%I
%S 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,
%T 28,56,70,56,28,8,1,9,36,84,126,126,84,36,9,1,10,45,120,210,252,210,
%U 120,45,10,1,11,55,165,330,462,462,330,165,55,11,1,12,66,220,495,792,924,792
%N Triangle read by rows, giving the numbers T(n,m) = binomial(n+1,m+1); or, Pascal's triangle A007318 with its lefthand edge removed.
%C T(n,m) is the number of mfaces of a regular nsimplex.
%C An nsimplex is the ndimensional 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 mplane contains more than (m + 1) of them. Such points are said to be in general position.
%C Reversing the rows gives A074909, which as a linear sequence is essentially the same as this.
%C From _Tom Copeland_, Dec 07 2007: (Start)
%C 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.
%C If T is the diagonallyshifted 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.
%C The infinitesimal matrix generator is given by A132681, so T = LM(1) of A132681 with inverse LM(1).
%C 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.
%C 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/(1x) ] / (1x)^2.
%C [O.g.f. for T ] = 1 / { [ 1  t * x/(1x) ] * (1x)^2 }. [ O.g.f. for row sums ] = 1 / { (1x) * (12x) }, giving A000225 (without a leading zero) for the row sums. Alternating sign row sums are all 1. (Sign correction noted by Vincent J. Matsko July 19 2015).
%C 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)
%C Given matrices A and B with A(n,k) = T(n,k)*a(nk) and B(n,k) = T(n,k)*b(nk), then A*B = C where C(n,k) = T(n,k)*[a(.)+b(.)]^(nk), 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
%C A007318*A097806 as infinite lower triangular matrices.  _Philippe Deléham_, Feb 08 2009
%C Riordan array (1/(1x)^2, x/(1x)).  _Philippe Deléham_, Feb 22 2012
%C The elements of the matrix inverse are T^(1)(n,k)=(1)^(n+k)*T(n,k).  _R. J. Mathar_, Mar 12 2013
%C Relation to Ktheory: 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
%C 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)^(p1)*sum(k=0,..,p1, T(p1,k)*(n2)^k) = [(n+1)^p(1)^p]/(n+2) = (1)^(p1)*sum(k=0,..,p1, (n1)^k). This is equal to (1)^(p1)*Phi(p,n1), 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
%C Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x1)^0 + A_1*(x1)^1 + A_2*(x1)^2 + ... + A_n*(x1)^n. This sequence gives A_0, ... A_n as the entries in the nth row of this triangle, starting at n = 0.  _Derek Orr_, Oct 14 2014
%C 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
%C From _Wolfdieter Lang_, Dec 10 2015: (Start)
%C A(r, n) = T(n+r2, r1) = risefac(n,r)/r! = binomial(n+r1, 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.
%C 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)
%H V. Buchstaber, <a href="http://www.mathnet.or.kr/mathnet/thesis_file/kaistbookupdated.pdf">Lectures on Toric Topology</a>, Trends in Mathematics  New Series, Information Center for Mathematical Sciences, Vol. 10, No. 1, 2008. pg. 7
%H Tom Copeland, <a href="http://mathoverflow.net/questions/82560/cyclotomicpolynomialsincombinatorics">Cyclotomic polynomials in combinatorics</a>
%H Tom Copeland, <a href="https://tcjpn.files.wordpress.com">Goin' with the Flow: Logarithm of the Derivative Operator</a> Part VI on simplices
%H D. Dugger, <a href="http://math.uoregon.edu/~ddugger/kgeom.pdf">A Geometric Introduction to KTheory</a> [From _Tom Copeland_, Apr 11 2014]
%H B. Grünbaum and G. C. Shephard, <a href="http://dx.doi.org/10.1112/blms/1.3.257">Convex polytopes</a>, Bull. London Math. Soc. (1969) 1 (3): 257300.
%H G. Hetyei, <a href="http://arxiv.org/abs/0909.4352">Meixner polynomials of the second kind and quantum algebras representing su(1,1)</a>, arXiv preprint arXiv:0909.4352 [math.QA], 2009, p. 4 (Added by Tom Copeland, Oct 01 2015).
