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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A119467 A masked Pascal triangle.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 0, 6, 0, 1, 0, 5, 0, 10, 0, 1, 1, 0, 15, 0, 15, 0, 1, 0, 7, 0, 35, 0, 21, 0, 1, 1, 0, 28, 0, 70, 0, 28, 0, 1, 0, 9, 0, 84, 0, 126, 0, 36, 0, 1, 1, 0, 45, 0, 210, 0, 210, 0, 45, 0, 1, 0, 11, 0, 165, 0, 462, 0, 330, 0, 55, 0, 1, 1, 0, 66, 0, 495, 0, 924
Offset: 0

Views

Author

Paul Barry, May 21 2006

Keywords

Comments

Row sums are A011782. Diagonal sums are F(n+1)*(1+(-1)^n)/2 (aerated version of A001519). Product by Pascal's triangle A007318 is A119468. Schur product of (1/(1-x),x/(1-x)) and (1/(1-x^2),x).
Exponential Riordan array (cosh(x),x). Inverse is (sech(x),x) or A119879. - Paul Barry, May 26 2006
Rows give coefficients of polynomials p_n(x) = Sum_{k=0..n} (k+1 mod 2)*binomial(n,k)*x^(n-k) having e.g.f. exp(x*t)*cosh(t)= 1*(t^0/0!) + x*(t^1/1!) + (1+x^2)*(t^2/2!) + ... - Peter Luschny, Jul 14 2009
Inverse of the coefficient matrix of the Swiss-Knife polynomials in ascending order of x^i (reversed and aerated rows of A153641). - Peter Luschny, Jul 16 2012
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*... is equal to A136630 but with the first row and column omitted. - Peter Bala, Jul 28 2014
The row polynomials SKv(n,x) = [(x+1)^n + (x-1)^n]/2 , with e.g.f. cosh(t)*exp(xt), are the umbral compositional inverses of the row polynomials of A119879 (basically the Swiss Knife polynomials SK(n,x) of A153641); i.e., umbrally SKv(n,SK(.,x)) = x^n = SK(n,SKv(.,x)). Therefore, this entry's matrix and A119879 are an inverse pair. Both sequences of polynomials are Appell sequences, i.e., d/dx P(n,x) = n * P(n-1,x) and (P(.,x)+y)^n = P(n,x+y). In particular, (SKv(.,0)+x)^n = SKv(n,x), reflecting that the first column has the e.g.f. cosh(t). The raising operator is R = x + tanh(d/dx); i.e., R SKv(n,x) = SKv(n+1,x). The coefficients of this operator are basically the signed and aerated zag numbers A000182, which can be expressed as normalized Bernoulli numbers. The triangle is formed by multiplying the n-th diagonal of the lower triangular Pascal matrix by the Taylor series coefficient a(n) of cosh(x). More relations for this type of triangle and its inverse are given by the formalism of A133314. - Tom Copeland, Sep 05 2015
The signed version of this matrix has the e.g.f. cos(t) e^{xt}, generating Appell polynomials that have only real, simple zeros and whose extrema are maxima above the x-axis and minima below and situated above and below the zeros of the next lower degree polynomial. The bivariate versions appear on p. 27 of Dimitrov and Rusev in conditions for entire functions that are cosine transforms of a class of functions to have only real zeros. - Tom Copeland, May 21 2020
The n-th row of the triangle is obtained by multiplying by 2^(n-1) the elements of the first row of the limit as k approaches infinity of the stochastic matrix P^(2k-1) where P is the stochastic matrix associated with the Ehrenfest model with n balls. The elements of a stochastic matrix P give the probabilities of arriving in a state j given the previous state i. In particular the sum of every row of the matrix must be 1, and so the sum of the terms of the n-th row of this triangle is 2^(n-1). Furthermore, by the properties of Markov chains, we can interpret P^(2k-1) as the (2k-1)-step transition matrix of the Ehrenfest model and its limit exists and it is again a stochastic matrix. The rows of the triangle divided by 2^(n-1) are the even rows (second, fourth, ...) and the odd rows (first, third, ...) of the limit matrix P^(2k-1). - Luca Onnis, Oct 29 2023

Examples

			Triangle begins
  1,
  0, 1,
  1, 0,  1,
  0, 3,  0,  1,
  1, 0,  6,  0,   1,
  0, 5,  0, 10,   0,   1,
  1, 0, 15,  0,  15,   0,   1,
  0, 7,  0, 35,   0,  21,   0,  1,
  1, 0, 28,  0,  70,   0,  28,  0,  1,
  0, 9,  0, 84,   0, 126,   0, 36,  0, 1,
  1, 0, 45,  0, 210,   0, 210,  0, 45, 0, 1
p[0](x) = 1
p[1](x) = x
p[2](x) = 1 + x^2
p[3](x) = 3*x + x^3
p[4](x) = 1 + 6*x^2 + x^4
p[5](x) = 5*x + 10*x^3 + x^5
Connection with A136630: With the arrays M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1        \/1        \      /1         \
|0 1      ||0 1      ||0 1      |      |0 1       |
|1 0 1    ||0 0 1    ||0 0 1    |... = |1 0  1    |
|0 3 0 1  ||0 1 0 1  ||0 0 0 1  |      |0 4  0 1  |
|1 0 6 0 1||0 0 3 0 1||0 0 1 0 1|      |1 0 10 0 1|
|...      ||...      ||...      |      |...       |
- _Peter Bala_, Jul 28 2014
		

References

  • Paul and Tatjana Ehrenfest, Über zwei bekannte Einwände gegen das Boltzmannsche H-Theorem, Physikalische Zeitschrift, vol. 8 (1907), pp. 311-314.

