cp's OEIS Frontend

Triangle T(n, k) read by rows; given by [0, 1, 0, 1, 0, 1, ...] DELTA [1, 1, 1, 1, 1, 1, 1, 1, 1, ...] where DELTA is Deléham's operator defined in A084938.

For k>0, T(n, k) = binomial(n-1, k-1)*binomial(n+k, k)/(n+1); T(0, 0) = 1 and T(n, 0) = 0 if n > 0. [corrected by Marko Riedel, May 04 2023]

Sum_{k>=0} T(n, k)*2^k = A107841(n). - Philippe Deléham, May 26 2005

Sum_{k>=0} T(n-k, k) = A005043(n). - Philippe Deléham, May 30 2005

T(n, k) = A108263(n+k, k). - Philippe Deléham, May 30 2005

Sum_{k=0..n} T(n,k)*x^k = A000007(n), A001003(n), A107841(n), A131763(n), A131765(n), A131846(n), A131926(n), A131869(n), A131927(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Nov 05 2007

Sum_{k=0..n} T(n,k)*5^k*(-2)^(n-k) = A152601(n). - Philippe Deléham, Dec 10 2008

Sum_{k=0..n} T(n,k)*(-1)^k*3^(n-k) = A154825(n). - Philippe Deléham, Jan 17 2009

Umbrally, P(n,t) = Lah[n-1,-t*a.]/n! = (1/n)*Sum_{k=1..n-1} binomial(n-2,k-1)a_k t^k/k!, where (a.)^k = a_k = (n-1+k)!/(n-1)!, the rising factorial, and Lah(n,t) = n!*Laguerre(n,-1,t) are the Lah polynomials A008297 related to the Laguerre polynomials of order -1. - Tom Copeland, Oct 04 2014

T(n, k) = binomial(n, k)*binomial(n+k, k-1)/n, for k >= 0; T(0, 0) = 1 (see Kreweras, p. 21). - Michel Marcus, Nov 22 2014

P(n,t) = Lah[n-1,-:Dt:]/n! t^(n-1) with (:Dt:)^k = (d/dt)^k t^k = k! Laguerre(k,0,-:tD:) with (:tD:)^j = t^j D^j. The normalized Laguerre polynomials of 0 order are given in A021009. - Tom Copeland, Aug 22 2016

a(n,k) = n!/(Product_{j=1..n} e(n,k,j)!) with the exponent e(n,k,j) of j in the k-th partition of n in the A-St ordering of the partitions of n. Exponents 0 can be omitted due to 0!=1.

From Tom Copeland, Sep 18 2011: (Start)

Raising and lowering operators are given for the partition polynomials formed from A130561 in the Copeland link in "Lagrange a la Lah Part I" on pp. 22-23.

An e.g.f. for the partition polynomials is on page 3:

exp[t*:c.*x/(1-c.*x):] = exp[t*(c_1*x + c_2*x^2 + c_3*x^3 + ...)] where :(...): denotes umbral evaluation of the enclosed expression and c. is an umbral coefficient. (End)

From Tom Copeland, Sep 07 2016: (Start)

The row partition polynomials of this array P(n,x_1,x_2,...,x_n), given in the Lang link, are n! * S(n,x_1,x_2,...,x_n), where S(n,x_1,...,x_n) are the elementary Schur polynomials, for which d/d(x_m) S(n,x_1,...,x_n) = S(n-m,x_1,...,x_(n-m)) with S(k,...) = 0 for k < 0, so d/d(x_m) P(n,x_1,...,x_n) = (n!/(n-m)!) P(n-m,x_1,...,x_(n-m)), confirming that the row polynomials form an Appell sequence in the indeterminate x_1 with P(0,...) = 1. See p. 127 of the Ernst paper for more on these Schur polynomials.

With the e.g.f. exp[t * P(.,x_1,x_2,..)] = exp(t*x_1) * exp(x_2 t^2 + x_3 t^3 + ...), the e.g.f. for the partition polynomials that form the umbral compositional inverse sequence U(n,x_1,...,x_n) in the indeterminate x_1 is exp[t * U(.,x_1,x_2,...)] = exp(t*x_1) exp[-(x_2 t^2 + x_3 t^3 + ...)]; therefore, U(n,x_1,x_2,...,x_n) = P(n,x_1,-x_2,.,-x_n), so umbrally P[n,P(.,x_1,-x_2,-x_3,...),x_2,x_3,...,x_n] = (x_1)^n = P[n,P(.,x_1,x_2,...),-x_2,-x_3,...,-x_n]. For example, P(1,x_1) = x_1, P2(x_1,x_2) = 2 x_2 + x_1^2, and P(3,x_1,x_2,x_3) = 6 x_3 + 6 x_2 x_1 + x_1^3, then P[3,P(.,x_1,-x_2,...),x_2,x_3] = 6 x_3 + 6 x_2 P(1,x_1) + P(3,x_1,-x_2,-x_3) = 6 x_3 + 6 x_2 x_1 + 6 (-x_3) + 6 (-x_2) x_1 + x_1^3 = x_1^3.