%H Justin Hughes, <a href="http://www.groupsstandrews.org/2013/slides/Hughes.pdf ">Representations Arising from an Action on Dneighborhoods of Cayley Graphs</a>, 2013; slides from a talk.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Simplex">Simplex</a>
%F 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
%F 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
%F T(n,k) = 2*T(n1,k) + T(n1,k1)  T(n2,k)  T(n2,k1), 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
%F T(n,k) = A193862(n,k)/2^k.  _Philippe Deléham_, Jan 29 2014
%F G.f.: 1/((1x)*(1xx*y)).  _Philippe Deléham_, Mar 13 2014
%F From _Tom Copeland_, Mar 26 2014: (Start)
%F [From Copeland's 2007 and 2008 comments]
%F A) O.g.f.: 1 / { [ 1  t * x/(1x) ] * (1x)^2 } (same as Deleham's).
%F 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).
%F C) O.g.f. of row e.g.f.s: Sum_{n>=0} L(n,t,1) x^n = exp[t*x/(1x)]/(1x)^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+ ... .
%F D) E.g.f. of row o.g.f.s: ((1+t)*exp((1+t)*x)exp(x))/t (same as Bala's).
%F E) E.g.f. for T(n,k)*a(nk): {(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.
%F (End)
%F From _Tom Copeland_, Apr 28 2014: (Start)
%F With different indexing
%F A) O.g.f. by row: [(1+t)^n1]/t.
%F B) O.g.f. of row o.g.f.s: {1/[1(1+t)*x]  1/(1x)}/t.
%F C) E.g.f. of row o.g.f.s: {exp[(1+t)*x]exp(x)}/t.
%F These generating functions are related to row e.g.f.s of A111492. (End)
%F From _Tom Copeland_, Sep 17 2014:
%F A) U(x,s,t)= x^2/[(1t*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)
%F B) dU/dt=x^2 dU/dx with U(x,s,0)= x^2/(1s*x) (Buchstaber, 2008).
%F C) U(x,s,t) = exp(t*x^2*d/dx)U(x,s,0) = U(x/(1t*x),s,0).
%F 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)
%F 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
%F 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 selfconvolutions of the Bernoulli numbers, through the e.g.f. exp[NB(.,x;m)t] = [t/(e^t1)]^(m+1) * e^(xt). Norlund gave the relation to the factorials (x1)!/(x1n)! = (x1) ... (xn) = NB(n,x;n), so T(n,m) = NB(m+1,n+2;m+1)/(m+1)!.  _Tom Copeland_, Oct 01 2015
%e The triangle T(n, k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 ...
%e 0: 1
%e 1: 2 1
%e 2: 3 3 1
%e 3: 4 6 4 1
%e 4: 5 10 10 5 1
%e 5: 6 15 20 15 6 1
%e 6: 7 21 35 35 21 7 1
%e 7: 8 28 56 70 56 28 8 1
%e 8: 9 36 84 126 126 84 36 9 1
%e 9: 10 45 120 210 252 210 120 45 10 1
%e 10: 11 55 165 330 462 462 330 165 55 11 1
%e 11: 12 66 220 495 792 924 792 495 220 66 12 1
%e ... reformatted by _Wolfdieter Lang_, Mar 23 2015
%e Production matrix begins
%e 2 1
%e 1 1 1
%e 1 0 1 1
%e 1 0 0 1 1
%e 1 0 0 0 1 1
%e 1 0 0 0 0 1 1
%e 1 0 0 0 0 0 1 1
%e 1 0 0 0 0 0 0 1 1
%e 1 0 0 0 0 0 0 0 1 1
%e  _Philippe Deléham_, Jan 29 2014
%p for i from 0 to 12 do seq(binomial(i, j)*1^(ij), j = 1 .. i) od;
%t Flatten[Table[CoefficientList[D[1/x ((x + 1) Exp[(x + 1) z]  Exp[z]), {z, k}] /. z > 0, x], {k, 0, 11}]]
%o (PARI) for(n=0, 20, for(k=0, n, print1(1/k!*sum(i=0, n, (prod(j=0, k1, ij))), ", "))) \\ _Derek Orr_, Oct 14 2014
%Y Cf. A007318, A014410, A228196.
%Y Cf. Column sequences: A000027, A000217, A000292, A000332, A000389, A000579  A000582, A001287, A001288, A010965  A011001, A017713  A017764.
%K easy,nonn,tabl,changed
%O 0,2
%A _Zerinvary Lajos_, Dec 02 2007
%E Edited by _Tom Copeland_ and _N. J. A. Sloane_, Dec 11 2007
