cp's OEIS Frontend

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

N(J(n,r)) = C(n,r)*S(n,r+1)*r! where S(n, r + 1) is a Stirling number of the second kind (given by A048993 with zeros removed); generating function = (x+1)^(n-1).

From Peter Bala, Oct 22 2008: (Start)

Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).

Let f(x) = 1 + a*x + a*x^2/2! + a*x^3/3! + ... . Then the e.g.f. for this table is I(f(x)) = 1 + a*x +(a + 2*a^2)*x^2/2! + (a + 9*a^2 + 6*a^3)*x^3/3! + (a + 28*a^2 + 72*a^3 + 24*a^4)*x^4/4! + ... . Note, if we take f(x) = 1 + a*x + a*x^2 + a*x^3 + ... then I(f(x)) is the o.g.f. of the Narayana triangle A001263. (End)

A generator for this array is given by the inverse, g(x,t), of f(x,t)= x/(1 + t * (e^x-1)). Then A248927 gives h(x,t)= x / f(x,t) = 1 + t*(e^x-1)= 1 + t * (x + x^2/2! + x^3/3! + ...) and g(x,t)= x * (1 + t * x + (t + 2 t^2) * x^2/2! + (t + 9 t^2 + 6 t^3) * x^3/3! + ...), so by Bala's arguments A248927 is a refinement of A141618 with row sums A000272. The connection to Narayana numbers is reflected in the relation between A248927 and A134264. See A145271 for more relations that g(x,t) and f(x,t) must satisfy. - Tom Copeland, Oct 17 2014

T(n,k) = C(n,k-1) * A028246(n,k) = C(n,k-1) * A019538(n,k)/k = A055302(n+1,n+1-k) / (n+1). - Tom Copeland, Oct 25 2014

E.g.f. is the series reversion of log(1 + x)/(1 + t*x) with respect to x. Cf. A198204. - Peter Bala, Oct 21 2015

From Peter Bala, Aug 25 2011: (Start)

The function sl(x) satisfies the differential equation sl''(x) = -2*sl^3(x) with initial conditions sl(0) = 0, sl'(0) = 1.

Recurrence relation:

a(n+2) = -2*sum {i+j+k = n} n!/(i!*j!*k!)*a(i)*a(j)*a(k).

The inverse of the sine lemniscate function may be defined as the algebraic integral

sl^(-1)(x) := Integral_{s=0..x} 1/sqrt(1-s^4) ds = x + x^5/10 + x^9/24 + 5*x^13/208 + ....

Series reversion produces the expansion

sl(x) = x - 12*x^5/5! + 3024*x^9/9! - 4390848*x^13/13! + ....

The coefficients in this expansion can be calculated using nested derivatives as follows (see [Dominici, Theorem 4.1]): Let f(x) = sqrt(1-x^4). Define the nested derivative D^n[f](x) by means of the recursion

D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0.

The coefficients in the expansion of D^n[f](x) in powers of f(x) are given in A145271. Then we have a(n) = D^(n-1)[f](0).

a(n) is divisible by 12^n and a(n)/12^n produces (a signed and aerated version of) A144853(n).

(End)

The function sl(x) satisfies the differential equation sl'(x)^2 + sl(x)^4 = 1 with initial conditions sl(0) = 0, sl'(0) = 1. - Michael Somos, Oct 12 2019

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.

If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.

T(j,k) is (j-1)! P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).

For example, from g(t) above, T(6,3) = 5! * [6!/(3!*2!)]/(2!)^2 = 1800 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = 5! * [6!/4!]/(2!*3!) = 300.

If the initial factorial and final denominator of T(n,k) are removed and the expression divided by j and the partitions reversed in order, then A134264 is obtained, a refinement of the Narayana numbers.

For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums are A000169(n) = n^(n-1) = n* A000272(n).

Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/[d{x/[h_0+h_1*x+ ...]/dx]. Then the partition polynomials above are given by (W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t)=exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)). See A145271.

With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*W(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 * W(d/dt) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).

Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)

Following the notes connected to the Lagrange reversion theorem in A248927, a generator for the n-th partition polynomial P_n of this entry is (d/dx)^(n-1) (h (x))^n, and -log(1-t*P.) = (t*Q.) / (1 - t*Q.), umbrally, where (Q.)^n = Q_n is the n-th partition polynomial of A248927. - Tom Copeland, Nov 25 2016

With h_0 = 1, the n-th partition polynomial is obtained as the n-th element (with initial index 0) of the first column of M^{n+1}, where M is the matrix with M_{i,j}= binomial(i,j) h_{i-j}, i.e., the lower triangular Pascal matrix with its n-th diagonal multiplied by h_n. This follows from the Lagrange inversion theorem and the relation between powers of matrices such as M and powers of formal Taylor series discussed in A133314. This is equivalent to repeated binomial convolution of the coefficients of the Taylor series with itself. - Tom Copeland, Nov 13 2019

T(n,k) = n*A248927(n,k). - Andrew Howroyd, Feb 02 2022

E.g.f.: ( tanh(arctanh(sqrt(2)) - sqrt(2)*x) )/sqrt(2) = sqrt(2)/2* (1 + (3-2*sqrt(2))* exp(2*sqrt(2)*x) )/( 1 - (3-2*sqrt(2))* exp(2*sqrt(2)*x) ).

Recurrence: a(n+1) = 2*(Sum_{k=0..n} binomial(n,k)*a(k)*a(n-k)) - 0^n.

a(2*n) = 2^n * A006154(2*n), n>0 (conjectured). - Ralf Stephan, Apr 29 2004

For n>0, a(n) = sqrt(2)^(3*n+1)*Sum_{k>=0} k^n/(1+sqrt(2))^(2*k). - Benoit Cloitre, Jan 12 2005

From Peter Bala, Jan 30 2011: (Start)

A finite sum equivalent to the previous formula of Benoit Cloitre is

a(n) = (2*sqrt(2))^(n-1)*Sum_{k = 1..n} k!*Stirling2(n,k)*w^(k-1), for n >= 1, with w = (sqrt(2) - 1)/2.

This formula can be used to prove congruences for a(n). For example, a(p) == (-1)^((p^2-1)/8) (mod p) for odd prime p.

For similar formulas for labeled plane and non-plane unary-binary trees see A080635 and A000111 respectively.

For a sequence of related polynomials see A185419. For a recursive table to calculate a(n) see A185420.

The e.g.f. A(x) satisfies the autonomous differential equation d/dx (A(x)) = 2*A(x)^2 - 1. (End)

From Peter Bala, Aug 26 2011: (Start)

The inverse function A(x)^(-1) of the generating function A(x) satisfies A(x)^(-1) = Integral_{t = 1..x} 1/(2*t^2 - 1) dt.

Let f(x) = 2*x^2 - 1. Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0 (see A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x)). Then by [Dominici, Theorem 4.1] we have a(n+1) = D^n[f](1).

For n >= 1 we have a(n) = (2 + sqrt(2))^(n-1)*A(n, 3 - 2*sqrt(2)), where {A(n, x)}n>=1 = [1, 1 + x, 1 + 4*x + x^2, 1 + 11*x + 11*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials (see A008292).

a(n+1) = (-1)^n*(sqrt(-2))^n * R(n, sqrt(-2)) where R(n, x) are the polynomials defined in A185896 (derivative polynomials associated with the function sec^2(x)). (End)

G.f.: 1 + x/G(0) where G(k) = 1 - 4*x*(k+1) - 2*x^2*(k+1)*(k+2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013

G.f.: 1 + x/(G(0) -x), where G(k) = 1 - x*(k+1) - 2*x*(k+1)/(1 - x*(k+2)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013

E.g.f.: sqrt(2)*( -1/2 + (3+2*sqrt(2))/(4 + 2*sqrt(2)- E(0) )), where E(k) = 2 + 2*sqrt(2)*x/( 2*k+1 - 2*sqrt(2)*x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 27 2013

a(n) ~ n! * 2^((3*n+1)/2) / (log(3+2*sqrt(2)))^(n+1). - Vaclav Kotesovec, Feb 25 2014

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.

If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.

T(j,k) is [(j-1)!/j]* P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).

For example, from g(t) above, T(6,3) = [5!/6][6!/(3!*2!)]/(2!)^2 = 300 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = [5!/6][6!/4!]/(2!*3!) = 50.

If the initial factorial and final denominator are removed and the partitions reversed in order, A134264 is obtained, a refinement of the Narayana numbers.

For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums of A248927 are A000169(n)/n = n^(n-2) = A000272(n).

Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/{d[x/[h_0+h_1*x+ ...]]/dx}. Then the partition polynomials above are given by (1/n)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t)= exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)). See A145271.

With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*W(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 * W(d/dt) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).

Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)

As noted in A248120 and A134264, this entry is given by the Hadamard product by partition of A134264 and A036038. For example, (1,4,2,6,1)*(1,4,6,12,24) = (1,16,12,72,24). - Tom Copeland, Nov 25 2016

T(n,k) = ((n-1)!)^2/((n-j)!*Product_{i>=1} s_i!*(i!)^s_i), where (1*s_1 + 2*s_2 + ... = n-1) is the k-th partition of n-1 and j = s_1 + s_2 ... is the number of parts. - Andrew Howroyd, Feb 02 2022

t(n,0)=n!; t(n,k)=tr(n,k)+tr(n,k-1), k<=n/2; t(n,floor((n+1)/2)-1)=tr(n,floor((n+1)/2)-1); tr(n,i)=((sum(j=0..2*i, binomial(j+n-2*i-1,n-2*i-1)*(j+n-2*i)!*2^(2*i-j)*(-1)^(j-i)*Stirling2(n,j+n-2*i)))). - Vladimir Kruchinin, May 27 2011

From Tom Copeland, Sep 30 2015: (Start)

Reversed rows signed and aerated are generated by [(1-x^2)D]^n x with D = d/dx, so exp(t(1-x^2)D) x = tanh(t + atanh(x)) is the e.g.f. of this reversed array (see A145271).

Reversed rows unsigned and aerated are generated by [(1+x^2)D]^n x, so exp(t(1+x^2)D) x = tan(t + atan(x)) = x + (1 +x^2)*t + (2x + 2x^3)*t^2/2! + (2 + 8x^2 + 6x^4)*t^3/3! + (16x + 40x^3 + 24x^5)*t^4/4! + ... is the e.g.f. for the matrix on p. 666 of the Knuth and Buckholtz link.

E.g.f. for this entry's aerated array 1 + (1 + x^2)*t + (2 + 2x^2)*t^2/2! + (6 + 8x^2 + 2x^4)*t^3/3! + (24 + 40^x^2 + 16x^4)*t^4/4! + ... = x * tan(t*x + atan(1/x)). (End)

From Fabián Pereyra, Apr 22 2022: (Start)

T(n,k) = (n-2k)*T(n-1,k) + (n-2k+2)*T(n-1,k-1).

E.g.f.: A(x,t) = sqrt(t)*(sqrt(t)*sin(x*sqrt(t))+cos(x*sqrt(t)))/ (sqrt(t)*cos(x*sqrt(t))-sin(x*sqrt(t))). (End)

T(n,k) = k*(2*n-k+1)!/[2^(n-k+1)*(n-k+1)!] (1 <= k <= n+1).

From Tom Copeland, Nov 11 2007: (Start)

Bivariate G.F.: exp[P(.,t)*x] = D_x {1 - [g(x)/(1+t*g(x))]} = 1 / {(1+g(x))*[1+t*g(x)]^2}, where g(x) = sqrt(1-2*x) - 1 and P(n,t) = Sum_{k=0..n} T(n,k) * t^k.

Also D_x g(x) = -(1-2*x)^(-1/2) = -exp[x*A001147(.)] = -exp[x *(2*(.)-1)!! ], so the coefficients of x^n/n! in the expansion of g(x) are -(2*(n-1)-1)!! = -A001147(n-1) for n > 0.

See A132382 for an array which is essentially the revert from which this G.f. may be derived and for connections to other arrays. (End)

E.g.f.: 1/(1 - x + x*sqrt(1-2*z)) = 1 + x*z + (x+2*x^2)*z^2/2! + (3*x+6*x^2+6*x^3)*z^3/3! + .... T(n,k) gives the number of plane recursive trees on n+2 nodes where the root has degree k (Bergeron et al., Corollary 5). - Peter Bala, Jul 09 2012

From Peter Bala, Jul 09 2014: (Start)

T(n,k) = k!*A001497(n,k) modulo offset differences.

The n-th row polynomial R(n,x) = (-1)^n/(x - 1)*( Sum_{k = 1..infinity} k*(k - 2)*...*(k - 2*n)*(x/(x - 1))^k ). Cf. the Dobinski-type formula for the row polynomials of A001497. (End)

From Tom Copeland, Aug 06 2016: (Start)

From the 2007 formulas above, an alternate g.f. for this entry is GF(x,t) = -g(x) / [1 + t*g(x)] = x + (1 + 2*t)*x^2/2! + (3 + 6*t + 6*t^2)*x^3/3! + ... with compositional inverse GFinv(x,t) = {1 - [1 - x / (1+t*x)]^2} / 2 = -(1/2)[x / (1+t*x)]^2 + x / (1+t*x) = Sum_{n>0} (-1)^(n+1) [(n-1)/2*t^(n-2) + t^(n-1)]*x^n, a series containing the Lah numbers A001286 when expressed as an e.g.f.