Crossrefs

From Peter Luschny, Jul 14 2009: (Start)
p[n](k), n=0,1,...
k= 0: 1, 0, 1, 0, 1, 0, ... A128174
k= 1: 1, 1, 2, 4, 8, 16, ... A011782
k= 2: 1, 2, 5, 14, 41, 122, ... A007051
k= 3: 1, 3, 10, 36, 136, ... A007582
k= 4: 1, 4, 17, 76, 353, ... A081186
k= 5: 1, 5, 26, 140, 776, ... A081187
k= 6: 1, 6, 37, 234, 1513, ... A081188
k= 7: 1, 7, 50, 364, 2696, ... A081189
k= 8: 1, 8, 65, 536, 4481, ... A081190
k= 9: 1, 9, 82, 756, 7048, ... A060531
k=10: 1, 10, 101, 1030, ... A081192
p[n](k), k=0,1,...
p[0]: 1,1,1,1,1,1, ....... A000012
p[1]: 0,1,2,3,4,5, ....... A001477
p[2]: 1,2,5,10,17,26, .... A002522
p[3]: 0,4,14,36,76,140, .. A079908 (End)

Programs

  • Haskell
    a119467 n k = a119467_tabl !! n !! k
    a119467_row n = a119467_tabl !! n
    a119467_tabl = map (map (flip div 2)) $
                   zipWith (zipWith (+)) a007318_tabl a130595_tabl
    -- Reinhard Zumkeller, Mar 23 2014
    
  • Magma
    /* As triangle */ [[Binomial(n, k)*(1 + (-1)^(n - k))/2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 26 2015
  • Maple
    # Polynomials: p_n(x)
    p := proc(n,x) local k, pow; pow := (n,k) -> `if`(n=0 and k=0,1,n^k);
    add((k+1 mod 2)*binomial(n,k)*pow(x,n-k),k=0..n) end;
    # Coefficients: a(n)
    seq(print(seq(coeff(i!*coeff(series(exp(x*t)*cosh(t),t,16),t,i),x,n),n=0..i)),i=0..8); # Peter Luschny, Jul 14 2009
  • Mathematica
    Table[Binomial[n, k] (1 + (-1)^(n - k))/2, {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 06 2015 *)
    n = 15; "n-th row"
    mat = Table[Table[0, {j, 1, n + 1}], {i, 1, n + 1}];
    mat[[1, 2]] = 1;
    mat[[n + 1, n]] = 1;
    For[i = 2, i <= n, i++, mat[[i, i - 1]] = (i - 1)/n ];
    For[i = 2, i <= n, i++, mat[[i, i + 1]] = (n - i + 1)/n];
    mat // MatrixForm;
    P2 = Dot[mat, mat];
    R1 = Simplify[
      Eigenvectors[Transpose[P2]][[1]]/
       Total[Eigenvectors[Transpose[P2]][[1]]]]
    R2 = Table[Dot[R1, Transpose[mat][[k]]], {k, 1, n + 1}]
    odd = R2*2^(n - 1) (* _Luca Onnis *)
  • Sage
    @CachedFunction
    def A119467_poly(n):
        R = PolynomialRing(ZZ, 'x')
        x = R.gen()
        return R.one() if n==0 else R.sum(binomial(n,k)*x^(n-k) for k in range(0,n+1,2))
    def A119467_row(n):
        return list(A119467_poly(n))
    for n in (0..10) : print(A119467_row(n)) # Peter Luschny, Jul 16 2012
    

Formula

G.f.: (1-x*y)/(1-2*x*y-x^2+x^2*y^2);
T(n,k) = C(n,k)*(1+(-1)^(n-k))/2;
Column k has g.f. (1/(1-x^2))*(x/(1-x^2))^k*Sum_{j=0..k+1} binomial(k+1,j)*sin((j+1)*Pi/2)^2*x^j.
Column k has e.g.f. cosh(x)*x^k/k!. - Paul Barry, May 26 2006
Let Pascal's triangle, A007318 = P; then this triangle = (1/2) * (P + 1/P). Also A131047 = (1/2) * (P - 1/P). - Gary W. Adamson, Jun 12 2007
Equals A007318 - A131047 since the zeros of the triangle are masks for the terms of A131047. Thus A119467 + A131047 = Pascal's triangle. - Gary W. Adamson, Jun 12 2007
T(n,k) = (A007318(n,k) + A130595(n,k))/2, 0<=k<=n. - Reinhard Zumkeller, Mar 23 2014

Extensions

Edited by N. J. A. Sloane, Jul 14 2009

A055137 Regard triangle of rencontres numbers (see A008290) as infinite matrix, compute inverse, read by rows.

Original entry on oeis.org

1, 0, 1, -1, 0, 1, -2, -3, 0, 1, -3, -8, -6, 0, 1, -4, -15, -20, -10, 0, 1, -5, -24, -45, -40, -15, 0, 1, -6, -35, -84, -105, -70, -21, 0, 1, -7, -48, -140, -224, -210, -112, -28, 0, 1, -8, -63, -216, -420, -504, -378, -168, -36, 0, 1, -9, -80, -315, -720
Offset: 0

Views

Author

Christian G. Bower, Apr 25 2000

Keywords

Comments

The n-th row consists of coefficients of the characteristic polynomial of the adjacency matrix of the complete n-graph.
Triangle of coefficients of det(M(n)) where M(n) is the n X n matrix m(i,j)=x if i=j, m(i,j)=i/j otherwise. - Benoit Cloitre, Feb 01 2003
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for rB(0,1). The e.g.f. for the row polynomials is exp(x*t) * exp(x) *(1-x). T(n,k) = Binomial(n,k)* s(n-k) where s = (1,0,-1,-2,-3,...) with an e.g.f. of exp(x)*(1-x) which is the reciprocal of the e.g.f. of A000166. - Tom Copeland, Sep 10 2008
Row sums are: {1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...}. - Roger L. Bagula, Feb 20 2009
T is related to an operational calculus connecting an infinitesimal generator for fractional integro-derivatives with the values of the Riemann zeta function at positive integers (see MathOverflow links). - Tom Copeland, Nov 02 2012
The submatrix below the null subdiagonal is signed and row reversed A127717. The submatrix below the diagonal is A074909(n,k)s(n-k) where s(n)= -n, i.e., multiply the n-th diagonal by -n. A074909 and its reverse A135278 have several combinatorial interpretations. - Tom Copeland, Nov 04 2012
T(n,k) is the difference between the number of even (A145224) and odd (A145225) permutations (of an n-set) with exactly k fixed points. - Julian Hatfield Iacoponi, Aug 08 2024

Examples

			1; 0,1; -1,0,1; -2,-3,0,1; -3,-8,-6,0,1; ...
(Bagula's matrix has a different sign convention from the list.)
From _Roger L. Bagula_, Feb 20 2009: (Start)
  { 1},
  { 0,   1},
  {-1,   0,    1},
  { 2,  -3,    0,    1},
  {-3,   8,   -6,    0,     1},
  { 4, -15,   20,  -10,     0,    1},
  {-5,  24,  -45,   40,   -15,    0,    1},
  { 6, -35,   84, -105,    70,  -21,    0,   1},
  {-7,  48, -140,  224,  -210,  112,  -28,   0,   1},
  { 8, -63,  216, -420,   504, -378,  168, -36,   0, 1},
  {-9,  80, -315,  720, -1050, 1008, -630, 240, -45, 0, 1}
(End)
R(3,x) = (-1)^3*Sum_{permutations p in S_3} sign(p)*(-x)^(fix(p)).
    p   | fix(p) | sign(p) | (-1)^3*sign(p)*(-x)^fix(p)
========+========+=========+===========================
  (123) |    3   |    +1   |      x^3
  (132) |    1   |    -1   |       -x
  (213) |    1   |    -1   |       -x
  (231) |    0   |    +1   |       -1
  (312) |    0   |    +1   |       -1
  (321) |    1   |    -1   |       -x
========+========+=========+===========================
                           | R(3,x) = x^3 - 3*x - 2
- _Peter Bala_, Aug 08 2011
		

References

  • Norman Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University Press, 1993. p. 17.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p.184 problem 3.

