column 4 to column 7), A062146 (alt. row sums), A062147 (row sums), A105278 (unsigned Lah numbers), A143491, A143494, A143498, A143499.

Programs

  • GAP
    T:=Flat(List([2..10],n->List([2..n],k->(Factorial(n-2)/Factorial(k-2))*Binomial(n+1,k+1)))); # Muniru A Asiru, Nov 27 2018
  • Maple
    T := (n, k) -> ((n-2)!/(k-2)!)*binomial(n+1, k+1):
    for n from 2 to 11 do seq(T(n, k), k = 2..n) od;
  • Mathematica
    T[n_, k_] := (n-2)!/(k-2)!*Binomial[n+1, k+1]; Table[T[n, k], {n,2,10}, {k,2,n}] // Flatten (* Amiram Eldar, Nov 27 2018 *)
  • Maxima
    create_list((n - 2)!/(k - 2)!*binomial(n + 1, k + 1), n, 2, 12, k, 2, n); /* Franck Maminirina Ramaharo, Nov 27 2018 */
    

Formula

T(n, k) = ((n - 2)!/(k - 2)!)*C(n+1, k+1), for n, k >= 2.
Recurrence: T(n, k) = (n + k - 1)*T(n-1, k) + T(n-1, k-1) for n, k >= 2, with the boundary conditions: T(n, k) = 0 if n < 2 or k < 2; T(2, 2) = 1.
E.g.f. for column k: Sum_{n>=k} T(n, k)*t^n/(n - 2)! = 1/(k - 2)!*t^k/(1 - t)^(k+2) for k >= 2.
E.g.f: Sum_{n=2..inf} Sum_{k=2..n} T(n, k)*x^k*t^n/(n - 2)! = (x*t)^2/(1 - t)^4* exp(x*t/(1 - t)) = (x*t)^2*(1 + (4 + x)*t + (20 + 10*x + x^2)*t^2/2! + ... ).
Generalized Lah identity: (x + 3)*(x + 4)*...*(x + n) = Sum_{k = 2..n} T(n, k)*(x - 1)*(x - 2)*...*(x - k + 2).
The polynomials 1/n!*Sum_{k=2..n+2} T(n+2, k)*(-x)^(k - 2) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,3,x). See A062137.
Array = A143491 * A143494 = abs(A008275) * (A007318)^2 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (4, 10, 18, 28, ...) on the main subdiagonal and zeros elsewhere.
After adding 1 to the head of the main diagonal and a zero to each of the subdiagonals, the n-th diagonal may be generated as coefficients of (1/n!) [D^(-1) tDt t^(-3)D t^3]^n exp(x*t), where D is the derivative w.r.t. t and D^(-1) t^j/j! = t^(j + 1)/(j + 1)!. E.g., n = 2 generates 20*x*t^3/3! + 90*x^2*t^4/4! + 252*x^3* t^5/5! + ... . For the general unsigned r-Lah number array, replace the threes by (2*r - 1) in the operator. The e.g.f. of the row polynomials is then exp[D^(-1) tDt t^(-(2*r-1))D t^(2*r - 1)] exp(x*t), with offset n = 0. - Tom Copeland, Sep 21 2008

A144696 Triangle of 2-Eulerian numbers.

Original entry on oeis.org

1, 1, 2, 1, 7, 4, 1, 18, 33, 8, 1, 41, 171, 131, 16, 1, 88, 718, 1208, 473, 32, 1, 183, 2682, 8422, 7197, 1611, 64, 1, 374, 9327, 49780, 78095, 38454, 5281, 128, 1, 757, 30973, 264409, 689155, 621199, 190783, 16867, 256
Offset: 2

