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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A133314 Coefficients of list partition transform: reciprocal of an exponential generating function (e.g.f.).

Original entry on oeis.org

1, -1, -1, 2, -1, 6, -6, -1, 8, 6, -36, 24, -1, 10, 20, -60, -90, 240, -120, -1, 12, 30, -90, 20, -360, 480, -90, 1080, -1800, 720, -1, 14, 42, -126, 70, -630, 840, -420, -630, 5040, -4200, 2520, -12600, 15120, -5040, -1, 16, 56, -168, 112, -1008, 1344, 70
Offset: 0

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Author

Tom Copeland, Oct 18 2007, Oct 29 2007, Nov 16 2007

Keywords

Comments

The list partition transform of a sequence a(n) for which a(0)=1 is illustrated by:
b_0 = 1
b_1 = -a_1
b_2 = -a_2 + 2 a_1^2
b_3 = -a_3 + 6 a_2 a_1 - 6 a_1^3
b_4 = -a_4 + 8 a_3 a_1 + 6 a_2^2 - 36 a_2 a_1^2 + 24 a_1^4
... .
The unsigned coefficients are A049019 with a leading 1. The sign is dependent on the partition as evident from inspection (replace a_n's by -1).
Expressed umbrally, i.e., with the umbral operation (a.)^n := a_n,
exp(a.x) exp(b.x) = exp[(a.+b.)x] = 1; i.e., (a.+b.)^n = 1 for n=0 and 0 for all other values of n.
Expressed recursively,
b_0 = 1, b_n = -Sum_{j=1..n} binomial(n,j) a_j b_{n-j}; which is conditionally self-inverse, i.e., the roles of a_k and b_k may be reversed with a_0 = b_0 = 1.
Expressed in matrix form, b_n form the first column of B = matrix inverse of A .
A = Pascal matrix diagonally multiplied by a_n, i.e., A_{n,k} = binomial(n,k)* a_{n-k}.
Some examples of reciprocal pairs of sequences under these operations are:
1) A084358 and -A000262 with the first term set to 1.
2) (1,-1,0,0,...) and (0!,1!,2!,3!,...) with the unsigned associated matrices A128229 and A094587.
3) (1,-1,-1,-1,...) and A000670.
5) (1,-2,-2,0,0,0,...) and (0! c_1,1! c_2,2! c_3,3! c_4,...) where c_n = A000129(n) with the associated matrices A110327 and A110330.
6) (1,-2,2,0,0,0,...) and (1!,2!,3!,4!,...).
7) Sequences of rising and signed lowering factorials form reciprocal pairs where a_n = (-1)^n m!/(m-n)! and b_n = (m-1+n)!/(m-1)! for m=0,1,2,... .
Denote the action of the list partition transform on the sequence a. or an invertible matrix M by LPT(a.) = b. or LPT(M)= M^(-1).
If the matrix equation M = exp(T) also holds, then exp[a.*T]*exp[b.*T] = exp[(a.+b.)*T] = I, the identity matrix, because (a.+b.)^n = delta_n, the Kronecker delta with delta_n = 1 and delta_n = 0 otherwise, i.e., (0)^n = delta_n.
Therefore, [exp(a.*T)]^(-1) = exp[b.*T] = exp[LPT(a.)*T] = LPT[exp(a.*T)].
The fundamental Pascal (A007318), unsigned Lah (A105278) and associated Laguerre matrices can be generated by exponentiation of special infinitesimal matrices (see A132440, A132710 and A132681) such that finding LPT(a.) amounts to multiplying the k'th diagonal of the fundamental matrices by a_k for every diagonal followed by matrix inversion and then extraction of the b_n factors from the first column (simplest for the Pascal formulas above).
Conversely, the inverses of matrices formed by diagonally multiplying the three fundamental matrices by a_k are given by diagonally multiplying the fundamental matrices by b_k.
If LPT(M) is defined differently as application of the top formula to a_n = M^n, then b_n = (-M)^n and the formalism could even be applied to more general sequences of matrices M., providing the reciprocal of exp[t*M.].
The group of fundamental lower triangular matrices M = exp(T) such that LPT[exp(a.*T)] = exp[LPT(a.)*T] = [exp[a.*T]]^(-1) are obtained by infinitesimal generator matrices of the form T =
0;
t(0), 0;
0, t(1), 0;
0, 0, t(2), 0;
0, 0, 0, t(3), 0;
... .
T^m has trivially vanishing terms except along the m'th subdiagonal, which is a sequence of generalized factorials:
[ t(0)*t(1)...t(m-2)*t(m-1), t(1)*t(2)...t(m-1)*t(m), t(2)*t(3)...t(m)*t(m+1), ... ].