Crossrefs

Cf. A005563, A005564 (absolute values of columns 1, 2).
Cf. A000312.

Programs

  • Mathematica
    M[n_] := Table[If[i == j, x, 1], {i, 1, n}, {j, 1, n}]; a = Join[{{1}}, Flatten[Table[CoefficientList[Det[M[n]], x], {n, 1, 10}]]] (* Roger L. Bagula, Feb 20 2009 *)
    t[n_, k_] := (k-n+1)*Binomial[n, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 29 2013, after Pari *)
  • PARI
    T(n,k)=(1-n+k)*if(k<0 || k>n,0,n!/k!/(n-k)!)

Formula

G.f.: (x-n+1)*(x+1)^(n-1) = Sum_(k=0..n) T(n,k) x^k.
T(n, k) = (1-n+k)*binomial(n, k).
k-th column has o.g.f. x^k(1-(k+2)x)/(1-x)^(k+2). k-th row gives coefficients of (x-k)(x+1)^k. - Paul Barry, Jan 25 2004
T(n,k) = Coefficientslist[Det[Table[If[i == j, x, 1], {i, 1, n}, {k, 1, n}],x]. - Roger L. Bagula, Feb 20 2009
From Peter Bala, Aug 08 2011: (Start)
Given a permutation p belonging to the symmetric group S_n, let fix(p) be the number of fixed points of p and sign(p) its parity. The row polynomials R(n,x) have a combinatorial interpretation as R(n,x) = (-1)^n*Sum_{permutations p in S_n} sign(p)*(-x)^(fix(p)). An example is given below.
Note: The polynomials P(n,x) = Sum_{permutations p in S_n} x^(fix(p)) are the row polynomials of the rencontres numbers A008290. The integral results Integral_{x = 0..n} R(n,x) dx = n/(n+1) = Integral_{x = 0..-1} R(n,x) dx lead to the identities Sum_{p in S_n} sign(p)*(-n)^(1 + fix(p))/(1 + fix(p)) = (-1)^(n+1)*n/(n+1) = Sum_{p in S_n} sign(p)/(1 + fix(p)). The latter equality was Problem B6 in the 66th William Lowell Putnam Mathematical Competition 2005. (End)
From Tom Copeland, Jul 26 2017: (Start)
The e.g.f. in Copeland's 2008 comment implies this entry is an Appell sequence of polynomials P(n,x) with lowering and raising operators L = d/dx and R = x + d/dL log[exp(L)(1-L)] = x+1 - 1/(1-L) = x - L - L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L) exp(L) x^n = (1-L) (x+1)^n = (x+1)^n - n (x+1)^(n-1) = (x+1-n)(x+1)^(n-1) = (x+s.)^n umbrally, where (s.)^n = s_n = P(n,0).
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
The exponential infinitesimal generator (infinigen) of this entry is the negated infinigen of A008290, the matrix (M) noted by Bala, related to A238363. Then e^M = [the lower triangular A008290], and e^(-M) = [the lower triangular A055137]. For more on the infinigens, see A238385. (End)
From the row g.f.s corresponding to Bagula's matrix example below, the n-th row polynomial has a zero of multiplicity n-1 at x = 1 and a zero at x = -n+1. Since this is an Appell sequence dP_n(x)/dx = n P_{n-1}(x), the critical points of P_n(x) have the same abscissas as the zeros of P_{n-1}(x); therefore, x = 1 is an inflection point for the polynomials of degree > 2 with P_n(1) = 0, and the one local extremum of P_n has the abscissa x = -n + 2 with the value (-n+1)^{n-1}, signed values of A000312. - Tom Copeland, Nov 15 2019
From Julian Hatfield Iacoponi, Aug 08 2024: (Start)
T(n,k) = A145224(n,k) - A145225(n,k).
T(n,k) = binomial(n,k)*(A003221(n-k)-A000387(n-k)). (End)

Extensions

Additional comments from Michael Somos, Jul 04 2002

A132013 T(n,j) for an iterated mixed order Laguerre transform. Coefficients of the normalized generalized Laguerre polynomials (-1)^n*n!*L(n,1-n,x).

Original entry on oeis.org

1, -1, 1, 0, -2, 1, 0, 0, -3, 1, 0, 0, 0, -4, 1, 0, 0, 0, 0, -5, 1, 0, 0, 0, 0, 0, -6, 1, 0, 0, 0, 0, 0, 0, -7, 1, 0, 0, 0, 0, 0, 0, 0, -8, 1, 0, 0, 0, 0, 0, 0, 0, 0, -9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -10, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -11, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -12, 1
Offset: 0

Views

Author

Tom Copeland, Oct 30 2007

Keywords

Comments

The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x).
1) b(0) = a(0), b(n) = a(n) - n*a(n-1) for n > 0
2) b(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}, umbrally
3) b(n) = n! Sum_{j=0..min(1,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),1-n)
5) B(x) = (1-xDx) A(x) = [1-x*Lag(1,-xD:,0)] A(x)
6) EB(x) = (1-x) EA(x),
where D is the derivative w.r.t. x and Lag(n,x,m) is the associated Laguerre polynomial of order m. These formulas are easily generalized for repeated applications of the operator.
c = (1,-1,0,0,0,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314. The reciprocal sequence is d = (0!,1!,2!,3!,4!,...).
Consequently, the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A094587, which has the property that the terms at and below TI(m,m) are the associated sequence under the list partition transform for the inverse for T^(m+1) for m=0,1,2,3,... .
Row sums of T = [formula 3 with all a(n) = 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = A024000 = (1,0,-1,-2,-3,...), with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x) * exp(x) = exp(x)*exp(c(.)*x) = exp[(1+c(.))*x].
Alternating row sums of T = [formula 3 with all a(n) = (-1)^n] = [finite differences of c] = [coefficients of B(x) with A(x) = 1/(1+x)] = (1,-2,3,-4,...), with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x) * exp(-x) = exp(-x)*exp(c(.)*x) = exp[-(1-c(.))*x].
An e.g.f. for the o.g.f.s for repeated applications of T on A(x) is given by
exp[t*(1-xDx)] A(x) = e^t * Sum_{n=0,1,...} (-t*x)^n * Lag(n,-:xD:,0) A(x)
= e^t * exp{[-t*u/(1+t*u)]*:xD:} / (1+t*u) A(x) (eval. at u=x)
= e^t * A[x/(1+t*x)]/(1+t*x) .
See A132014 for more notes on repeated applications.