Views

Author

Peter Bala, Sep 19 2008

Keywords

Comments

Let [n] denote the ordered set {1,2,...,n}. The symmetric group S_n consists of the injective mappings p:[n] -> [n]. Such a permutation p has an excedance at position i, 1 <= i < n, if p(i) > i. One well-known interpretation of the Eulerian numbers A(n,k) is that they count the permutations in the symmetric group S_n with k excedances. The triangle of Eulerian numbers is A008292 (but with an offset of 1 in the column numbering). We generalize this definition to restricted permutations as follows.
Let r be a nonnegative integer and let Permute(n,n-r) denote the set of injective maps p:[n-r] -> [n], which we think of as permutations of n numbers taken n-r at a time. Clearly, |Permute(n,n-r)| = n!/r!. We say that p has an excedance at position i, 1 <= i <= n-r, if p(i) > i. Then the r-Eulerian number, denoted by A(r;n,k), is defined as the number of permutations in Permute(n,n-r) having k excedances. Thus the current array of 2-Eulerian numbers gives the number of permutations in Permute(n,n-2) with k excedances. See the example section below for some numerical examples.
Clearly A(0;n,k) = A(n,k). The case r = 1 also produces the ordinary Eulerian numbers A(n,k). There is an obvious bijection from Permute(n,n) to Permute(n,n-1) that preserves the number of excedances of a permutation. Consequently, the 1-Eulerian numbers are equal to the 0-Eulerian numbers: A(1;n,k) = A(0;n,k) = A(n,k).
For other cases of r-Eulerian numbers see A144697 (r = 3), A144698 (r = 4) and A144699 (r = 5). There is also a concept of r-Stirling numbers of the first and second kinds - see A143491 and A143494. If we multiply the entries of the current array by a factor of 2 and then reverse the rows we obtain A120434.
An alternative interpretation of the current array due to [Strosser] involves the 2-excedance statistic of a permutation (see also [Foata & Schutzenberger, Chapitre 4, Section 3]). We define a permutation p in Permute(n,n-2) to have a 2-excedance at position i (1 <=i <= n-2) if p(i) >= i + 2.
Given a permutation p in Permute(n,n-2), define ~p to be the permutation in Permute(n,n-2) that takes i to n+1 - p(n-i-1). The map ~ is a bijection of Permute(n,n-2) with the property that if p has (resp. does not have) an excedance in position i then ~p does not have (resp. has) a 2-excedance at position n-i-1. Hence ~ gives a bijection between the set of permutations with k excedances and the set of permutations with (n-k) 2-excedances. Thus reading the rows of this array in reverse order gives a triangle whose entries count the permutations in Permute(n,n-2) with k 2-excedances.
Example: Represent a permutation p:[n-2] -> [n] in Permute(n,n-2) by its image vector (p(1),...,p(n-2)). In Permute(10,8) the permutation p = (1,2,4,7,10,6,5,8) does not have an excedance in the first two positions (i = 1 and 2) or in the final three positions (i = 6, 7 and 8). The permutation ~p = (3,6,5,1,4,7,9,10) has 2-excedances only in the first three positions and the final two positions.
From Peter Bala, Dec 27 2019: (Start)
This is the array A(1,1,3) in the notation of Hwang et al. (p. 25), where the authors remark that the r-Eulerian numbers were first studied by Shanlan Li (Duoji Bilei, Ch. 4), who gave the summation formulas
Sum_{i = 2..n+1} (i-1)*C(i,2) = C(n+3,4) + 2*C(n+2,4)
Sum_{i = 2..n+1} (i-1)^2*C(i,2) = C(n+4,5) + 7*C(n+3,5) + 4*C(n+2,5)
Sum_{i = 2..n+1} (i-1)^3*C(i,2) = C(n+5,6) + 18*C(n+4,6) + 33*C(n+3,6) + 8*C(n+2,6). (End)

Examples

			The triangle begins
===========================================
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
2..|..1
3..|..1.....2
4..|..1.....7.....4
5..|..1....18....33.....8
6..|..1....41...171...131....16
7..|..1....88...718..1208...473....32
8..|..1...183..2682..8422..7197..1611....64
...
Row 4 = [1,7,4]: We represent a permutation p:[n-2] -> [n] in Permute(n,n-2) by its image vector (p(1),...,p(n-2)). Here n = 4. The permutation (1,2) has no excedances; 7 permutations have a single excedance, namely, (1,3), (1,4), (2,1), (3,1), (3,2), (4,1) and (4,2); the remaining 4 permutations, (2,3), (2,4), (3,4) and (4,3) each have two excedances.
		

References

  • J. Riordan. An introduction to combinatorial analysis. New York, J. Wiley, 1958.
  • R. Strosser. Séminaire de théorie combinatoire, I.R.M.A., Université de Strasbourg, 1969-1970.
  • Li, Shanlan (1867). Duoji bilei (Series summation by analogies), 4 scrolls. In Zeguxizhai suanxue (Mathematics from the Studio Devoted to the Imitation of the Ancient Chinese Tradition) (Jinling ed.), Volume 4.
  • Li, Shanlan (2019). Catégories analogues d’accumulations discrètes (Duoji bilei), traduit et commenté par Andrea Bréard. La Bibliothèque Chinoise. Paris: Les Belles Lettres.