Therefore the principal submatrices of T (given by setting t(j) = 0 for j > n-1) are nilpotent with at least [Tsub_n]^(n+1) = 0.
The general group of matrices GM[a.] = exp[a.*T] can also be obtained through diagonal multiplication of M = exp(T) by the sequence a_n, as in the Pascal matrix example above and their inverses by diagonal multiplication by b. = LPT(a.).
Weighted-mappings interpretation for the top partition equation:
Given n pre-nodes (Pre) and k post-nodes (Post), each Pre is connected to only one Post and each Post has at least one Pre connected to it (surjections or onto functions/maps). Weight each Post by -a_m where m is the number of connections to the Post.
Weight each map by the product of the Post weights and multiply by the number of maps that share the same connectivity. Sum over the possible mappings for n Pre. The result is b_n.
E.g., b_3 = [ 3 Pre to 1 Post ] + [ 3 Pre to 2 Post ] + [ 3 Pre to 3 Post ]
= [1 map with 1 Post with 3 connections] + [ 6 maps with 1 Post with 2 connections and 1 Post with 1 connection] + [6 maps with 3 Post with 1 connection each]
= -a_3 + 6 * [-a_2*(-a_1)] + 6 * [-a_1*(-a_1)*(-a_1)].
See A263633 for the complementary formulation for the reciprocal of o.g.f.s rather than e.g.f.s and computations of these partition polynomials as Gram determinants. - Tom Copeland, Dec 04 2016
The coefficients of the partition polynomials enumerate the faces of the convex, bounded polytopes called permutohedra, and the absolute value of the sum of the coefficients gives the Euler characteristic of unity for each polytope; i.e., the absolute value of the sum of each row of the array is unity. In addition, the signs of the faces alternate with dimension, and the coefficients of faces with the same dimension for each polytope have the same sign. - Tom Copeland, Nov 13 2019
With the fundamental matrix chosen to be the lower triangular Pascal matrix M, the matrix MA whose n-th diagonals are multiplied by a_n (i.e., MA_{i,j} = PM_{i,j} * a_{i-j}) gives a matrix representation of the e.g.f. associated to the Appell polynomial sequence defined by e^{a.t}e^{xt}= e^{(a.+x)t} = e^{A.(x)t} where umbrally (A.(x))^n = A_n(x) = (a. + x)^n = sum_{k=0..n} binomial(n,k) a_k x^{n-k} are the associated Appell polynomials. Left multiplication of the column vector (1,x,x^2,..) by MA gives the Appell polynomial sequence, and multiplication of the two e.g.f.s e^{a.t} and e^{b.t} corresponds to multiplication of their respective matrix representations MA and MB. Forming the reciprocal of an e.g.f. corresponds to taking the matrix inverse of its matrix representation as noted above. A263634 gives an associated modified Pascal matrix representation of the raising operator for the Appell sequence. - Tom Copeland, Nov 13 2019
The diagonal of MA consists of all ones. Let MAN be the truncated square submatrix of MA containing the coefficients of the first N Appell polynomials A_k=(a.+x)^k = Sum(j=0 to k) MAN(k,j) x^j. Then by the Cayley-Hamilton theorem (I-MAN)^N = 0; therefore, MAN^(-1) = Sum(k=1 to N) binomial(N,k) (-MAN)^{k-1} = MBN, the inverse of MAN, containing the coefficients of the first N rows of the Appell polynomials B_k(x) = (b. + x)^k = Sum(j=0 to k) MBN(k,j) x^j, which are the umbral compositional inverses of the Appell row polynomials A_k(x) of MAN; that is, A_k(B.(x)) = x^k = B_k(A.(x)), where, e.g., (A.(x))^k = A_k(x). - Tom Copeland, May 13 2020
The use of the term 'list partition transform' resulted from one of my first uses of these partition polynomials in relating A000262 to A084358 with their simple e.g.f.s. Other appropriate names would be the permutohedra polynomials since they are refined Euler characteristics of the permutohedra or the reciprocal polynomials since they give the multiplicative inverses of e.g.f.s with a constant of 1. - Tom Copeland, Oct 09 2022