Examples

			First few rows of the triangle are
   1;
  -1,  1;
   0, -2,  1;
   0,  0, -3,  1;
   0,  0,  0, -4,  1;
   0,  0,  0,  0, -5,  1;
   0,  0,  0,  0,  0, -6,  1;
   0,  0,  0,  0,  0,  0, -7,  1;
		

Crossrefs

Programs

  • Maple
    c := n -> `if`(n=0,1,`if`(n=1,-1,0)):
    T := (n,k) -> binomial(n,k)*c(n-k); # Peter Luschny, Nov 14 2016
  • Mathematica
    Table[PadLeft[{-n, 1}, n+1], {n, 0, 13}] // Flatten (* Jean-François Alcover, Apr 29 2014 *)
  • PARI
    row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 1-n)); \\ Michel Marcus, Jul 26 2021

Formula

T(n,k) = binomial(n,k)*c(n-k), with the sequence c defined in the comments.
E.g.f.: exp(x*y)(1-x), which implies the row polynomials form an Appell sequence. More relations can be found in A132382. - Tom Copeland, Dec 03 2013
From Tom Copeland, Apr 21 2014: (Start)
Change notation letting L(n,m,x) = Lag(n,x,m).
Row polynomials: (-1)^n*n!*L(n,1-n,x) = -x^(n-1)*L(1,n-1,x) =
(-1)^n*(1/(1-n)!)*K(-n,1-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
For the row polynomials, the lowering operator = d/dx and the raising operator = x - 1/(1-D).
T = I - A132440 = 2*I - exp[A238385-I] = signed exp[A238385-I], where I = identity matrix.
Operationally, (-1)^n*n!*L(n,1-n,-:xD:) = -x^(n-1)*:Dx:^n*x^(1-n) = (-1)^n*x^(-1)*:xD:^n*x = (-1)^n*n!*binomial(xD+1,n) = (-1)^n*n!*binomial(1,n)*K(-n,1-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
The unsigned row polynomials have e.g.f. (1+t)e^(xt) = exp(t*p.(x)), umbrally, and p_n(x) = (1+D) x^n. With q_n(x) the row polynomials of A094587, p_n(x) = u_n(q.(v.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form unsigned [T] = [p] = [u]*[q]*[v]. Conversely, q_n(x) = v_n (p.(u.(x))). - Tom Copeland, Nov 10 2016
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^k*binomial(n,k)*Lag(k,1,x). - Peter Bala, Jul 25 2021

Extensions

Title modified by Tom Copeland, Apr 21 2014

A248727 A046802(x,y) --> A046802(x,y+1), transform of e.g.f. for the graded number of positroids of the totally nonnegative Grassmannians G+(k,n); enumerates faces of the stellahedra.

Original entry on oeis.org

1, 2, 1, 5, 5, 1, 16, 24, 10, 1, 65, 130, 84, 19, 1, 326, 815, 720, 265, 36, 1, 1957, 5871, 6605, 3425, 803, 69, 1, 13700, 47950, 65646, 44240, 15106, 2394, 134, 1, 109601, 438404, 707840, 589106, 267134, 63896, 7094, 263, 1
Offset: 0

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Author

Tom Copeland, Oct 12 2014

Keywords

Comments

This is a transform of A046802 treating it as an array of h-vectors, so y is replaced by (y+1) in the e.g.f. for A046802.
An e.g.f. for the reversed row polynomials with signs is given by exp(a.(0;t)x) = [e^{(1+t)x} [1+t(1-e^(-x))]]^(-1) = 1 - (1+2t)x + (1+5t+5t^2)x^2/2! + ... . The reciprocal is an e.g.f. for the reversed face polynomials of the simplices A074909, i.e., exp(b.(0;t)x) = e^{(1+t)x} [1+t(1-e^(-x))] = 1 + (1+2t)x +(1+3t+3t^2) x^2/2! + ... , so the relations of A133314 apply between the two sets of polynomials. In particular, umbrally [a.(0;t)+b.(0;t)]^n vanishes except for n=0 for which it's unity, implying the two sets of Appell polynomials formed from the two bases, a_n(z;t) = (a.(0;t)+z)^n and b_n(z;t) = (b.(0;t) + z)^n, are an umbral compositional inverse pair, i.e., b_n(a.(x;t);t)= x^n = a_n(b.(x;t);t). Raising operators for these Appell polynomials are related to the polynomials of A028246, whose reverse polynomials are given by A123125 * A007318. Compare: A248727 = A007318 * A123125 * A007318 and A046802 = A007318 * A123125. See A074909 for definitions and related links. - Tom Copeland, Jan 21 2015
The o.g.f. for the umbral inverses is Og(x) = x / (1 - x b.(0;t)) = x / [(1-tx)(1-(1+t)x)] = x + (1+2t) x^2 + (1+3t+3t^2) x^3 + ... . Its compositional inverse is an o.g.f for signed A033282, the reverse f-polynomials for the simplicial duals of the Stasheff polytopes, or associahedra of type A, Oginv(x) =[1+(1+2t)x-sqrt[1+2(1+2t)x+x^2]] / (2t(1+t)x) = x - (1+2t) x^2 + (1+5t+5t^2) x^3 + ... . Contrast this with the o.g.f.s related to the corresponding h-polynomials in A046802. - Tom Copeland, Jan 24 2015
Face vectors, or coefficients of the face polynomials, of the stellahedra, or stellohedra. See p. 59 of Buchstaber and Panov. - Tom Copeland, Nov 08 2016
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016

Examples

			The triangle T(n, k) starts:
n\k    0     1     2     3     4    5   6  7 ...
1:     1
2:     2     1
3:     5     5     1
4:    16    24    10     1
5:    65   130    84    19     1
6:   326   815   720   265    36    1
7:  1957  5871  6605  3425   803   69   1
8: 13700 47950 65646 44240 15106 2394 134  1
... reformatted, _Wolfdieter Lang_, Mar 27 2015
		

