cp's OEIS Frontend

Number of arrangements of {1, 2, ..., n, n + 1} containing the element 1. - Emeric Deutsch, Oct 11 2001

From Thomas Wieder, Oct 21 2004: (Start)

"Also the number of hierarchies with unlabeled elements and labeled levels where the levels are permuted.

"Let l_x denote level x, e.g. l_2 is level 2. Let * denote an element. Then l_1*l_2***l_3** denotes a hierarchy of n = 6 unlabeled elements with one element on level 1, three elements on level 2 and 2 elements on level 3.

"E.g. for n=3 one has a(3) = 11 possible hierarchies: l_1***, l_1**l_2*, l_1*l_2**, l_2**l_1*, l_2*l_1**, l_1*l_2*l_3*, l_3*l_1*l_2*, l_2*l_3*l_1*, l_1*l_3*l_2*, l_2*l_1*l_3*, l_3*l_2*l_1*. See A064618 for the number of hierarchies with labeled elements and labeled levels." (End)

Polynomials in A010027 evaluated at 2.

Also the permanent of any n X n cofactor of an n+1 X n+1 version of J+I other than an n X n version of J + I (that is, a (1, 2) matrix with n - 1 2s, at most one per row and column). - D. G. Rogers, Aug 27 2006

a(n) = number of partitions of [n+1] into lists of sets that are both non-nesting and non-crossing. Non-nesting means that no set is contained in the span (interval from min to max) of another. For example, a(1) counts 12, 1-2, 2-1 and a(2) counts 123, 1-23, 23-1, 3-12, 12-3, 1-2-3, 1-3-2, 2-1-3, 2-3-1, 3-1-2, 3-2-1. - David Callan, Sep 20 2007

Row sums of triangle A137594. - Gary W. Adamson, Jan 28 2008

From Peter Bala, Jul 10 2008: (Start)

a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). See A000522 for further properties of difference divisibility sequences.

a(n) equals the sum of the lengths of the paths between a pair of distinct vertices of the complete graph K_(n + 2) on n + 2 vertices [Hassani]. For example, for the complete graph K_4 with vertex set {A,B,C,D} the 5 paths between A and B are AB of length 1, ACB and ADB, both of length 2 and ACDB and ADCB, both of length 3. The sum of the lengths is 1 + 2 + 2 + 3 + 3 = 11 = a(2).

The number of paths between 2 distinct vertices of K_n is equal to A000522(n - 2); the number of simple cycles through a vertex of K_n equals A038154(n - 1).

Recurrence relation: a(0) = 1, a(1) = 3, a(n) = (n+2)*a(n - 1) - (n - 1)*a(n - 2) for n >= 2. The sequence b(n) := n*n! = A001563(n) satisfies the same recurrence with the initial conditions b(0) = 0, b(1) = 1. This leads to the finite continued fraction expansion a(n)/b(n) = 3 - 1/(4 - 2/(5 - 3/(6 - ... - (n - 1)/(n + 2)))), n >= 1.

Limit_{n->oo} a(n)/b(n) = e = 3 - 1/(4 - 2/(5 - 3/(6 - ... - n/((n + 3) - ...)))).

For n >= 1, a(n) = b(n)*(3 - Sum_{k=2..n} 1/(k!*(k - 1)*k)) (see the formula by Deutsch) since the rhs satisfies the above recurrence with the same initial conditions. Hence e = 3 - Sum_{k>=2} 1/(k!*(k - 1)*k).

For sequences satisfying the more general recurrence a(n) = (n + 1 + r)*a(n-1) - (n-1)*a(n-2), which yield series acceleration formulas for e/r! that involve the Poisson-Charlier polynomials c_r(-n; -1), refer to A000522 (r=0), A082030 (r=2), A095000 (r=3) and A095177 (r=4). (End)

Binomial transform of n! Offset 1. a(3) = 11. - Al Hakanson (hawkuu(AT)gmail.com), May 18 2009