Examples

			Table starts:
[0] [ 1]
[1] [-1]
[2] [-1,  2]
[3] [-1,  6, -6]
[4] [-1,  8,  6, -36,  24]
[5] [-1, 10, 20, -60, -90,  240, -120]
[6] [-1, 12, 30, -90,  20, -360,  480, -90, 1080, -1800, 720]
		

Crossrefs

Programs

  • Mathematica
    b[0] = 1; b[n_] := b[n] = -Sum[Binomial[n, j]*a[j]*b[n-j], {j, 1, n}];
    row[0] = {1}; row[n_] := Coefficient[b[n], #]& /@ (Times @@ (a /@ #)&) /@ IntegerPartitions[n];
    Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 23 2014 *)
  • Sage
    def A133314_row(n): return [(-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in Partitions(n)]
    for n in (0..10): print(A133314_row(n)) # Peter Luschny, Sep 18 2015

Formula

b_{n-1} = (1/n)(d/da(1))p_n[a_1, a_2, ..., a_n] where p_n are the row partition polynomials of the cumulant generator A127671. - Tom Copeland, Oct 13 2012
(E.g.f. of matrix B) = (e.g.f. of b)·exp(xt) = exp(b.t)·exp(xt) = exp(xt)/exp(a.t) = (e.g.f. of A^(-1)) and (e.g.f. of matrix A) = exp(a.t)·exp(xt) = exp(xt)/exp(b.t) = (e.g.f. of B^(-1)), where the umbral evaluation of exp(b.t) = Sum{n >= 0} (b.t)^n / n! = Sum_{n >= 0} b_n t^n / n! is understood in the denominator. These e.g.f.s define Appell sequences of polynomials. - Tom Copeland, Mar 22 2014
Sum of the n-th row is (-1)^n. - Peter Luschny, Sep 18 2015
The unsigned coefficients for the partitions a_2*a_1^n for n >= 0 are the Lah numbers A001286. - Tom Copeland, Aug 06 2016
G.f.: 1 / (1 + Sum_{n > 0} a_n x^n/n!) = 1 / exp(a.x). - Tom Copeland, Oct 18 2016
Let a_1 = 1 + x + B_1 = x + 1/2 and a_n = B_n = (B.)^n, where B_n are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted signed polynomials of A019538: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... , p_n(x) = n * b_{n-1}. - Tom Copeland, Oct 18 2016
With a_n = 1/(n+1), b_n = B_n, the Bernoulli numbers. - Tom Copeland, Nov 08 2016
Indeterminate substitutions as illustrated in A356145 lead to [E] = [L][P] = [P][E]^(-1)[P] = [P][RT] and [E]^(-1) = [P][L] = [P][E][P] = [RT][P], where [E] contains the refined Eulerian partition polynomials of A145271; [E]^(-1), A356145, the inverse set to [E]; [P], the permutohedra polynomials of this entry; [L], the classic Lagrange inversion polynomials of A134685; and [RT], the reciprocal tangent polynomials of A356144. Since [L]^2 = [P]^2 = [RT]^2 = [I], the substitutional identity, [L] = [E][P] = [P][E]^(-1) = [RT][P], [RT] = [E]^(-1)[P] = [P][L][P] = [P][E], and [P] = [L][E] = [E][RT] = [E]^(-1)[L] = [RT][E]^(-1). - Tom Copeland, Oct 05 2022