Crossrefs

Cf. A000079 (right diagonal), A001710 (row sums).
Cf. A000182 (related to alt. row sums).

Programs

  • Magma
    m:=2; [(&+[(-1)^(k-j)*Binomial(n+1,k-j)*Binomial(j+m,m-1)*(j+1)^(n-m+1): j in [0..k]])/m: k in [0..n-m], n in [m..m+10]]; // G. C. Greubel, Jun 04 2022
    
  • Maple
    with(combinat):
    T:= (n,k) -> 1/2!*add((-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2), j = 0..k):
    for n from 2 to 10 do
    seq(T(n,k),k = 0..n-2)
    end do;
  • Mathematica
    T[n_, k_]:= 1/2!*Sum[(-1)^(k-j)*Binomial[n+1, k-j]*(j+1)^(n-1)*(j+2), {j, 0, k}];
    Table[T[n, k], {n,2,10}, {k,0,n-2}]//Flatten (* Jean-François Alcover, Oct 15 2019 *)
  • SageMath
    m=2 # A144696
    def T(n,k): return (1/m)*sum( (-1)^(k-j)*binomial(n+1,k-j)*binomial(j+m,m-1)*(j+1)^(n-m+1) for j in (0..k) )
    flatten([[T(n,k) for k in (0..n-m)] for n in (m..m+10)]) # G. C. Greubel, Jun 04 2022

Formula

T(n,k) = (1/2!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(j+1)^(n-1)*(j+2);
T(n,n-k) = (1/2!)*Sum_{j = 2..k} (-1)^(k-j)*binomial(n+1,k-j)*j^(n-1)*(j-1).
Recurrence relations:
T(n,k) = (k+1)*T(n-1,k) + (n-k)*T(n-1,k-1) with boundary conditions T(n,0) = 1 for n >= 2, T(2,k) = 0 for k >= 1. Special cases: T(n,n-2) = 2^(n-2); T(n,n-3) = A066810(n-1).
E.g.f. (with suitable offsets): (1/2)*[(1 - x)/(1 - x*exp(t - t*x))]^2 = 1/2 + x*t + (x + 2*x^2)*t^2/2! + (x + 7*x^2 + 4*x^3)*t^3/3! + ... .
The row generating polynomials R_n(x) satisfy the recurrence R_(n+1)(x) = (n*x+1)*R_n(x) + x*(1-x)*d/dx(R_n(x)) with R_2(x) = 1. It follows that the polynomials R_n(x) for n >= 3 have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
The (n+1)-th row generating polynomial = (1/2!)*Sum_{k = 1..n} (k+1)!*Stirling2(n,k) *x^(k-1)*(1-x)^(n-k).
For n >= 2,
(1/2)*(x*d/dx)^(n-1) (1/(1-x)^2) = x/(1-x)^(n+1) * Sum_{k = 0..n-2} T(n,k)*x^k,
(1/2)*(x*d/dx)^(n-1) (x^2/(1-x)^2) = 1/(1-x)^(n+1) * Sum_{k = 2..n} T(n,n-k)*x^k,
1/(1-x)^(n+1)*Sum_{k = 0..n-2} T(n,k)*x^k = (1/2!) * Sum_{m = 0..inf} (m+1)^(n-1)*(m+2)*x^m,
1/(1-x)^(n+1)*Sum_{k = 2..n} T(n,n-k)*x^k = (1/2!) * Sum_{m = 2..inf} m^(n-1)*(m-1)*x^m.
Worpitzky-type identities:
Sum_{k = 0..n-2} T(n,k)*binomial(x+k,n) = (1/2!)*x^(n-1)*(x - 1);
Sum_{k = 2..n} T(n,n-k)*binomial(x+k,n) = (1/2!)*(x + 1)^(n-1)*(x + 2).
Relation with Stirling numbers (Frobenius-type identities):
T(n+1,k-1) = (1/2!) * Sum_{j = 0..k} (-1)^(k-j)*(j+1)!* binomial(n-j,k-j)*Stirling2(n,j) for n,k >= 1;
T(n+1,k-1) = (1/2!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+1)!* binomial(n-j,k)*S(2;n+2,j+2) for n,k >= 1 and
T(n+2,k) = (1/2!) * Sum_{j = 0..n-k} (-1)^(n-k-j)*(j+2)!* binomial(n-j,k)*S(2;n+2,j+2) for n,k >= 0, where S(2;n,k) denotes the 2-Stirling numbers A143494(n,k).
The row polynomials of this array are related to the Eulerian polynomials. For example, 1/2*x*d/dx [x*(x + 4*x^2 + x^3)/(1-x)^4] = x^2*(1 + 7*x + 4*x^2)/(1-x)^5 and 1/2*x*d/dx [x*(x + 11*x^2 + 11*x^3 + x^4)/(1-x)^5] = x^2*(1 + 18*x + 33*x^2 + 8*x^3)/(1-x)^6.
Row sums A001710. Alternating row sums [1, -1, -2, 8, 16, -136, -272, 3968, 7936, ... ] are alternately (signed) tangent numbers and half tangent numbers - see A000182.
Sum_{k = 0..n-2} 2^k*T(n,k) = A069321(n-1). Sum_{k = 0..n-2} 2^(n-k)*T(n,k) = 4*A083410(n-1).
For n >=2, the shifted row polynomial t*R(n,t) = (1/2)*D^(n-1)(f(x,t)) evaluated at x = 0, where D is the operator (1-t)*(1+x)*d/dx and f(x,t) = (1+x*t/(t-1))^(-2). - Peter Bala, Apr 22 2012