From the Appell formalism, umbrally [P(.,0,x_2,x_3,...) + y]^n = P(n,y,x_2,x_3,...,x_n).

The indeterminates of the partition polynomials can also be extracted using the Faber polynomials of A263916 with -n * x_n = F(n,S(1,x_1),...,S(n,x_1,...,x_n)) = F(n,P(1,x_1),...,P(n,x_1,...,x_n)/n!). Compare with A263634.

Also P(n,x_1,...,x_n) = ST1(n,x_1,2*x_2,...,n*x_n), where ST1(n,...) are the row partition polynomials of A036039.

(End)

T(n,k) = C(n,k)*C(2*n-k,n+1)/n (0 <= k <= n-1).

G.f.: G(t,z) = (1-2*z-t*z-sqrt(1-4*z-2*t*z+t^2*z^2))/(2*(1+t)*z).

Equals N * P, where N = the Narayana triangle (A001263) and P = Pascal's triangle, as infinite lower triangular matrices. A126182 = P * N. - Gary W. Adamson, Nov 30 2007

G.f.: 1/(1-x-(x+xy)/(1-xy/(1-(x+xy)/(1-xy/(1-(x+xy)/(1-xy/(1-.... (continued fraction). - Paul Barry, Feb 06 2009

Let h(t) = (1-t)^2/(1+(u-1)*(1-t)^2) = 1/(u + 2*t + 3*t^2 + 4*t^3 + ...), then a signed (n-1)-th row polynomial of A126216 is given by u^(2n-1)*(1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=2. The power series expansion of h(t) is related to A181289 (cf. A086810). - Tom Copeland, Oct 09 2011

From Tom Copeland, Oct 10 2011: (Start)

With polynomials

P(0,t) = 0

P(1,t) = 1

P(2,t) = 1

P(3,t) = 2 + t

P(4,t) = 5 + 5 t + t^2

P(5,t) = 14 + 21 t + 9 t^2 + t^3

The o.g.f. A(x,t) = (1+x*t-sqrt((1-x*t)^2-4x))/(2(1+t)), and

B(x,t) = x - x^2/(1-t*x) = x - x^2 - ((t*x)^3 + (t*x)^4 + ...)/t^2 is the compositional inverse in x. [series corrected by Tom Copeland, Dec 10 2019]

Let h(x,t) = 1/(dB/dx) = (1-tx)^2/(1-(t+1)(2x-tx^2)) = 1/(1 - 2x - 3tx^2 + 4t^2x^3 + ...). Then P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A = exp(x*h(u,t)*d/du) u, evaluated at u=0, and dA/dx = h(A(x,t),t). (End)

From Tom Copeland, Dec 09 2019: (Start)

The polynomials in my 2011 formula entry above evaluate to an aerated, alternating sign sequence of the Catalan numbers A000108 with t = -2. The first few are P(2,-2) = 1, P(3,-2) = 0, P(4,t) = -1, P(5,-2) = 0, P(6,-2) = 2, P(7,-2) = 0, P(8,-2) = -5, P(9,-2) = 0, P(10,-2) = 14.

Generalizing the relations between w = theta and u = phi in Mizera on pp. 32-34, modify the inverse pair above to w = i * B(-i*u,t) = u + i * u^2/(1+i*t*u), where i is the imaginary number, and u = i*A(-i*w,t) = i*(1 - i*w*t - sqrt((1 + i*w*t)^2 + i*4*w))/(2(1+t)). Then the expression for V'(w) in Mizera generalizes to V'(w) = -i*(w - u) and reduces to V'(w) = (1 - sqrt(1-4 w^2))/2 when evaluated at t = -2, which is an o.g.f. for A126120. Cf. also A086810. (End)

Sum_{k = 0..n-1} (-1)^k*T(n,k)*binomial(x + 2*n - k, 2*n - k) = ( (x + 1) * ( Product_{k = 2..n} (x + k)^2 ) * (x + n + 1) )/(n!*(n + 1)!) for n >= 1. Cf. A243660 and A243661. - Peter Bala, Oct 08 2022

a(n) = A048996(n) * A036038(n);

a(n) = A036040(n) * factorial(A036043(n)).