Extensions

More terms from Jean-François Alcover, Apr 23 2014

A135278 Triangle read by rows, giving the numbers T(n,m) = binomial(n+1, m+1); or, Pascal's triangle A007318 with its left-hand edge removed.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 10, 5, 1, 6, 15, 20, 15, 6, 1, 7, 21, 35, 35, 21, 7, 1, 8, 28, 56, 70, 56, 28, 8, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 12, 66, 220, 495, 792, 924, 792
Offset: 0

Views

Author

Zerinvary Lajos, Dec 02 2007

Keywords

Comments

T(n,m) is the number of m-faces of a regular n-simplex.
An n-simplex is the n-dimensional analog of a triangle. Specifically, a simplex is the convex hull of a set of (n + 1) affinely independent points in some Euclidean space of dimension n or higher, i.e., a set of points such that no m-plane contains more than (m + 1) of them. Such points are said to be in general position.
Reversing the rows gives A074909, which as a linear sequence is essentially the same as this.
From Tom Copeland, Dec 07 2007: (Start)
T(n,k) * (k+1)! = A068424. The comment on permuted words in A068424 shows that T is related to combinations of letters defined by connectivity of regular polytope simplexes.
If T is the diagonally-shifted Pascal matrix, binomial(n+m, k+m), for m=1, then T is a fundamental type of matrix that is discussed in A133314 and the following hold.
The infinitesimal matrix generator is given by A132681, so T = LM(1) of A132681 with inverse LM(-1).
With a(k) = (-x)^k / k!, T * a = [ Laguerre(n,x,1) ], a vector array with index n for the Laguerre polynomials of order 1. Other formulas for the action of T are given in A132681.
T(n,k) = (1/n!) (D_x)^n (D_t)^k Gf(x,t) evaluated at x=t=0 with Gf(x,t) = exp[ t * x/(1-x) ] / (1-x)^2.
[O.g.f. for T ] = 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 }. [ O.g.f. for row sums ] = 1 / { (1-x) * (1-2x) }, giving A000225 (without a leading zero) for the row sums. Alternating sign row sums are all 1. [Sign correction noted by Vincent J. Matsko, Jul 19 2015]
O.g.f. for row polynomials = [ (1+q)**(n+1) - 1 ] / [ (1+q) -1 ] = A(1,n+1,q) on page 15 of reference on Grassmann cells in A008292. (End)
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. The e.g.f. for the row polynomials of A is {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. - Tom Copeland, Aug 21 2008
A007318*A097806 as infinite lower triangular matrices. - Philippe Deléham, Feb 08 2009
Riordan array (1/(1-x)^2, x/(1-x)). - Philippe Deléham, Feb 22 2012
The elements of the matrix inverse are T^(-1)(n,k)=(-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 12 2013
Relation to K-theory: T acting on the column vector (-0,d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168 and also A104712, A111492, and A238363. - Tom Copeland, Apr 11 2014
Number of walks of length p>0 between any two distinct vertices of the complete graph K_(n+2) is W(n+2,p)=(-1)^(p-1)*Sum_{k=0..p-1} T(p-1,k)*(-n-2)^k = ((n+1)^p - (-1)^p)/(n+2) = (-1)^(p-1)*Sum_{k=0..p-1} (-n-1)^k. This is equal to (-1)^(p-1)*Phi(p,-n-1), where Phi is the cyclotomic polynomial when p is an odd prime. For K_3, see A001045; for K_4, A015518; for K_5, A015521; for K_6, A015531; for K_7, A015540. - Tom Copeland, Apr 14 2014
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-1)^0 + A_1*(x-1)^1 + A_2*(x-1)^2 + ... + A_n*(x-1)^n. This sequence gives A_0, ..., A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
See A074909 for associations among this array, the Bernoulli polynomials and their umbral compositional inverses, and the face polynomials of permutahedra and their duals (cf. A019538). - Tom Copeland, Nov 14 2014
From Wolfdieter Lang, Dec 10 2015: (Start)
A(r, n) = T(n+r-2, r-1) = risefac(n,r)/r! = binomial(n+r-1, r), for n >= 1 and r >= 1, gives the array with the number of independent components of a symmetric tensors of rank r (number of indices) and dimension n (indices run from 1 to n). Here risefac(n, k) is the rising factorial.
As(r, n) = T(n+1, r+1) = fallfac(n, r)/r! = binomial(n, r), r >= 1 and n >= 1 (with the triangle entries T(n, k) = 0 for n < k) gives the array with the number of independent components of an antisymmetric tensor of rank r and dimension n. Here fallfac is the falling factorial. (End)
The h-vectors associated to these f-vectors are given by A000012 regarded as a lower triangular matrix. Read as bivariate polynomials, the h-polynomials are the complete homogeneous symmetric polynomials in two variables, found in the compositional inverse of an e.g.f. for A008292, the h-vectors of the permutahedra. - Tom Copeland, Jan 10 2017
For a correlation between the states of a quantum system and the combinatorics of the n-simplex, see Boya and Dixit. - Tom Copeland, Jul 24 2017