From A145271, with K(x,t) = 1 / dGinv(x,t)/dx = 1 + (1+2*t) x + (1+t+t^2) x^2 + x^3 / [1-(1-t)*x], then [K(x,t) d/dx]^n x evaluated at x=0 gives the n-th row polynomial of this entry.

Since the reciprocal of Bala's e.g.f. above generates a shifted, signed A001147, for the polynomials P(n,t) generated by Bala's e.g.f., umbrally (P(.,t) + a.)^n = 0 for n > 0 with a_0 = 1 and a_n = -t * A001147(n-1) for n > 0. E.g., (P(.,t) + a.)^2 = a_0 * P(2,t) + 2 a_1 * P(1,t) + a_2 * P(0,t) = 1 * (t + 2*t^2) + 2 * -t * t + -t * 1 = 0. (End)

From Peter Bala, Apr 16 2017: (Start)

T(n,k) = k*T(n-1,k-1) + (2*n - k)*T(n-1,k).

E.g.f.: t*x*c(x/2)/(1 - t*x*c(x/2)) = t*x + (t + 2*t^2)*x^2/2! + (3*t + 6*t^2 + 6*t^2)*x^3/3! + ..., where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. for the Catalan numbers A000108. Note that the related g.f. t*x*c(x)/(1 - t*x*c(x)) is the o.g.f. for A033184 (essentially the same as the Riordan array A106566) and enumerates the number of vertices labeled k on the n_th level of the Catalan tree (k >= 1, n >= 0). (End)

T(n,k) = 2^k*binomial(n+1,k)binomial(n+1-k,(n-k)/2)/(n+1) if n-k is even; otherwise, T(n,k) = 0. G.f. G=G(t,z) satisfies G=1+2tzG+z^2*G^2.

T(n,k) = 2^k*A097610(n,k). - Philippe Deléham, Aug 17 2006

From Tom Copeland, Feb 09 2016: (Start)

The following is from the formalism in A097610 with h1 = 2t, h2 = 1, and MT(n,h1,h2) = MT(n,2t,1) and with OG(x,t) defined above.

E.g.f.: M(x,t) = e^(2tx) AC(x) = exp[x MT(.,2t,1)] = exp[x P(.,t)], where AC(x) = I_1(2x)/x = Sum_{n>=0} x^(2n)/(n!(n+1)!) = exp(c.x) is the e.g.f. of A126120.

P(n,t) = MT(n,2t,1) = (c. + 2t)^n = Sum_{k=0..n} binomial(n,k) c(n-k) (2t)^k with c(k) = A126120(k). P(n,t+s) = (c. + 2t + 2s)^n = (P(.,t) + 2s)^n.

P(n,t) = t^n FC(n,c./t) = t^n (2 + c./t)^n, where FC(n,t) = (2 + t)^n are the face polynomials (vectors) of the hypercubes of A038207, i.e., the row polynomials of this entry can be obtained as the umbral composition of the reverse face polynomials with the aerated Catalan numbers of A000108.

The lowering and raising operators for the row polynomials P(n,t) of this entry are L = (1/2) d/dt = (1/2) D and R = 2t + dlog{AC(L)}/dL = 2t + Sum_{n>=0} b(n) L^(2n+1)/(2n+1)! = 2t + L - L^3/3! + 5 L^5/5! - ... with b(n) = (-1)^n A180874(n+1).

Let CP(n,t) = P(n+1,t) with CP(0,t) = 0. Then the infinitesimal generator for CP(n,t) is g(x) d/dx with g(x) = 1 /[dOGinv(x,t)/dx] = x^2 / [(OGinv(x,t))^2 (1 - x^2)] = (1 + 2t x + x^2)^2 / (1 - x^2) so that [g(x)d/dx]^n/n! x evaluated at x = 0 gives the row polynomial CP(n,t), i.e., exp[x g(u)d/du] u |_(u=0) = OG(x,t) = 1 /[1 - x P(.,t)]. Cf. A145271.

g(x) = 1 + 4t x + (3+4t) x^2 + 8t x^3 + 4(1+t^2) x^4 + 8t x^5 + 4(1+t^2) x^6 + 8t x^7 + ... has the repeating coefficients of the vector V = (1, 4t, 3+4t, 8t, 4(1+t^2), 8t, 4(1+t^2), 8t, ...). Form the lower triangular matrix U with all ones on the diagonal and below. Multiply the n-th diagonal of U by V(n), giving the matrix VU with VU(n,k) = V(n-k). Then (1,0,0,0,..) [VU * DM]^n/n! (0,1,0,0,..)^T = CP(n,t) = P(n-1,t) for n>0 with DM being the matrix A218272 representing differentiation of a power series.

(End)