Equals eigensequence of a triangle with (1, 2, 3, ...) as the right border and the rest 1's; equivalent to a(n) = [n terms of the sequence (1, 1, 1, ...) followed by (n + 1)] dot [(n + 1) terms of the sequence (1, 1, 3, 11, 245, ...)]. Example: 261 = a(4) = (1, 1, 1, 1, 5) dot (1, 1, 3, 11, 49) = 1 + 1 + 3 + 11 + 245 = 261. - Gary W. Adamson, Jul 24 2010

Number of nonnegative integers which use each digit at most once in base n+1. - Franklin T. Adams-Watters, Oct 02 2011

a(n) is the number of permutations of {1,2,...,n+2} in which there is an increasing contiguous subsequence (ascending run) beginning with 1 and ending with n+2. The number of such permutations with exactly k, 0<=k<=n, elements between the 1 and (n+2) is given by A132159(n,k) whose row sums equal this sequence. See example. - Geoffrey Critzer, Feb 15 2013

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016

Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...

Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007

From Tom Copeland, Nov 01 2007: (Start)

T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.

b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).

1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,

2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)

3) b(n) = Sum_{j=0..n} (n!/j!) a(j)

4) B(x) = (1-xDx)^(-1) A(x), formally

5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)

6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)

7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).

c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)

From Peter Bala, Jul 10 2008: (Start)

This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:

n\k|0.....................1...............2.......3......4

----------------------------------------------------------

0..|1.....................................................

1..|a....................1................................

2..|a(a+b)...............2a..............1................

3..|a(a+b)(a+2b).........3a(a+b).........3a........1......

4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1

...

The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.

Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.

When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).

We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).

An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).

Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)

(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009

Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019

The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009

T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011

T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011

T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013

For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016

See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016

Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016

The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017

Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017

If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

a(n) is also the sum of the positions of the right-to-left minima in all permutations of [n]. Example: a(3)=26 because the positions of the right-to-left minima in the permutations 123,132,213,231,312 and 321 are 123, 13, 23, 3, 23 and 3, respectively and 1 + 2 + 3 + 1 + 3 + 2 + 3 + 3 + 2 + 3 + 3 = 26. - Emeric Deutsch, Sep 22 2008

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=2) ~ exp(-x)/x^2*(1 - 5/x + 26/x^2 - 154/x^3 + 1044/x^4 - 8028/x^5 + 69264/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009

a(n) is the total number of cycles (excluding fixed points) in all permutations of [n+1]. - Olivier Gérard, Oct 23 2012; Dec 31 2012

A length n sequence is formed by randomly selecting (one-by-one) n real numbers in (0,1). a(n)/(n+1)! is the expected value of the sum of the new maximums in such a sequence. For example for n=3: If we select (in this order): 0.591996, 0.646474, 0.163659 we would add 0.591996 + 0.646474 which would be a bit above the average of a(3)/4! = 26/24. - Geoffrey Critzer, Oct 17 2013

Let M(t) = exp(t*T) = lim_{n->oo} (1 + t*T/n)^n.

Pascal matrix = [ binomial(n,k) ] = M(1) = exp(T), truncating the series gives the n X n submatrices.

Inverse Pascal matrix = M(-1) = exp(-T) = matrix for inverse binomial transform.

A(j) = T^j / j! equals the matrix [binomial(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 Pascal triangle. Hence the A(j)'s form a linearly independent basis for all matrices of the form [binomial(n,k) * d(n-k)] which include as a subset the invertible associated matrices of the list partition transform (LPT) of A133314.

For sequences with b(0) = 1, umbrally,

M[b(.)] = exp(b(.)*T) = [ binomial(n,k) * b(n-k) ] = matrices associated to b by LPT.

[M[b(.)]]^(-1) = exp(c(.)*T) = [ binomial(n,k) * c(n-k) ] = matrices associated to c, where c = LPT(b) . Or,

[M[b(.)]]^(-1) = exp[LPT(b(.))*T] = LPT[M(b(.))] = M[LPT(b(.))]= M[c(.)].

This is related to xDx, the iterated Laguerre transform and the general Euler transformation of a sequence through the comments in A132013 and A132014 and the relation [Sum_{k=0..n} binomial(n,k) * b(n-k) * d(k)] = M(b)*d, (n-th term). See also A132382.