A143492 Unsigned 3-Stirling numbers of the first kind.

Original entry on oeis.org

1, 3, 1, 12, 7, 1, 60, 47, 12, 1, 360, 342, 119, 18, 1, 2520, 2754, 1175, 245, 25, 1, 20160, 24552, 12154, 3135, 445, 33, 1, 181440, 241128, 133938, 40369, 7140, 742, 42, 1, 1814400, 2592720, 1580508, 537628, 111769, 14560, 1162, 52, 1, 19958400
Offset: 3

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

See A049458 for a signed version of this array. The unsigned 3-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1, 2 and 3 belong to distinct cycles. This is the case r = 3 of the unsigned r-Stirling numbers of the first kind. For other cases see abs(A008275) (r = 1), A143491 (r = 2) and A143493 (r = 4). See A143495 for the corresponding 3-Stirling numbers of the second kind. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related 3-Lah numbers see A143498.
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^3, -log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-3*t), 1-exp(-t))). See the e.g.f given below. Compare also with the e.g.f. for the signed version A049458. - Wolfdieter Lang, Oct 10 2011
With offset n=0 and k=0 : triangle T(n,k), read by rows, given by (3,1,4,2,5,3,6,4,7,5,8,6,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 31 2011

Examples

			Triangle begins
n\k|.....3.....4.....5.....6.....7.....8
========================================
3..|.....1
4..|.....3.....1
5..|....12.....7.....1
6..|....60....47....12.....1
7..|...360...342...119....18.....1
8..|..2520..2754..1175...245....25.....1
...
T(5,4) = 7. The permutations of {1,2,3,4,5} with 4 cycles such that 1, 2 and 3 belong to different cycles are: (14)(2)(3)(5), (15)(2)(3)(4), (24)(1)(3)(5), (25)(1)(3)(4), (34)(1)(2)(5), (35)(1)(2)(4) and (45)(1)(2)(3).
		

Crossrefs

Cf. A001710 - A001714 (column 3 - column 7), A001715 (row sums), A008275, A049458 (signed version), A143491, A143493, A143495, A143498.