Examples

			The triangle T(n, k) begins:
   n\k  0  1   2   3   4   5   6   7   8  9 10 11 ...
   0:   1
   1:   2  1
   2:   3  3   1
   3:   4  6   4   1
   4:   5 10  10   5   1
   5:   6 15  20  15   6   1
   6:   7 21  35  35  21   7   1
   7:   8 28  56  70  56  28   8   1
   8:   9 36  84 126 126  84  36   9   1
   9:  10 45 120 210 252 210 120  45  10  1
  10:  11 55 165 330 462 462 330 165  55 11  1
  11:  12 66 220 495 792 924 792 495 220 66 12  1
  ... reformatted by _Wolfdieter Lang_, Mar 23 2015
Production matrix begins
   2   1
  -1   1   1
   1   0   1   1
  -1   0   0   1   1
   1   0   0   0   1   1
  -1   0   0   0   0   1   1
   1   0   0   0   0   0   1   1
  -1   0   0   0   0   0   0   1   1
   1   0   0   0   0   0   0   0   1   1
- _Philippe Deléham_, Jan 29 2014
From _Wolfdieter Lang_, Nov 08 2018: (Start)
Recurrence [_Philippe Deléham_]: T(7, 3) = 2*35 + 35 - 15 - 20 = 70.
Recurrence from Riordan A- and Z-sequences: [1,1,repeat(0)] and [2, repeat(-1, +1)]: From Z: T(5, 0) = 2*5 - 10 + 10 - 5 + 1 = 6. From A: T(7, 3) = 35 + 35 = 70.
Boas-Buck column k=3 recurrence: T(7, 3) = (5/4)*(1 + 5 + 15 + 35) = 70. (End)
		

Crossrefs

Programs

  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*1^(i-j), j = 1 .. i) od;
  • Mathematica
    Flatten[Table[CoefficientList[D[1/x ((x + 1) Exp[(x + 1) z] - Exp[z]), {z, k}] /. z -> 0, x], {k, 0, 11}]]
    CoefficientList[CoefficientList[Series[1/((1 - x)*(1 - x - x*y)), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    for(n=0, 20, for(k=0, n, print1(1/k!*sum(i=0, n, (prod(j=0, k-1, i-j))), ", "))) \\ Derek Orr, Oct 14 2014
    
  • Sage
    Trow = lambda n: sum((x+1)^j for j in (0..n)).list()
    for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