If b(n,x) is a binomial type Sheffer sequence, then M[b(.,x)]*s(y) = s(x+y) when s(y) = (s(0,y),s(1,y),s(2,y),...) is an array for a Sheffer sequence with the same delta operator as b(n,x) and [M[b(.,x)]]^(-1) is given by the formulas above with b(n) replaced by b(n,x) as b(0,x)=1 for a binomial-type Sheffer sequence.

T = I - A132013 and conversely A132013 = I - T, which is the matrix representation for the iterated mixed order Laguerre transform characterized in A132013 (and A132014).

(I-T)^m generates the group [A132013]^m for m = 0,1,2,... discussed in A132014.

The inverse is 1/(I-T) = I + T + T^2 + T^3 + ... = [A132013]^(-1) = A094587 with the associated sequence (0!,1!,2!,3!,...) under the LPT.

And 1/(I-T)^2 = I + 2*T + 3*T^2 + 4*T^3 + ... = [A132013]^(-2) = A132159 with the associated sequence (1!,2!,3!,4!,...) under the LPT.

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 * a(n-1),

2) B(x) = xDx A(x)

3) B(x) = x * Lag(1,-:xD:) A(x)

4) EB(x) = x * EA(x) where D is the derivative w.r.t. x, (:xD:)^j = x^j*D^j and Lag(n,x) is the Laguerre polynomial.

So the exponentiated operator can be characterized as

5) exp(t*T) A(x) = exp(t*xDx) A(x) = [Sum_{n=0,1,...} (t*x)^n * Lag(n,-:xD:)] A(x) = [exp{[t*u/(1-t*u)]*:xD:} / (1-t*u) ] A(x) (eval. at u=x) = A[x/(1-t*x)]/(1-t*x), a generalized Euler transformation for an o.g.f.,

6) exp(t*T) EA(x) = exp(t*x)*EA(x) = exp[(t+a(.))*x], gen. Euler trf. for an e.g.f.

7) exp(t*T) * a = M(t) * a = [Sum_{k=0..n} binomial(n,k) * t^(n-k) * a(k)].

The umbral extension of formulas 5, 6 and 7 gives formally

8) exp[c(.)*T] A(x) = exp(c(.)*xDx) A(x) = [Sum_{n>=0} (c(.)*x)^n * Lag(n,-:xD:)] A(x) = [exp{[c(.)*u/(1-c(.)*u)]*:xD:} / (1-c(.)*u) ] A(x) (eval. at u=x) = A[x/(1-c(.)*x)]/(1-c(.)*x), where the umbral evaluation should be applied only after a power series in c is obtained,

9) exp[c(.)*T] EA(x) = exp(c(.)*x)*EA(x) = exp[(c(.)+a(.))*x]

10) exp[c(.)*T] * a = M[c(.)] * a = [Sum_{k=0..n} binomial(n,k) * c(n-k) * a(k)] .

The n X n principal submatrix of T is nilpotent, in particular, [Tsub_n]^(n+1) = 0, n=0,1,2,3,....

Note (xDx)^n = x^n D^n x^n = x^n n! (:Dx:)^n/n! = x^n n! Lag(n,-:xD:).

The operator xDx is an important, classical operator explored by among others Dattoli, Al-Salam, Carlitz and Stokes and even earlier investigators.

For a recent treatment of xDx, DxD and more general operators see the paper "Laguerre-type derivatives: Dobinski relations and combinatorial identities". - Karol A. Penson, Sep 15 2009

See Copeland's link for generalized Laguerre functions and connection to fractional differ-integrals in exercises through (:Dx:)^a/a!=(D^a x^a)/a!. - Tom Copeland, Nov 17 2011

From Tom Copeland, Apr 25 2014: (Start)

Conjugation or "similarity" transformations of [dP]=A132440 have an operator interpretation (cf. A074909 and A238363):

In general, select two operators A and B such that A^n = F1(n,B) and B^n = F2(n,A); then A^n = F1(n,F2(.,A)) and B^n = F2(n,F1(.,B)), evaluated umbrally, i.e., F1(n,F2(.,x))=F2(n,F1(.,x))=x^n, implying the polynomials F1 and F2 are an umbral compositional inverse pair.