Crossrefs

Programs

  • Mathematica
    (* t = A046802 *) t[, 1] = 1; t[n, n_] = 1; t[n_, 2] = 2^(n - 1) - 1; t[n_, k_] = Sum[((i - k + 1)^i*(k - i)^(n - i - 1) - (i - k + 2)^i*(k - i - 1)^(n - i - 1))*Binomial[n - 1, i], {i, 0, k - 1}]; T[n_, j_] := Sum[Binomial[k, j]*t[n + 1, k + 1], {k, j, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 23 2015, after Tom Copeland *)

Formula

Let M(n,k)= sum{i=0,..,k-1, C(n,i)[(i-k)^i*(k-i+1)^(n-i)- (i-k+1)^i*(k-i)^(n-i)]} with M(n,0)=1. Then M(n,k)= A046802(n,k), and T(n,j)= sum(k=j,..,n, C(k,j)*M(n,k)) for j>0 with T(n,0)= 1 + sum(k=1,..,n, M(n,k)) for n>0 and T(0,0)=1.
E.g.f: y * exp[x*(y+1)]/[y+1-exp(x*y)].
Row sums are A007047. Row polynomials evaluated at -1 are unity. Row polynomials evaluated at -2 are A122045.
First column is A000522. Second column appears to be A036918/2 = (A001339-1)/2 = n*A000522(n)/2.
Second diagonal is A052944. (Changed from conjecture to fact on Nov 08 2016.)
The raising operator for the reverse row polynomials with row signs is R = x - (1+t) - t e^(-D) / [1 + t(1-e^(-D))] evaluated at x = 0, with D = d/dx. Also R = x - d/dD log[exp(a.(0;t)D], or R = - d/dz log[e^(-xz) exp(a.(0;t)z)] = - d/dz log[exp(a.(-x;t)z)] with the e.g.f. defined in the comments and z replaced by D. Note that t e ^(-D) / [1+t(1-e^(-D))] = t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... is an e.g.f. for the signed reverse row polynomials of A028246. - Tom Copeland, Jan 23 2015
Equals A007318*(padded A090582)*A007318*A097808 = A007318*(padded (A008292*A007318))*A007318*A097808 = A007318*A130850 = A007318*(mirror of A028246). Padded means in the same way that A097805 is padded A007318. - Tom Copeland, Nov 14 2016
Umbrally, the row polynomials are p_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A130850. - Tom Copeland, Nov 16 2016
From the previous umbral statement, OP(x,d/dy) y^n = (y + q.(x))^n, where OP(x,y) = exp[y * q.(x)] = x/((1+x)*exp(-x*y) - 1), the e.g.f. of A130850, so OP(x,d/dy) y^n evaluated at y = 1 is p_n(x), the n-th row polynomial of this entry, with offset 0. - Tom Copeland, Jun 25 2018
Consolidating some formulas in this entry and A046082, in umbral notation for concision, with all offsets 0: Let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of this entry (A248727, the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020

Extensions

Title expanded by Tom Copeland, Nov 08 2016

A133932 Coefficients of a partition transform for Lagrange inversion of -log(1 - u(.)*t), complementary to A134685 for an e.g.f.

Original entry on oeis.org

1, -1, 3, -2, -15, 20, -6, 105, -210, 40, 90, -24, -945, 2520, -1120, -1260, 420, 504, -120, 10395, -34650, 25200, 18900, -2240, -15120, -9072, 1260, 2688, 3360, -720, -135135, 540540, -554400, -311850, 123200, 415800, 166320, -50400, -56700, -120960, -75600, 18144, 20160, 25920, -5040
Offset: 1

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Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = -log(1 - u(.)*t) = Sum_{n>=1} (u_n / n) * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -1 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - 2 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 20 (1')(2')(3') - 6 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 210 (1') (2')^2 (3') + 40 (1')^2 (3')^2 + 90 (1')^2 (2') (4') - 24 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 2520 (1') (2')^3 (3') - 1120 (1')^2 (2') (3')^2 - 1260 (1')^2 (2')^2 (4') + 420 (1')^3 (3')(4') + 504 (1')^3 (2')(5') - 120 (1')^4 (6') ] * t^6 / 6!
See A134685 for more information.
From Tom Copeland, Sep 28 2016: (Start)
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 34650 (1')(2')^4(3') + (1')^2 [25200 (2')^2(3')^2 + 18900 (2')^3(4')] - (1')^3 [2240 (3')^3 + 15120 (2')(3')(4') + 9072 (2')^2(5')] + (1')^4 [1260 (4')^2 + 2688 (3')(5') + 3360 (2')(6')] - 720 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 540540 (1')(2')^5(3') - (1')^2 [554400 (2')^3(3')^2 + 311850 (2')^4(4')] + (1')^3 [123200 (2')(3')^3 + 415800 (2')^2(3')(4') + 166320 (2')^3(5')] - (1')^4 [50400 (3')^2(4') + 56700 (2')(4')^2 + 120960 (2')(3')(5') + 75600 (2')^2(6')] + (1')^5 [18144 (4')(5') + 20160 (3')(6') + 25920 (2')(7')] - 5040 (1')^6(8')] * t^8 / 8! (End)

Examples

			From _Tom Copeland_, Sep 18 2014: (Start)
Let f(x) = log((1-ax)/(1-bx))/(b-a) = -log(1-u.*x) = x + (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 + (a^3+a^2b+ab^2+a^3)x^4/4 + ... . with (u.)^n = u_n = h_(n-1)(a,b) the complete homogeneous polynomials in two indeterminates.
Then the inverse g(x) = (e^(ax)-e^(bx))/(a*e^(ax)-b*e^(bx)) = x - (a+b)x^2/2! + (a^2+4ab+b^2)x^3/3! - (a^3+11a^2b+11ab^2+b^3)x^4/4! + ... , where the bivariate polynomials are the Eulerian polynomials of A008292.
The inversion formula gives, e.g., P(3,x) = 3(u_2)^2 - 2u_3 = 3(h_1)^2 - 2h_2 = 3(a+b)^2 - 2(a^2 + ab + b^2) = a^2 + 4ab + b^2. (End)
		

Crossrefs

Cf. A145271 (A111999, A007318) = (reduced array, associated g(x)).