Formula

T(n, k) = Sum_{j=k..n} binomial(j,k) = binomial(n+1, k+1), n >= k >= 0, else 0. (Partial sum of column k of A007318 (Pascal), or summation on the upper binomial index (Graham et al. (GKP), eq. (5.10). For the GKP reference see A007318.) - Wolfdieter Lang, Aug 22 2012
E.g.f.: 1/x*((1 + x)*exp(t*(1 + x)) - exp(t)) = 1 + (2 + x)*t + (3 + 3*x + x^2)*t^2/2! + .... The infinitesimal generator for this triangle has the sequence [2,3,4,...] on the main subdiagonal and 0's elsewhere. - Peter Bala, Jul 16 2013
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Dec 27 2013
T(n,k) = A193862(n,k)/2^k. - Philippe Deléham, Jan 29 2014
G.f.: 1/((1-x)*(1-x-x*y)). - Philippe Deléham, Mar 13 2014
From Tom Copeland, Mar 26 2014: (Start)
[From Copeland's 2007 and 2008 comments]
A) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^2 } (same as Deleham's).
B) The infinitesimal generator for T is given in A132681 with m=1 (same as Bala's), which makes connections to the ubiquitous associated Laguerre polynomials of integer orders, for this case the Laguerre polynomials of order one L(n,-t,1).
C) O.g.f. of row e.g.f.s: Sum_{n>=0} L(n,-t,1) x^n = exp[t*x/(1-x)]/(1-x)^2 = 1 + (2+t)x + (3+3*t+t^2/2!)x^2 + (4+6*t+4*t^2/2!+t^3/3!)x^3+ ... .
D) E.g.f. of row o.g.f.s: ((1+t)*exp((1+t)*x)-exp(x))/t (same as Bala's).
E) E.g.f. for T(n,k)*a(n-k): {(a+t) exp[(a+t)x] - a exp(a x)}/t, umbrally. For example, for a(k)=2^k, the e.g.f. for the row o.g.f.s is {(2+t) exp[(2+t)x] - 2 exp(2x)}/t.
(End)
From Tom Copeland, Apr 28 2014: (Start)
With different indexing
A) O.g.f. by row: [(1+t)^n-1]/t.
B) O.g.f. of row o.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
C) E.g.f. of row o.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
These generating functions are related to row e.g.f.s of A111492. (End)
From Tom Copeland, Sep 17 2014: (Start)
A) U(x,s,t)= x^2/[(1-t*x)(1-(s+t)x)] = Sum_{n >= 0} F(n,s,t)x^(n+2) is a generating function for bivariate row polynomials of T, e.g., F(2,s,t)= s^2 + 3s*t + 3t^2 (Buchstaber, 2008).
B) dU/dt=x^2 dU/dx with U(x,s,0)= x^2/(1-s*x) (Buchstaber, 2008).
C) U(x,s,t) = exp(t*x^2*d/dx)U(x,s,0) = U(x/(1-t*x),s,0).
D) U(x,s,t) = Sum[n >= 0, (t*x)^n L(n,-:xD:,-1)] U(x,s,0), where (:xD:)^k=x^k*(d/dx)^k and L(n,x,-1) are the Laguerre polynomials of order -1, related to normalized Lah numbers. (End)
E.g.f. satisfies the differential equation d/dt(e.g.f.(x,t)) = (x+1)*e.g.f.(x,t) + exp(t). - Vincent J. Matsko, Jul 18 2015
The e.g.f. of the Norlund generalized Bernoulli (Appell) polynomials of order m, NB(n,x;m), is given by exponentiation of the e.g.f. of the Bernoulli numbers, i.e., multiple binomial self-convolutions of the Bernoulli numbers, through the e.g.f. exp[NB(.,x;m)t] = (t/(e^t - 1))^(m+1) * e^(xt). Norlund gave the relation to the factorials (x-1)!/(x-1-n)! = (x-1) ... (x-n) = NB(n,x;n), so T(n,m) = NB(m+1,n+2;m+1)/(m+1)!. - Tom Copeland, Oct 01 2015
From Wolfdieter Lang, Nov 08 2018: (Start)
Recurrences from the A- and Z- sequences for the Riordan triangle (see the W. Lang link under A006232 with references), which are A(n) = A019590(n+1), [1, 1, repeat (0)] and Z(n) = (-1)^(n+1)*A054977(n), [2, repeat(-1, 1)]:
T(0, 0) = 1, T(n, k) = 0 for n < k, and T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and T(n, k) = T(n-1, k-1) + T(n-1, k), for n >= m >= 1.
Boas-Buck recurrence for columns (see the Aug 10 2017 remark in A036521 also for references):
T(n, k) = ((2 + k)/(n - k))*Sum_{j=k..n-1} T(j, k), for n >= 1, k = 0, 1, ..., n-1, and input T(n, n) = 1, for n >= 0, (the BB-sequences are alpha(n) = 2 and beta(n) = 1). (End)
T(n, k) = [x^k] Sum_{j=0..n} (x+1)^j. - Peter Luschny, Jul 09 2019