One such pair are the Bell polynomials Bell(n,x) and falling factorials (x)_n with Bell(n,:xD:)=(xD)^n and (xD)_n=:xD:^n (cf. A074909). Another are the Laguerre polynomials LN(n,x)= n!*Lag(n,x) (A021009), which are umbrally self-inverse, with LN(n,-:xD:)=:Dx:^n and LN(n,:Dx:)= (-:xD:)^n with :Dx:^n=D^n*x^n.

Evaluating, for n>=0, the operator derivative d(B^n)/dA = d(F2(n,A))/dA in the basis B^n, i.e., with A^n finally replaced by F1(n,B), or A^n=F1(.,B)^n=F1(n,B), is equivalent to the matrix conjugation

A) [F2]*[dP]*[F1]

B) = [F2]*[dP]*[F2]^(-1)

C) = [F1]^(-1)*[dP]*[F1],

where [F1] is the lower triangular matrix with the n-th row the coefficients of F1(n,x) and analogously for [F2].

So, given the row vector Rv=(c0 c1 c2 c3 ...) and the column vector Cv(x)=(1 x x^2 x^3 ...)^Transpose, form the power series V(x)=Rv*Cv(x).

D) dV(B)/dA = Rv * [F2]*[dP]*[F1] * Cv(B).

E) With A=D and B=D, F1(n,x)=F2(n,x)=x^n and [F1]=[F2]=I. Then d(B^n)/dA = d(D^n)/dD = n * D^(n-1); therefore, consistently [F2]*[dP]*[F1] = [dP] and dV(D)/dD = Rv * [dP] * Cv(D). (End)

Let f(t) = u(t) - u(0) = Sum_{n>=1} u_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) * [ -2 (2') ] * t^2 / 2!

P(3,t) = (1')^(-5) * [ 12 (2')^2 - 6 (1')(3') ] * t^3 / 3!

P(4,t) = (1')^(-7) * [ -120 (2')^3 + 120 (1')(2')(3') - 24 (1')^2 (4') ] * t^4 / 4!

P(5,t) = (1')^(-9) * [ 1680 (2')^4 - 2520 (1') (2')^2 (3') + 360 (1')^2 (3')^2 + 720 (1')^2 (2') (4') - 120 (1')^3 (5') ] * t^5 / 5!

P(6,t) = (1')^(-11) * [ -30240 (2')^5 + 60480 (1') (2')^3 (3') - 20160 (1')^2 (2') (3')^2 - 20160 (1')^2 (2')^2 (4') + 5040 (1')^3 (3')(4') + 5040 (1')^3 (2')(5') - 720 (1')^4 (6') ] * t^6 / 6!

P(7,t) = (1')^(-13) * [ 665280 (2')^6 - 1663200 (1')(2')^4(3') + (1')^2 [907200 (2')^2(3')^2 + 604800 (2')^3(4')] - (1')^3 [60480 (3')^3 + 362880 (2')(3')(4') + 181440 (2')^2(5')] + (1')^4 [20160 (4')^2 + 40320 (3')(5') + 40320 (2')(6')] - 5040 (1')^5(7')] * t^7 / 7!

P(8,t) = (1')^(-15) * [ -17297280 (2')^7 + 51891840 (1')(2')^5(3') - (1')^2 [39916800 (2')^3(3')^2 + 19958400 (2')^4(4')] + (1')^3 [6652800 (2')(3')^3 + 19958400 (2')^2(3')(4') + 6652800 (2')^3(5')] - (1')^4 [1814400 (3')^2(4') + 1814400 (2')(4')^2 + 3628800 (2')(3')(5') + 1814400 (2')^2(6')] + (1')^5 [362880 (4')(5') + 362880 (3')(6') + 362880 (2')(7')] - 40320 (1')^6(8')] * t^8 / 8!

...

See A134685 for more information.