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k/(k+1), {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [ 2^e(2) (e(2))! * 3^e(3) (e(3))! * ... n^e(n) * (e(n))! ].
From Tom Copeland, Sep 06 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u./(1-u.t)]
= 1/((u_1) + (u_2)*t + (u_3)*t^2 + (u_4)*t^3 + ...),
where Ev denotes umbral evaluation.
Then for the partition polynomials of A133932,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (End)
From Tom Copeland, Oct 20 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)] = exp[x*h(y)d/dy] exp(t*y) eval. at y=0, the raising/creation and lowering/annihilation operators defined by R PS(n,t)=PS(n+1,t) and L PS(n,t)= n*PS(n-1,t) are
R = t*h(d/dt) = t* 1/[(u_1) + (u_2)*d/dt + (u_3)*(d/dt)^2 + ...], and
L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2/2 + (u_3)*(d/dt)^3/3 + ....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2 + u_3 * x^2/3 + ... + u_n * x^(n-1)/n]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
From Tom Copeland, Sep 19 2016: (Start)
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix A007318 by f_m = (m-1)! u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) f_{n+1-k}, or equivalently, multiply the diagonals of A094587 by u_m. Then P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314.
With u_1 = 1, the first column of UP^(-1) with u_1 = 1 (with initial indices [0,0]) is composed of the row polynomials n! * OP_n(-u_2,...,-u_(n+1)), where OP_n(x[1],...,x[n]) are the row polynomials of A263633 for n > 0 and OP_0 = 1, which are related to those of A133314 as reciprocal o.g.f.s are related to reciprocal e.g.f.s; e.g., UP^(-1)[0,0] = 1, Up^(-1)[1,0] = -u_2, UP^(-1)[2,0] = 2! * (-u_3 + u_2^2) = 2! * OP_2(-u_2,-u_3).
Also, P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. (End)
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -24 u5, b2 = 90 u2 u4 + 40 u3^2, b3 = -210 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers's equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2, ..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1, u_1=1, u_2, ..., u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 210 u2^2 u3 + 1 * 90 u2 u4 + 1 * 40 u3^2 - 0 * -24 u5 = 315 u2^4 - 420 u2^2 u3 + 90 u2 u4 + 40 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 20 u2 u3 - 6 u4) + 6!/(3!*3!) * (3 u2^2 - 2 u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the refined Stirling polynomials of the first kind A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 08 2024

A263633 Irregular triangle read by rows: row n gives coefficients of n-th ordinary Bell polynomial B_n(x_1, x_2, ...) with monomials sorted into graded lexicographic order.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 2, 2, 1, 1, 5, 4, 6, 3, 6, 1, 2, 2, 1, 1, 1, 6, 5, 10, 4, 12, 4, 3, 6, 3, 3, 2, 2, 2, 1, 1, 7, 6, 15, 5, 20, 10, 4, 12, 6, 12, 1, 3, 6, 6, 3, 3, 2, 2, 2, 1, 1, 1, 8, 7, 21, 6, 30, 20, 5, 20, 10, 30, 5, 4, 12, 12, 12, 12, 4, 3, 6, 6, 3, 3, 6, 1, 2, 2, 2, 2, 1
Offset: 1

Views

Author

N. J. A. Sloane, Oct 28 2015

Keywords

Comments

"Ordinary" here means in contrast to "exponential", cf. A178867 (see Comtet).
Graded lexicographic order with x[1] > x[2] > ... > x[n] means that monomials are compared first by their total degree, with ties broken by lexicographic order. These monomials correspond to integer partitions.
Row sums are powers of 2. Numbers of terms in rows are partition numbers A000041.
OP_n(-a_1,..,-a_n) = EP_n(a_1,2!*a_2,..,n!*a_n) / n!, where OP_n(a_1,..,a_n) are the partition polynomials of this entry and EP_n, the polynomials of A133314; i.e., the sequences are related as reciprocal o.g.f.s are to reciprocal e.g.f.s. The polynomials play a role in expansion of the iterated Lie derivative (g(x) D_x)^n) formalism for the compositional inversion sketched in A133932. With x[n] = t, the array reduces to the Pascal matrix A007318. - Tom Copeland, Sep 19 2016
The signed row partition polynomials can be generated by the Gram determinants of equation 2.23 on page 133 of the Verde-Star paper. E.g., h_3 = -b_1^3 + 2 b_1 b_2 - b_3 corresponds to the third row. The connection to A133314 is obtained by substituting a(k) = k!*b_k = -k!*x[k] and b(k) = k!*h_k in A133314 to compute reciprocals of o.g.f.s rather than e.g.f.s. - Tom Copeland, Dec 04 2016
For a relation to lambda operations in K-theory on vector bundles, see p. 218 of Dugger. - Tom Copeland, Jul 25 2017
Since E(x) = (1+x_1*x)(1+x_2*x)...(1+x_m*x) is the o.g.f. for the elementary symmetric polynomials e_n(x_1,x_2,...,x_m) and the o.g.f. for the complete symmetric polynomials h_n(x_1,x_2,...,x_m) is H(x) = 1 / E(-x), this entry's partition polynomials with correct signs give either sequence in terms of the other. - Tom Copeland, Jan 29 2018
A133314 has an interpretation as weighted surjective mappings. With the connections of this mapping colored and permuted to give mappings distinguished by the order of the colorings (an induced linear ordering by color of the connecting arrows), the signed partition polynomials of this entry, multiplied by n!, are generated. - Tom Copeland, Sep 10 2020

Examples

			The first few polynomials are:
1, x[1]
2, x[1]^2 + x[2]
3, x[1]^3 + 2*x[1]*x[2] + x[3]
4, x[1]^4 + 3*x[1]^2*x[2] + 2*x[1]*x[3] + x[2]^2 + x[4]
5, x[1]^5 + 4*x[1]^3*x[2] + 3*x[1]^2*x[3] + 3*x[1]*x[2]^2 + 2*x[1]*x[4] + 2*x[2]*x[3] + x[5]
6, x[1]^6 + 5*x[1]^4*x[2] + 4*x[1]^3*x[3] + 6*x[1]^2*x[2]^2 + 3*x[1]^2*x[4] + 6*x[1]*x[2]*x[3] + x[2]^3 + 2*x[1]*x[5] + 2*x[2]*x[4] + x[3]^2 + x[6]
...
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 136, 309.

Crossrefs

For triangle of coefficients of exponential Bell polynomials see A178867.