Programs

  • Maple
    with combinat: T := (n, k) -> (n-3)! * add(binomial(n-j-1,2)*abs(stirling1(j,k-3))/j!,j = k-3..n-3): for n from 3 to 12 do seq(T(n, k), k = 3..n) end do;

Formula

T(n,k) = (n-3)! * Sum_{j = k-3 .. n-3} C(n-j-1,2)*|Stirling1(j,k-3)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 3, with boundary conditions: T(n,2) = T(2,n) = 0, for all n; T(3,3) = 1; T(3,k) = 0 for k > 3.
Special cases:
T(n,3) = (n-1)!/2! for n >= 3.
T(n,4) = (n-1)!/2!*(1/3 + ... + 1/(n-1)) for n >= 3.
T(n,k) = Sum_{3 <= i_1 < ... < i_(n-k) < n} (i_1*i_2* ...*i_(n-k)). For example, T(6,4) = Sum_{3 <= i < j < 6} (i*j) = 3*4 + 3*5 + 4*5 = 47.
Row g.f.: Sum_{k = 3..n} T(n,k)*x^k = x^3*(x+3)*(x+4)* ... *(x+n-1).
E.g.f. for column (k+3): Sum_{n = k..inf} T(n+3,k+3)*x^n/n! = 1/k!*1/(1-x)^3 * (log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+3) = Sum_{n = 0..inf} Sum_{k = 0..n} T(n+3,k+3)*x^k*t^n/n! = 1 + (3+x)*t/1! + (12+7*x+x^2)*t^2/2! + ....
This array is the matrix product St1 * P^2, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/3! ( A001715 ). The alternating row sums are (n-2)!.
If we define f(n,i,a) = sum(binomial(n,k)*Stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n,i) = |f(n,i,3)|, for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008

A143493 Unsigned 4-Stirling numbers of the first kind.

Original entry on oeis.org

1, 4, 1, 20, 9, 1, 120, 74, 15, 1, 840, 638, 179, 22, 1, 6720, 5944, 2070, 355, 30, 1, 60480, 60216, 24574, 5265, 625, 39, 1, 604800, 662640, 305956, 77224, 11515, 1015, 49, 1, 6652800, 7893840, 4028156, 1155420, 203889, 22680, 1554, 60, 1, 79833600
Offset: 4

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

See A049459 for a signed version of the array. The unsigned 4-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1, 2, 3 and 4 belong to distinct cycles. This is the case r = 4 of the unsigned r-Stirling numbers of the first kind. For other cases see abs(A008275) (r = 1), A143491 (r = 2) and A143492 (r = 3). See A143496 for the corresponding triangle of 4-Stirling numbers of the second kind.
The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related 4-Lah numbers see A143499.
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^4,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-4*t),1-exp(-t))). See the e.g.f given below. Compare also with the e.g.f. for the signed version A049459. - Wolfdieter Lang, Oct 10 2011
With offset n=0 and k=0: triangle T(n,k), read by rows, given by (4,1,5,2,6,3,7,4,8,5,9,6,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 31 2011

Examples

			Triangle begins
  n\k|     4     5     6     7     8     9
  ========================================
  4  |     1
  5  |     4     1
  6  |    20     9     1
  7  |   120    74    15     1
  8  |   840   638   179    22     1
  9  |  6720  5944  2070   355    30     1
  ...
T(6,5) = 9. The 9 permutations of {1,2,3,4,5,6} with 5 cycles such that 1, 2, 3 and 4 belong to different cycles are: (1,5)(2)(3)(4)(6), (1,6)(2)(3)(4)(5), (2,5)(1)(3)(4)(6), (2,6)(1)(3)(4)(5), (3,5)(1)(2)(4)(6), (3,6)(1)(2)(4)(5), (4,5)(1)(2)(3)(6), (4,6)(1)(2)(3)(5) and (5,6)(1)(2)(3)(4).
		

Crossrefs

Cf. A001715 - A001719 (column 4 - column 8), A001720 (row sums), A008275, A049459 (signed version), A143491, A143492, A143496, A143499.

Programs

  • Maple
    with combinat: T := (n, k) -> (n-4)! * add(binomial(n-j-1,3)*abs(stirling1(j,k-4))/j!,j = k-4..n-4): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;