Extensions

Edited by Tom Copeland and N. J. A. Sloane, Dec 11 2007

A104712 Pascal's triangle, with the first two columns removed.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715
Offset: 2

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Author

Gary W. Adamson, Mar 19 2005

Keywords

Comments

A000295 (Eulerian numbers) gives the row sums.
Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle.
From Peter Luschny, Apr 10 2011: (Start)
A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:
(a) Each partition has at least one singleton block;
(c) k is the size of the largest block of the partition;
(b) m = n - k + 1 is the number of parts of the partition.
Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)).
For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11.
This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts
1;
1, 3;
1, 6, 4;
1, 10, 10, 5;
1, 15, 20, 15, 6;
...
(End)
Diagonal sums are A001924(n+1). - Philippe Deléham, Jan 11 2014
Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014

Examples

			The triangle a(n, k) begins:
  n\k  2   3   4    5    6    7    8   9  10 11 12 13
  2:   1
  3:   3   1
  4:   6   4   1
  5:  10  10   5    1
  6:  15  20  15    6    1
  7:  21  35  35   21    7    1
  8:  28  56  70   56   28    8    1
  9:  36  84 126  126   84   36    9   1
  10: 45 120 210  252  210  120   45  10   1
  11: 55 165 330  462  462  330  165  55  11  1
  12: 66 220 495  792  924  792  495 220  66 12  1
  13: 78 286 715 1287 1716 1716 1287 715 286 78 13  1
... reformatted. - _Wolfdieter Lang_, Mar 20 2015
		

Crossrefs

Cf. A000295, A007318, A008292, A104713, A027641/A027642 (first Bernoulli numbers B-), A164555/A027642 (second Bernoulli numbers B+), A176327/A176289.

Programs

  • Magma
    /* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018
  • Mathematica
    t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)
  • PARI
    for(n=2, 10, for(k=2,n, print1(binomial(n,k), ", "))) \\ G. C. Greubel, May 15 2018
    

Formula

T(n,k) = binomial(n,k), for 2 <= k <= n.
From Peter Bala, Jul 16 2013: (Start)
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^3, x/(1 - x)).
O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t)) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)
As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
From Tom Copeland, Apr 11 2014: (Start)
A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!.
B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }
C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....
D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)
O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014
Reverse T = [St2]*dP*[St1]- dP = [St2]*(exp(x*M)-I)*[St1]-(exp(x*M)-I) with top two rows of zeros removed, [St1]=padded A008275 just as [St2]=A048993=padded A008277, dP= A132440, M=A238385-I, and I=identity matrix. Cf. A238363. - Tom Copeland, Apr 26 2014
O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015

Extensions

Edited and extended by David Wasserman, Jul 03 2007

A132710 Infinitesimal generator for a diagonally-shifted Lah matrix, unsigned A105278, related to n! Laguerre(n,-x,1).