A111785 is obtained from A133437 by dividing through the bracketed terms of the P(n,t) by n! and unsigned A111785 is a refinement of A033282 and A126216. - Tom Copeland, Sep 28 2008

For relation to the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see the Loday and McCammond links. E.g., P(5,t) = (1')^(-9) * [ 14 (2')^4 - 21 (1') (2')^2 (3') + 6 (1')^2 (2') (4')+ 3 (1')^2 (3')^2 - 1 (1')^3 (5') ] * t^5 is related to the 3-D associahedron with 14 vertices (0-D faces), 21 edges (1-D faces), 6 pentagons (2-D faces), 3 rectangles (2-D faces), 1 3-D polytope (3-D faces). Summing over faces of the same dimension gives A033282 or A126216. - Tom Copeland, Sep 29 2008

A relation between this Lagrange inversion for an o.g.f. and partition polynomials formed from the "refined Lah numbers" A130561 is presented in the link "Lagrange a la Lah" along with umbral binary tree representations.

With f(x,t) = x + t*Sum_{n>=2} u_n*x^n, the compositional inverse in x is related to the velocity profile of particles governed by the inviscid Burgers's, or Hopf, eqn. See A001764 and A086810. - Tom Copeland, Feb 15 2014

Newton was aware of this power series expansion for series reversion. See the Ferraro reference p. 75 eqn. 52. - Tom Copeland, Jan 22 2017

The coefficients of the partition polynomials divided by the associated factorial enumerate the faces of the convex, bounded polytopes called the associahedra, and the absolute value of the sum of the renormalized 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 either n! (unnormalized) or unity (normalized). 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 u_1 = 1 and the other u_n replaced by suitably signed partition polynomials of A263633, the partition polynomials enumerating the refined noncrossing partitions of A134264 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Nov 16 2019

Relations between associahedra and oriented n-simplices are presented by Halvorson and by Street. - Tom Copeland, Dec 08 2019

Let f(x;t,n) = x - t * x^(n+1), giving u_1 = (1') = 1 and u_(n+1) = (n+1) = -t. Then inverting in x with t a parameter gives finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). Comparing this with the same result in A134264 gives relations between the faces of associahedra and noncrossing partitions (and other combinatorial constructs related to these inversion formulas and those listed in A145271). - Tom Copeland, Jan 27 2020

From Tom Copeland, Mar 24 2020: (Start)

There is a mapping between the faces of K_n, the associahedron of dimension (n-1), and polygon dissections. The dissecting noncrossing diagonals (i.e., nonintersecting in the interior) form subpolygons. Assign the indeterminate x_k to a subpolygon where k = (number of vertices of the subpolygon) - 1. Multiply the x_k together to form the monomials for the inversion formula.

For the 3-dimensional associahedron K_4, the fundamental polygon is the hexagon, which can be dissected into pentagons, associated to x_4; tetragons, to x_3; and triangles, to x_2; for example, there are six distinguished partitions of the hexagon into one triangle and one pentagon, sharing two vertices, associated to the monomial 6 * x_2 * x_4 since the unshared vertex of the triangle can be moved consecutively from one vertex of the hexagon to the next. This term corresponds to 720 (1')^2 (2') (4') / 5! in P(5,t) above, denumerating the six pentagonal facets of K_4. (End)

Let D=d/dx and [A,B]=A·B-B·A. Then each row corresponds to the coefficients of the operators :xD:^k = x^k D^k in the expansion of the commutator [log(D),:xD:^n]=[-log(x),:xD:^n]=sum(k=0 to n-1, a(n,k) :xD:^k). The e.g.f. is derived from [log(D), exp(t:xD:)]=[-log(x), exp(t:xD:)]= log(1+t)exp(t:xD:), using the shift property exp(t:xD:)f(x)=f((1+t)x).

The reversed unsigned array is A111492.

See the mathoverflow link and link therein to an associated mathstackexchange question for other formulas for log(D). In addition, R_x = log(D) = -log(x) + c - sum[n=1 to infnty, (-1)^n 1/n :xD:^n/n!]=

-log(x) + Psi(1+xD) = -log(x) + c + Ein(:xD:), where c is the Euler-Mascheroni constant, Psi(x), the digamma function, and Ein(x), a breed of the exponential integrals (cf. Wikipedia). The :xD:^k ops. commute; therefore, the commutator reduces to the -log(x) term.