Formula

T(n,k) = (n-4)! * Sum_{j = k-4 .. n-4} C(n-j-1,3)*|stirling1(j,k-4)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 4, with boundary conditions: T(n,3) = T(3,n) = 0 for all n; T(4,4) = 1; T(4,k) = 0 for k > 4.
Special cases:
T(n,4) = (n-1)!/3!.
T(n,5) = (n-1)!/3!*(1/4 + ... + 1/(n-1)).
T(n,k) = sum {4 <= i_1 < ...< i_(n-k) < n} (i_1*i_2* ...*i_(n-k)). For example, T(7,5) = Sum_{4 <= i < j < 7} (i*j) = 4*5 + 4*6 + 5*6 = 74.
Row g.f.: Sum_{k = 4..n} T(n,k)*x^k = x^4*(x+4)*(x+5)* ... *(x+n-1).
E.g.f. for column (k+4): Sum_{n = k..inf} T(n+4,k+4)*x^n/n! = 1/k!*1/(1-x)^4 * (log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+4) = Sum_{n = 0..inf} Sum_{k = 0..n} T(n+4,k+4)*x^k*t^n/n! = 1 + (4+x)*t/1! + (20+9*x+x^2)*t^2/2! + .... This array is the matrix product St1 * P^3, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/4! ( A001720 ). The alternating row sums are (n-2)!/2!.
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n+4,i) = |f(n,i,3)|, for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008

A074060 Graded dimension of the cohomology ring of the moduli space of n-pointed stable curves of genus 0 satisfying the associativity equations of physics (also known as the WDVV equations).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 16, 16, 1, 1, 42, 127, 42, 1, 1, 99, 715, 715, 99, 1, 1, 219, 3292, 7723, 3292, 219, 1, 1, 466, 13333, 63173, 63173, 13333, 466, 1, 1, 968, 49556, 429594, 861235, 429594, 49556, 968, 1, 1, 1981, 173570, 2567940, 9300303, 9300303, 2567940, 173570, 1981, 1
Offset: 3

Views

Author

Margaret A. Readdy, Aug 16 2002

Keywords

Comments

Combinatorial interpretations of Lagrange inversion (A134685) and the 2-Stirling numbers of the first kind (A049444 and A143491) provide a combinatorial construction for A074060 (see first Copeland link). For relations of A074060 to other arrays see second Copeland link page 19. - Tom Copeland, Sep 28 2008
These Poincare polynomials for the compactified moduli space of rational curves are presented on p. 5 of Lando and Zvonkin as well as those for the non-compactified Poincare polynomials of A049444 in factorial form. - Tom Copeland, Jun 13 2021

Examples

			Viewed as a triangular array, the values are
  1;
  1,   1;
  1,   5,   1;
  1,  16,  16,   1;
  1,  42, 127,  42,   1; ...
		

Crossrefs

Cf. A074059. 2nd diagonal is A002662.

Programs

  • Maple
    DA:=((1+t)*A(u,t)+u)/(1-t*A(u,t)): F:=0: for k from 1 to 10 do F:=map(simplify,int(series(subs(A(u,t)=F,DA),u,k),u)); od: # Eric Rains, Apr 02 2005
  • Mathematica
    DA = ((1+t) A[u, t] + u)/(1 - t A[u, t]); F = 0;
    Do[F = Integrate[Series[DA /. A[u, t] -> F, {u, 0, k}], u], {k, 1, 10}];
    (cc = CoefficientList[#, t]; cc Denominator[cc[[1]]])& /@ Drop[ CoefficientList[F, u], 2] // Flatten (* Jean-François Alcover, Oct 15 2019, after Eric Rains *)

Formula

Define offset to be 0 and P(n,t) = (-1)^n Sum_{j=0..n-2} a(n-2,j)*t^j with P(1,t) = -1 and P(0,t) = 1, then H(x,t) = -1 + exp(P(.,t)*x) is the compositional inverse in x about 0 of G(x,t) in A049444. H(x,0) = exp(-x) - 1, H(x,1) = -1 + exp( 2 + W( -exp(-2) * (2-x) ) ) and H(x,2) = 1 - (1+2*x)^(1/2), where W is a branch of the Lambert function such that W(-2*exp(-2)) = -2. - Tom Copeland, Feb 17 2008
Let offset=0 and g(x,t) = (1-t)/((1+x)^(t-1)-t), then the n-th row polynomial of the table is given by [(g(x,t)*D_x)^(n+1)]x with the derivative evaluated at x=0. - Tom Copeland, Jun 01 2008
With the notation in Copeland's comments, dH(x,t)/dx = -g(H(x,t),t). - Tom Copeland, Sep 01 2011
The term linear in x of [x*g(d/dx,t)]^n 1 gives the n-th row polynomial with offset 1. (See A134685.) - Tom Copeland, Oct 21 2011

Extensions

More terms from Eric Rains, Apr 02 2005
Showing 1-10 of 14 results. Next