Original entry on oeis.org

0, 2, 0, 0, 6, 0, 0, 0, 12, 0, 0, 0, 0, 20, 0, 0, 0, 0, 0, 30, 0, 0, 0, 0, 0, 0, 42, 0, 0, 0, 0, 0, 0, 0, 56, 0, 0, 0, 0, 0, 0, 0, 0, 72, 0, 0, 0, 0, 0, 0, 0, 0, 0, 90, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 110, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 132, 0
Offset: 0

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Author

Tom Copeland, Nov 15 2007, Nov 16 2007, Nov 27 2007

Keywords

Comments

Analogous to the infinitesimal generators of A132681 and A132792.
The matrix T begins
0;
2, 0;
0, 6, 0;
0, 0, 12 0;
0, 0, 0, 20, 0;
Along the nonvanishing diagonal the n-th term is (n+2)*(n+1).
Let LM(t) = exp(t*T) = lim_{n->infinity} (1 + t*T/n)^n.
Shifted Lah matrix = [bin(n+1,k+1)*(n)!/(k)! ] = LM(1) = exp(T). Truncating the series gives the n X n submatrices. In fact, the submatrices of T are nilpotent with [Tsub_n]^(n+1) = 0 for n=0,1,2,....
Inverse shifted Lah matrix = LM(-1) = exp(-T)
Umbrally shifted Lah[b(.)] = exp(b(.)*T) = [ binomial(n+1,k+1)*(n)!/(k)! * b(n-k) ]
A(j) = T^j / j! equals the matrix [binomial(n+1,k+1)*(n)!/(k)! * delta(n-k-j)] where delta(n) = 1 if n=0 and vanishes otherwise (Kronecker delta); i.e. A(j) is a matrix with all the terms 0 except for the j-th lower (or main for j=0) diagonal which equals that of the Lah matrix. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n+1,k+1) * (n)! / (k)! * d(n-k)].
For sequences with b(0) = 1, umbrally,
LM[b(.)] = exp(b(.)*T) = [ bin(n+1,k+1)*(n)!/(k)! * b(n-k) ] .
[LM[b(.)]]^(-1) = exp(c(.)*T) = [ bin(n+1,k+1)*(n)!/(k)! * c(n-k) ] where c = LPT(b) with LPT the list partition transform of A133314. Or,
[LM[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[LM(b(.))] = LM[LPT(b(.))] = LM[c(.)] .
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) = 0, b(n) = (n+1)*(n) * a(n-1),
2) B(x) = x * D^2 * x^2 A(x)
3) B(x) = x * 2 *Lag(2,-:xD:,0) A(x)
4) EB(x) = D * x^2 EA(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.
The exponentiated operator can be characterized (with loose notation) as
5) exp(t*T) * a = LM(t) * a = [sum(k=0,...,n) bin(n+1,k+1) * n!/k! t^(n-k) * a(k)] = [ t^n * n! * Lag(n,-a(.)/t,1) ], a vector array.
With t=1 and a(k) = (-x)^k, then LM(1) * a = [ n! * Laguerre(n,x,1) ], a vector array with index n .
6) exp(t*T) EA(x) = EB(x) = EA[ x / (1-x*t) ] / (1-x*t)^2

Programs

  • Mathematica
    Table[PadLeft[{n*(n-1), 0}, n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 30 2014 *)

Formula

Given a polynomial sequence p_n(x) with p_0(x)=1 and the lowering and raising operators L and R defined by L P_n(x) = n * P_(n-1)(x) and R P_n(x) = P_(n+1)(x), the matrix T represents the action of R*L^2*R^2 in the p_n(x) basis. For p_n(x) = x^n, L = D = d/dx and R = x. For p_n(x) = x^n/n!, L = DxD and R = D^(-1). - Tom Copeland, Oct 25 2012
Showing 1-4 of 4 results.