A lowering operator for the unsigned multinomials in the brackets in the example is [d/du(1) 1/POP] where u(1) is treated as a continuous variable and POP is an operator that pulls off the number of parts of a partition ignoring u(0), e.g., [d/du(1) 1/POP][ u(0)u(2) + 2 u(1)^2 ] = (1/2) 2*2 u(1) = 2*u(1), analogous to the prototypical delta operator (d/dz) z^n = n z^(n-1). - Tom Copeland, Oct 04 2008

From the matrix formulation with M_m,k = 1/(m-k)!; g(x) = exp[ u(.) x]; an orthonormal vector basis x_1, ..., x_n and En(x^k) = x_k for k <= n and zero otherwise, for j=0 to n-1 the j-th signed row multinomial is given by the wedge product of x_1 with the wedge product (-1)^j * j! * u(0)^(-n) * Wedge{ En[x g(x), x^2 g(x), ..., x^(j) g(x), ~, x^(j+2) g(x), ..., x^n g(x)] } where Wedge{a,b,c} = a v b v c (the usual wedge symbol is inverted here to prevent confusion with the power notation, see Mathworld) and the (j+1)-th element is omitted from the product. Tom Copeland, Oct 06 2008 [Changed an x^n to x^(n-1) and "inner product of x_1" to "wedge". - Tom Copeland, Feb 03 2010]

For row n >= 1 the row polynomial in the variables g[1], ..., g[n] is P(n) = (1/(n+1)!)*(d^n/dx^n)(1/G(x)^(n+1))|{x=0}. P(0):=1. (d^k/dx^k)G(x)|{x=0} = k!*g[k], k>=1; G(0)=1.

a(n, k) is the coefficient in P(n) of g[1]^e(k, 1)*g[2]^e(k, 2)*..*g[n]^e(k, n) with the k-th partition of n written as [1^e(k, 1), 2^e(k, 2), ..., n^e(k, n)] in Abramowitz-Stegun order (e(k, j) >= 0; if e(k, j)=0 then j^0 is not recorded).

T(n,k) = (-1)^j*(n+j)!/((n+1)!*Product_{i>=1} s_i!), where (1*s_1 + 2*s_2 + ... = n) is the k-th partition of n and j = s_1 + s_2 ... is the number of parts. - Andrew Howroyd, Feb 01 2022

T(n,k) = Sum_{j=0..k} (-1)^j*2^(k-j)*binomial(k,j)*binomial(n+k-j-1, 2*k-1) (0 <= k <= n).

G.f.: G(t,x) = (1-x)^2/((1-x)^2 - t*x*(2-x)).

G.f. of column k = x^k*(2-x)^k/(1-x)^{2k} (k>=1) (we have a Riordan array).

Recurrences satisfied by the numbers u_{n,k}=T(n,k) can be found in the Castiglione et al. reference.

Sum_{k=0..n} k*T(n,k) = A181290(n).

T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=0, T(1,1)=2, T(2,0)=0, T(1,1)=3, T(2,2)=4, T(n,k)=0, if k < 0 or if k > n. - Philippe Deléham, Nov 29 2013

a(n, m) = binomial(n+1-m, m)*C(n-m) if 0 <= m <= floor((n+1)/2), otherwise 0, with C(n) := A000108(n) (Catalan).

G.f. for column m=1, 2, ...: (x^(2*m-1))*C(m-1)/(1-4*x)^((2*m-1)/2); m=0: c(x), g.f. for A000108 (Catalan).

G.f. for row polynomials p(n, x): c(z) + x*z*c(x*(z^2)/(1-4*z))/sqrt(1-4*z) = (1-sqrt(1-4*z*(1+x*z)))/(2*z), where c(x) is the g.f. of A000108 (Catalan).

G.f. for triangle: (1 - sqrt(1 - 4*x (1 + y*x)))/(2*x). - Geoffrey Critzer, Jul 24 2020

The alternating row sums are {1,repeat 0} = {A000007(n)}{n>=0}. From the second comment p(n, x=-1) = K(1,0; n), with g.f. K(1,0; x) = 1. Or from the g.f. of the triangle with y = -1. Thanks to Aaron Li for asking for a proof. - _Wolfdieter Lang, Jan 21 2023

