A112493 Triangle read by rows, T(n, k) = Sum_{j=0..n} C(n-j, n-k)*E2(n, j), where E2 are the second-order Eulerian numbers A201637, for n >= 0 and 0 <= k <= n.
1, 1, 1, 1, 4, 3, 1, 11, 25, 15, 1, 26, 130, 210, 105, 1, 57, 546, 1750, 2205, 945, 1, 120, 2037, 11368, 26775, 27720, 10395, 1, 247, 7071, 63805, 247555, 460845, 405405, 135135, 1, 502, 23436, 325930, 1939630, 5735730, 8828820, 6756750, 2027025, 1
Offset: 0
Examples
Triangle starts: [1] [1, 1] [1, 4, 3] [1, 11, 25, 15] [1, 26, 130, 210, 105] [1, 57, 546, 1750, 2205, 945] ... The e.g.f. of [0,0,1,7,25,65,...], the k=3 column of A008278, but with offset n=0, is exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!). Third row [1,4,3]: There are three plane increasing trees on 3 vertices. The number of colors are shown to the right of a vertex. ................................................... ....1o.(1+t)...........1o.t*(1+t).....1o.t*(1+t)... ....|................. /.\............/.\.......... ....|................ /...\........../...\......... ....2o.(1+t)........2o.....3o......3o....2o........ ....|.............................................. ....|.............................................. ....3o............................................. ................................................... The total number of trees is (1+t)^2 + t*(1+t) + t*(1+t) = 1+4*t+3*t^2 = R(2,t).
Links
- F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
- D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA], 2005.
- Wolfdieter Lang, First ten rows.
- MathOverflow, Recursion for row polynomials of A112493, (2025).
- Andrew Elvey Price and Alan D. Sokal, Phylogenetic trees, augmented perfect matchings, and a Thron-type continued fraction (T-fraction) for the Ward polynomials, arXiv:2001.01468 [math.CO], 2020.
Crossrefs
Programs
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Maple
T := (n, k) -> add(combinat:-eulerian2(n, j)*binomial(n-j, n-k), j=0..n): seq(seq(T(n, k), k=0..n), n=0..9); # Peter Luschny, Apr 11 2016
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Mathematica
max = 11; f[x_, t_] := -1 - (1 + t)/t*ProductLog[-t/(1 + t)*Exp[(x - t)/(1 + t)]]; coes = CoefficientList[ Series[f[x, t], {x, 0, max}, {t, 0, max}], {x, t}]* Range[0, max]!; Table[coes[[n, k]], {n, 0, max}, {k, 1, n - 1}] // Flatten (* Jean-François Alcover, Nov 22 2012, from e.g.f. *)
Formula
a(k, m) = 0 if k < m, a(k, -1):=0, a(0, 0)=1, a(k, m)=(m+1)*a(k-1, m) + (k+m-1)*a(k-1, m-1) else.
From Peter Bala, Sep 30 2011: (Start)
E.g.f.: A(x,t) = -1-((1+t)/t)*LambertW(-(t/(1+t))*exp((x-t)/(1+t))) = x + (1+t)*x^2/2! + (1+4*t+3*t^2)*x^3/3! + .... A(x,t) is the inverse function of (1+t)*log(1+x)-t*x.
A(x,t) satisfies the partial differential equation (1-x*t)*dA/dx = 1 + A + t*(1+t)*dA/dt. It follows that the row generating polynomials R(n,t) satisfy the recurrence R(n+1,t) =(n*t+1)*R(n,t) + t*(1+t)*dR(n,t)/dt. Cf. A054589 and A075856. The polynomials t/(1+t)*R(n,t) are the row polynomials of A134991.
The generating function A(x,t) satisfies the autonomous differential equation dA/dx = (1+A)/(1-t*A). Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the row generating polynomials R(n,t): R(n,t) counts plane increasing trees on n+1 vertices where the non-leaf vertices of outdegree k come in t^(k-1)*(1+t) colors. An example is given below. Cf. A006351, which corresponds to the case t = 1. Applying [Dominici, Theorem 4.1] gives the following method for calculating the row polynomials R(n,t): Let f(x) = (1+x)/(1-x*t). Then R(n,t) = (f(x)*d/dx)^n(f(x)) evaluated at x = 0. (End)
Sum_{j=0..n} T(n-j,j) = A000110(n). - Alois P. Heinz, Jun 20 2022
From Mikhail Kurkov, Apr 01 2025: (Start)
E.g.f.: B(y) = -w/(x*(1+w)) where w = LambertW(-x/(1+x)*exp((y-x)/(1+x))) satisfies the first-order ordinary differential equation (1+x)*B'(y) = B(y)*(1+x*B(y))^2, hence row polynomials are P(n,x) = P(n-1,x) + x*Sum_{j=0..n-1} binomial(n, j)*P(j,x)*P(n-j-1,x) for n > 0 with P(0,x) = 1 (see MathOverflow link).
Conjecture: row polynomials are P(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling1(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(j+k)/((n+j-k)!*(i-j)!*k!). (End)
Conjecture: g.f. satisfies 1/(1 - x - x*y/(1 - 2*x - 2*x*y/(1 - 3*x - 3*x*y/(1 - 4*x - 4*x*y/(1 - 5*x - 5*x*y/(1 - ...)))))) (see A383019 for conjectures about combinatorial interpretation and algorithm for efficient computing). - Mikhail Kurkov, Apr 21 2025
Extensions
New name from Peter Luschny, Apr 11 2016
Comments