Also the n-th row corresponds to the expansion of d[(xD)!/(xD-n)!]/d(xD) = d[:xD:^n]/d(xD) in the operators :xD:^k, or, equivalently, the coefficients of x in d[z!/(z-n)!]/dz=d[St1(n,z)]]/dz evaluated umbrally with z=St2(.,x), i.e., z^n replaced by St2(n,x), where St1(n,x) and St2(n,x) are the signed and unsigned Stirling polynomials of the first (A008275) and second (A008277) kinds. The derivatives of the unsigned St1 are A028421. See examples. This formalism follows from the relations between the raising and lowering operators presented in the MathOverflow link and the Pincherle derivative. The results can be generalized through the operator relations in A094638, which are related to the celebrated Witt Lie algebra and pseudodifferential operators / symbols, to encompass other integral arrays.

A002741(n)*(-1)^(n+1) (row sums), A002104(n)*(-1)^(n+1) (alternating row sums). Column sequences: A133942(n-1), A001048(n-1), A238474, ... - Wolfdieter Lang, Mar 01 2014

Add an additional head row of zeros to the lower triangular array and denote it as T (with initial indexing in columns and rows being 0). Let dP = A132440, the infinitesimal generator for the Pascal matrix, and I, the identity matrix, then exp(T)=I+dP, i.e., T=log(I+dP). Also, (T_n)^n=0, where T_n denotes the n X n submatrix, i.e., T_n is nilpotent of order n. - Tom Copeland, Mar 01 2014

Any pair of lowering and raising ops. L p(n,x) = n·p(n-1,x) and R p(n,x) = p(n+1,x) satisfy [L,R]=1 which implies (RL)^n = St2(n,:RL:), and since (St2(·,u))!/(St2(·,u)-n)!= u^n, when evaluated umbrally, d[(RL)!/(RL-n)!]/d(RL) = d[:RL:^n]/d(RL) is well-defined and gives A238363 when the LHS is reduced to a sum of :RL:^k terms, exactly as for L=d/dx and R=x above. (Note that R_x above is a raising op. different from x, with associated L_x=-xD.) - Tom Copeland, Mar 02 2014

For relations to colored forests, disposition of flags on flagpoles, and the colorings of the vertices of the complete graphs K_n, encoded in their chromatic polynomials, see A130534. - Tom Copeland, Apr 05 2014

The unsigned triangle, omitting the main diagonal, gives A211603. See also A092271. Related to the infinitesimal generator of A008290. - Peter Bala, Feb 13 2017

Let b(n) = LPT[ A001147 ] = -A001147(n-1) for n > 0 and 1 for n=0, where LPT represents the action of the list partition transform described in A133314.

Then T(n,k) = binomial(n,k) * b(n-k) .

Form the matrix of polynomials TB(n,k,t) = T(n,k) * t^(n-k) = binomial(n,k) * b(n-k) * t^(n-k) = binomial(n,k) * Pb(n-k,t),

beginning as

1;

-1, 1;

-1*t, -2, 1;

-3*t^2, -3*t, -3, 1;

-15*t^3, -12*t^2, -6*t, -4, 1;

-105*t^4, -75*t^3, -30*t^2, -10*t, -5, 1;

Let Pc(n,t) = LPT(Pb(.,t)).

Then [TB(t)]^(-1) = TC(t) = [ binomial(n,k) * Pc(n-k,t) ] = LPT(TB),

whose first column is

Pc(0,t) = 1

Pc(1,t) = 1

Pc(2,t) = 2 + t

Pc(3,t) = 6 + 6*t + 3*t^2

Pc(4,t) = 24 + 36*t + 30*t^2 + 15*t^3

Pc(5,t) = 120 + 240*t + 270*t^2 + 210*t^3 + 105*t^4.

The coefficients of these polynomials are given by the reverse of A102625 with the highest order coefficients given by A001147 with an additional leading 1.