Programs

  • Maple
    with(Groebner):
    A263633_row := proc(n) local EE,t1,t2,Q,F,X,p,L,q,c,r;
    EE := add(x[i]*t^i, i=1..2*n);
    t1 := 1/(1-EE):
    t2 := series(t1, t, 2*n):
    Q := k -> expand(coeff(t2, t, k));
    X := seq(x[i], i=1..n);
    p := Q(n);
    L := [];
    while p <> 0 do
       r := LeadingTerm(p, grlex(X));
       c := r[1]; q := r[2];
       p := p - c*q;
       L := [op(L), c];
    od;
    L end:
    for n from 1 to 8 do A263633_row(n) od; # Program expanded by Peter Luschny, Sep 26 2016

Formula

G.f.: 1/(1-Sum_{i >= 1} x_i*t^i) = 1 + Sum_{n >= 1} B_n(x_1, x_2,...)*t^n. [Comtet, p. 136, Eq. [3o'].]

Extensions

More terms and some edits by Peter Luschny, Sep 26 2016

A084358 Lists of sets of lists.

Original entry on oeis.org

1, 1, 5, 37, 363, 4441, 65133, 1114009, 21771851, 478658101, 11692343253, 314170940293, 9209104364331, 292435635165649, 10000637145321917, 366427621403088433, 14321135069200849515, 594696814358067968461, 26147933188037724372069
Offset: 0

Views

Author

N. J. A. Sloane, Jun 22 2003

Keywords

Comments

This sequence and -A000262 with the first term set to 1 form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 21 2007

References

  • T. S. Motzkin, Sorting numbers ...: for a link to an annotated scanned version of this paper see A000262.
  • T. S. Motzkin, Sorting numbers for cylinders and other classification numbers, in Combinatorics, Proc. Symp. Pure Math. 19, AMS, 1971, pp. 167-176.

Programs

  • Magma
    m:=25; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1/(2-Exp(x/(1-x))))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, May 16 2018
  • Maple
    with(combstruct); SeqSetSeqL := [T, {T=Sequence(S), S=Set(U,card >= 1), U=Sequence(Z,card >=1)},labeled]; [seq(count(%,size=j),j=1..12)];
  • Mathematica
    With[{nn=20},CoefficientList[Series[1/(2-Exp[x/(1-x)]),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Apr 16 2013 *)
  • PARI
    x='x+O('x^30); Vec(serlaplace(1/(2-exp(x/(1-x))))) \\ G. C. Greubel, May 16 2018
    

Formula

a(n) = n!*Lag{n,(.)!*Lag[.,P(.,2),0],-1} = P(n,2) - n*P(n-1,2) umbrally, where P(j,t) are the polynomials in A131758 and Lag(n,x,a) are the associated Laguerre polynomials of order a; that is, the sequence is given by an iterated combinatorial Laguerre transform, of mixed order, of a set of polynomials related to the polylogarithms, which reduces to a simple finite difference. - Tom Copeland, Sep 30 2007
E.g.f.: 1/(2-exp(x/(1-x))). Lah transform of preferential arrangements: Sum_{k=0..n} n!/k!*binomial(n-1, k-1)*A000670(k). - Vladeta Jovovic, Sep 28 2003
a(n) ~ n! * (1+log(2))^(n-1) / (2*(log(2))^(n+1)). - Vaclav Kotesovec, Oct 08 2013

A132014 T(n,j) for double application of an iterated mixed order Laguerre transform: Coefficients of Laguerre polynomial (-1)^n*n!*L(n,2-n,x).

Original entry on oeis.org

1, -2, 1, 2, -4, 1, 0, 6, -6, 1, 0, 0, 12, -8, 1, 0, 0, 0, 20, -10, 1, 0, 0, 0, 0, 30, -12, 1, 0, 0, 0, 0, 0, 42, -14, 1, 0, 0, 0, 0, 0, 0, 56, -16, 1, 0, 0, 0, 0, 0, 0, 0, 72, -18, 1, 0, 0, 0, 0, 0, 0, 0, 0, 90, -20, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 110, -22, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 132, -24, 1
Offset: 0

Views

Author

Tom Copeland, Oct 30 2007, Nov 05 2007, Nov 11 2007

Keywords

Comments

The matrix operation b = T*a can be characterized in several ways in terms of the coefficients a(n) and b(n), their o.g.f.s A(x) and B(x), or e.g.f.s EA(x) and EB(x).
1) b(0) = a(0), b(1) = a(n) - 2 a(0), b(n) = a(n) - 2n a(n-1) + n(n-1) a(n-2) for n > 0.
2) b(n) = n! Lag{n,(.)!*Lag[.,a1(.),0],-1}, umbrally, where a1(n) = n! Lag{n,(.)!*Lag[.,a(.),0],-1}.
3) b(n) = n! Sum_{j=0..min(2,n)} (-1)^j * binomial(n,j)*a(n-j)/(n-j)!
4) b(n) = (-1)^n n! Lag(n,a(.),2-n)
5) B(x) = (1-xDx)^2 A(x)
6) B(x) = Sum_{j=0..2} {(-1)^j * binomial(2,j)*j!*x^j*Lag(j,-:xD:,0)} A(x)
where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x,m) is the associated Laguerre polynomial of order m.
7) EB(x) = (1-x)^2 EA(x)
8) T = S^2 = A132013^2 = A094587^(-2) = A132159^(-1).
c = (1,-2,2,0,0,...) is the sequence associated to T under the list partition transform and associated operations described in A133314. c are also the coefficients in formula 6. Thus T(n,k) = binomial(n,k)*c(n-k).
The reciprocal sequence to c is d = (1!,2!,3!,4!,...), so the inverse of T is TI(n,k) = binomial(n,k)*d(n-k) = A132159.
These formulas are easily generalized for m applications of the basic operator n! Lag[n,(.)!*Lag[.,a(.),0],-1] by replacing 2 with m in formulas 3, 4, 5, 6 and 7.
The generalized c are given by the generalized coefficients of 6, i.e.,
c(n) = (-1)^n * binomial(m,n)*n! = (-1)^n * m!/(m-n)!.
The generalized d are given by the array at and below the term SI(m-1,m-1) in SI(n,k) = binomial(n,k) * (n-k)!, the inverse of S; i.e.,
d(n) = SI(m-1+n,m-1) = binomial(m-1+n,m-1) * n! = (m-1+n)!/(m-1)!.
As an aside, this shows that the signed falling factorials and the rising factorials form reciprocal pairs under the list partition transform of A133314.
Row sums of T = [formula 3 with all a(n) = 1] = [binomial transform of c] = [coefficients of B(x) with A(x) = 1/(1-x)] = (1,-1,-1,1,5,11,19,...),
with e.g.f. = [EB(x) with EA(x) = exp(x)] = (1-x)^2 * exp(x) = exp(x)*exp(c(.)*x) = exp[(1+c(.))*x].
Alternating row sums of T = [formula 3 with all a(n) = (-1)^n] = [finite differences of c] = [coefficients of B(x) with A(x) = 1/(1+x)] = (1,-3,7,-13,21,-31,...) = (-1)^n A002061(n+1),
with e.g.f. = [EB(x) with EA(x) = exp(-x)] = (1-x)^2 * exp(-x) = exp(- x)*exp(c(.)*x) = exp[-(1-c(.))*x].
See A132159 for a relation to the Poisson-Charlier polynomials. - Tom Copeland, Jan 15 2016