The series expansion of f(x) = (1 + 2sx - sqrt(1 + 4sx + 4d^2x^2))/(2x) at x = 0 is (s^2 - d^2) x + (2 d^2s - 2 s^3) x^2 + (d^4 - 6 d^2 s^2 + 5 s^4) x^3 + (-6 d^4 s + 20 d^2 s^3 - 14 s^5) x^4 + ..., containing the coefficients of this array. With s = (a+b)/2 and d = (a-b)/2, then f(x)/ab = g(x) = (1 + (a+b)x - sqrt((1+(a+b)x)^2 - 4abx^2))/(2abx) = x - (a + b) x^2 + (a^2 + 3 a b + b^2) x^3 - (a^3 + 6 a^2 b + 6 a b^2 + b^3) x^4 + ..., containing the Narayana polynomials of A001263, which can be simply transformed into A033282. The compositional inverse about the origin of g(x) is g^(-1)(x) = x/((1-ax)(1-bx)) = x/((1-(s+d)x)(1-(s-d)x)) = x + (a + b) x^2 + (a^2 + a b + b^2) x^3 + (a^3 + a^2 b + a b^2 + b^3) x^4 + ..., containing the complete homogeneous symmetric polynomials h_n(a,b) = (a^n - b^n)/(a-b), which are the polynomials of A034867 when expressed in s and d, e.g., ((s + d)^7 - (s - d)^7)/(2 d) = d^6 + 21 d^4 s^2 + 35 d^2 s^4 + 7 s^6. A133437 and A134264 for compositional inversion of o.g.f.s can be used to relate the sets of polynomials above. - Tom Copeland, Nov 28 2023

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

See the fhat[n] formula explained above, and the W. Lang link for more details.

Denote this set of partition polynomials by [RT], the permutahedra polynomials of A133314 by [P], the refined Eulerian polynomials of A145271 by [E], and the Lagrange inversion polynomials of A134685 for e.g.f.s by [L]. Let the typically noncommutative product of two sets, e.g., [P][E], represent the substitution of the polynomials of [E] for the indeterminates of [P], i.e., a composition at the level of the indeterminates (see A356145 for examples). Let [I] be the substitutional identity transformation, and mark the substitutional inverse with the superscript -1. Then the following relations hold.

[RT] = [P][E] = [P][L][P] = [P]^{-1}[L][P] = [P][L][P]^{-1} since [P] is an involution, i.e., [P]^2 = [I], or [P] = [P]^{-1}, so [RT] and [L] are conjugate duals.

[RT]^{-1} = ([P][E])^{-1} = [E]^{-1}[P] = ([P][L][P])^{-1} = [P][L][P] = [RT], with [E]^{-1} = A356145, since [L] and [P] are involutions, so is [RT], i.e., [RT]^2 = [I].

RT_n(a_1,a_2,...,a_n) = D_{x=0}^n 1 / [ D_x f^{(-1)}(x)] for which D_x is the derivative w.r.t. x and the indeterminates are defined by 1 / [D_x f(x)] = 1 + a_1 x + a_2 x^2/2! + a_3 x^3/3! + ... with f(x) and f^{(-1)}(x) a compositional inverse pair of formal Taylor series, or e.g.f.s. This is the analytic equivalent of the algebraic relation [RT] = [P][E]. In words, the partition polynomials of row n (initial row is 0) is the n-th coefficient of the formal Taylor series of the reciprocal of the derivative of the compositional inverse of a function in terms of the Taylor series coefficients of the reciprocal of the derivative of that function. Note the correspondence with the analytic interpretation of [E]^{-1} of A356145, consistent with the algebraic identities above.

RT_n(a_1,a_2,...,a_n) = D_{x=0}^n f'(f^{(-1)}(x)) also, by the inverse function theorem, where the prime denotes differentiation with respect to the argument of the function.

With all a_k = (-1)^k, RT_0 = RT_1 = 1, otherwise RT_n = 0. This is determined with f(x) = e^{x}-1 and f^{(-1)}(x) = log(1+x).

With all a_k = 1, RT_0 = 1, RT_1 = -1, otherwise RT_n=0. This is determined with f(x) =1-e^{-x} and f^{(-1)}(x) = -log(1-x).

With all a_k = -1, RT_0 = 1 and RT_n = 2^(n-1) otherwise. This is determined with f(x) = (x - log(2-e^x))/2 and f^{(-1)}(x) = x - log(cosh(x)). (Careful, these are not the row sums of the absolute values of the numerical coefficients, which for the first ten polynomials are 1, 1, 2, 4, 8, 18, 40, 122, 446, and 2428.)

With a_k = k! 2^k, RT_0 = 1 and RT_n = -2*(2(n-1))! / (n-1)! = -2*n!*A000108(n-1) otherwise. This is determined with f(x) = x - x^2 and f^{(-1)}(x) = (1 - sqrt(1-4x))/2. Similar relations hold for the Fuss-Catalan sequences with f(x) = x - x^{m+1} for m > 1.