Note this is not the complete matrix TC. The complete matrix is formed by multiplying along the diagonal of the lower triangular Pascal matrix by these polynomials, embedding trees of coefficients in the matrix.

exp[Pb(.,t)*x] = 1 + [(1-2t*x)^(1/2) - 1] / (t-0) = [1 + a finite diff. of [(1-2t*x)^(1/2)] with step t] = e.g.f. of the first column of TB.

exp[Pc(.,t)*x] = 1 / { 1 + [(1-2t*x)^(1/2) - 1] / t } = 1 / exp[Pb(.,t)*x) = e.g.f. of the first column of TC.

TB(t) and TC(t), being inverse to each other, are the generators of an Abelian group.

TB(0) and TC(0) are generators for a subgroup representing the iterated Laguerre operator described in A132013 and A132014.

Let sb(t,m) and sc(t,m) be the associated sequences under the LPT to TB(t)^m = B(t,m) and TC(t)^m = C(t,m).

Let Esb(t,m) and Esc(t,m) be e.g.f.'s for sb(t,m) and sc(t,m), rB(t,m) and rC(t,m) be the row sums of B(t,m) and C(t,m) and aB(t,m) and aC(t,m) be the alternating row sums.

Then B(t,m) is the inverse of C(t,m), Esb(t,m) is the reciprocal of Esc(t,m) and sb(t,m) and sc(t,m) form a reciprocal pair under the LPT. Similar relations hold among the row sums and the alternating sign row sums and associated quantities.

All the group members have the form B(t,m) * C(u,p) = TB(t)^m * TC(u)^p = [ binomial(n,k) * s(n-k) ]

with associated e.g.f. Es(x) = exp[m * Pb(.,t) * x] * exp[p * Pc(.,u) * x] for the first column of the matrix, with terms s(n), so group multiplication is isomorphic to matrix multiplication and to multiplication of the e.g.f.'s for the associated sequences (see examples).

These results can be extended to other groups of integer-valued arrays by replacing the 2 by any natural number in the expression for exp[Pb(.,t)*x].

More generally,

[ G.f. for M = Product_{i=0..j} B[s(i),m(i)] * C[t(i),n(i)] ]

= exp(u*x) * Product_{i=0..j} { exp[m(i) * Pb(.,s(i)) * x] * exp[n(i) * Pc(.,t(i)) * x] }

= exp(u*x) * Product_{i=0..j} { 1 + [ (1 - 2*s(i)*x)^(1/2) - 1 ] / s(i) }^m(i) / { 1 + [ (1 - 2*t(i)*x)^(1/2) - 1 ] / t(i) }^n(i)

= exp(u*x) * H(x)

[ E.g.f. for M ] = I_o[2*(u*x)^(1/2)] * H(x).

M is an integer-valued matrix for m(i) and n(i) positive integers and s(i) and t(i) integers. To invert M, change B to C in Product for M.

H(x) is the e.g.f. for the first column of M and diagonally multiplying the Pascal matrix by the terms of this column generates M. See examples.

The G.f. for M, i.e., the e.g.f. for the row polynomials of M, implies that the row polynomials form an Appell sequence (see Wikipedia and Mathworld). - Tom Copeland, Dec 03 2013

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.

Limit to which the columns of array A093966 converge.

Number of objects in all permutations of n objects taken 1,2,...,n at a time. Example: a(2)=6 because the permutations of {a,b} taken 1 and 2 at a time are: a,b,ab and ba, containing altogether 1+1+2+2=6 objects. a(n)=Sum(k*A008279(n,k),k=1..n). - Emeric Deutsch, Aug 16 2006

The number of sequences -where each member is an element in a set consisting of n elements- such that the last member is a repetition of a former member. Example: Set of possible members: {l,r}. Sequences such that the last member is a repetition of a former member: l,l; r,r; l,r,l; l,r,r; r,l,l; r,l,r. a(n)=Sum(k*A008279(n,k),k=1..n). [From Franz Fritsche (ff(AT)simple-line.de), Feb 22 2009]

The total number of elements in all ascending runs (including runs of length 1) over all permutations of {1,2,...,n}. a(2) = 6 because in the permutations [1,2] and [2,1] there are 4 runs of length 1 and 1 run of length 2. a(n) = Sum_{k>=1} A132159(n,k)*k. - Geoffrey Critzer, Feb 24 2014

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