Examples

			First few rows of the triangle are
   1;
  -2,   1;
   2,  -4,   1;
   0,   6,  -6,   1;
   0,   0,  12,  -8,   1;
   0,   0,   0,  20, -10,   1;
		

Crossrefs

Programs

  • Mathematica
    m = 12; s = Exp[x*y]*(1 - x)^2 + O[x]^(m + 2) + O[y]^(m + 2); T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]*n!; T[0, 0] = 1; Table[T[n, k], {n, 0, m}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 09 2015 *)
  • PARI
    row(n) = Vecrev((-1)^n*n!*pollaguerre(n, 2-n)); \\ Michel Marcus, Feb 06 2021

Formula

T(n,k) = binomial(n,k)*c(n-k).
E.g.f. for row polynomials: exp(x*y)(1-x)^2. Implies the row polynomials form an Appell sequence (see Wikipedia). - Tom Copeland, Dec 03 2013
From Tom Copeland, Apr 21 2014: (Start)
Change notation letting L(n,m,x) = Lag(n,x,m).
Row polynomials: (-1)^n*n!*L(n,2-n,x) = (-1)^n*(-x)^(n-2)*2!*L(2,n-2,x) =
(-1)^n*(2!/(2-n)!)*K(-n,2-n+1,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
For the row polynomials, the lowering operator = d/dx and the raising operator = x - 2/(1-D).
T = (I - A132440)^2 = [2*I - exp(A238385-I)]^2 = signed exp[2*(A238385-I)], where I = identity matrix.
Operationally, (-1)^n*n!*L(n,2-n,-:xD:) = (-1)^n*x^(n-2)*:Dx:^n*x^(2-n) = (-1)^n*x^(-2)*:xD:^n*x^2 = (-1)^n*n!*binomial(xD+2,n) = (-1)^n*n!*binomial(2,n)*K(-n,2-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706. (End)
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,2,x). - Peter Bala, Jul 25 2021

Extensions

Title modified by Tom Copeland, Apr 21 2014
Missing term -18 inserted in 10th row by Jean-François Alcover, Jul 09 2015

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

A129652 Exponential Riordan array [e^(x/(1-x)),x].

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 13, 9, 3, 1, 73, 52, 18, 4, 1, 501, 365, 130, 30, 5, 1, 4051, 3006, 1095, 260, 45, 6, 1, 37633, 28357, 10521, 2555, 455, 63, 7, 1, 394353, 301064, 113428, 28056, 5110, 728, 84, 8, 1, 4596553, 3549177, 1354788, 340284, 63126, 9198, 1092, 108, 9, 1
Offset: 0

Views

Author

Paul Barry, Apr 26 2007

Keywords

Comments

Satisfies the equation e^[x/(1-x),x] = e*[e^(x/(1-x)),x].
Row sums are A052844.
Antidiagonal sums are A129653.

Examples

			Triangle begins:
     1;
     1,    1;
     3,    2,    1;
    13,    9,    3,   1;
    73,   52,   18,   4,  1;
   501,  365,  130,  30,  5, 1;
  4051, 3006, 1095, 260, 45, 6, 1;
  ...
		

Crossrefs

Cf. A000262 (column 0), A052844 (row sums).
T(2n,n) gives A350461.

Programs

  • Maple
    A129652 := (n, k) -> (-1)^(k-n+1)*binomial(n,k)*KummerU(k-n+1, 2, -1);
    seq(seq(round(evalf(A129652(n,k),99)),k=0..n),n=0..9); # Peter Luschny, Sep 17 2014
    # second Maple program:
    b:= proc(n) option remember; `if`(n=0, [1$2], add((p-> p+
         [0, p[1]*x^j])(b(n-j)*binomial(n-1, j-1)*j!), j=1..n))
        end:
    T:= n-> (p-> seq(coeff(p, x, i)/i!, i=0..n))(b(n)[2]):
    seq(T(n), n=0..10);  # Alois P. Heinz, Feb 21 2022
  • Mathematica
    T[n_, k_] := If[k==n, 1, n!/k! Sum[Binomial[n-k-1, j]/(j+1)!, {j, 0, n-k-1}]];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 14 2019 *)

Formula

Number triangle T(n,k)=(n!/k!)*sum{i=0..n-k, C(n-k-1,i)/(n-k-i)!}
From Peter Bala, May 14 2012 : (Start)
Array is exp(S*(I-S)^(-1)) where S is A132440 the infinitesimal generator for Pascal's triangle.
Column 0 is A000262.
T(n,k) = binomial(n,k)*A000262(n-k).
So T(n,k) gives the number of ways to choose a subset of {1,2,...,n} of size k and then arrange the remaining n-k elements into a set of lists. (End)
T(n,k) = (-1)^(k-n+1)*C(n,k)*KummerU(k-n+1, 2, -1). - Peter Luschny, Sep 17 2014
From Tom Copeland, Mar 11 2016: (Start)
The row polynomials P_n(x) form an Appell sequence with e.g.f. e^(t*P.(x)) = e^[t / (1-t)] e^(x*t), so the lowering and raising operators are L = d/dx = D and the R = x + 1 / (1-D)^2 = x + 1 + 2 D + 3 D^2 + ..., satisfying L P_n(x) = n * P_(n-1)(x) and R P_n(x) = P_(n+1)(x).
(P.(x) + y)^n = Sum_{k=0..n} binomial(n,k) P_k(x) * y^(n-k) = P_n(x+y).
The Appell polynomial umbral compositional inverse sequence has the e.g.f. e^(t*Q.(x)) = e^[-t / (1-t)] e^(x*t) (see A111884 and A133314), so Q_n(P.(x)) = P_n(Q.(x)) = x^n. The lower triangular matrices for the coefficients of these two Appell sequences are a multiplicative inverse pair.
(End)
Sum_{k=0..n} (-1)^k * T(n,k) = A052845(n). - Alois P. Heinz, Feb 21 2022
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