A185415 Table of coefficients of a polynomial sequence of binomial type related to A080635.

1, 0, 1, 2, 0, 1, 0, 8, 0, 1, 18, 0, 20, 0, 1, 0, 148, 0, 40, 0, 1, 378, 0, 658, 0, 70, 0, 1, 0, 5040, 0, 2128, 0, 112, 0, 1, 14562, 0, 33992, 0, 5628, 0, 168, 0, 1, 0, 277164, 0, 158480, 0, 12936, 0, 240, 0, 1

Offset: 1

Author

Peter Bala, Jan 27 2011

Keywords

• nonn
• easy
• tabl

Comments

Define a sequence of polynomials P(n,x) by means of the recurrence relation

(1)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}

with starting value P(0,x) = 1. The first few polynomials are

P(1,x) = x

P(2,x) = x^2

P(3,x) = x*(x^2+2),

P(4,x) = x^2*(x^2+8),

P(5,x) = x*(x^4+20*x^2+18).

This triangle lists the coefficients of these polynomials in ascending powers of x. The triangle has links with A080635, which gives the number of ordered increasing 0-1-2 trees on n nodes (plane unary-binary trees in the notation of [BERGERON et al.]). The number of forests of k such trees on n nodes is given by the formula

... 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).

See A185422.

We also have A080635(n) = P(n,1), which can be used to calculate the terms of A080635 - see A185416.

For similarly defined polynomial sequences for other families of trees see A147309 and A185419. See also A185417.

Exponential Riordan array [(3/2)*(1-sqrt(3)*tan((Pi+3*sqrt(3)*x)/6))/(-1+2*sin((Pi-6*sqrt(3)*x)/6)), log((1/2)*(1+sqrt(3)*tan(sqrt(3)*x/2+Pi/6)))]. Production matrix is the exponential Riordan array [2*cosh(x)-1,x] beheaded. A185422*A008277^{-1}.

Examples

			Triangle begins:
  n\k|....1......2......3......4......5......6......7......8
  ==========================================================
  ..1|....1
  ..2|....0......1
  ..3|....2......0......1
  ..4|....0......8......0......1
  ..5|...18......0.....20......0......1
  ..6|....0....148......0.....40......0......1..
  ..7|..378......0....658......0.....70......0......1
  ..8|....0...5040......0...2128......0....112......0......1
		

References

• F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, in Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1922, pp. 24-48.

Links

• G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
• F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees

Crossrefs

Cf. A080635, A147309, A185417, A185419, A185422, A185423.

Programs

Maple

#A185415
P := proc(n,x)
description 'polynomial sequence P(n,x)'
if n = 0
return 1
else return
x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
end proc:
with(PolynomialTools):
for n from 1 to 10
CoefficientList(P(n,x),x);
end do;

Mathematica

p[0][x_] = 1; p[n_][x_] := p[n][x] = x*(p[n-1][x-1] - p[n-1][x] + p[n-1][x+1]) // Expand; row[n_] := CoefficientList[ p[n][x], x]; Table[row[n] // Rest, {n, 1, 10}] // Flatten (* Jean-François Alcover, Sep 11 2012 *)

Formula

GENERATING FUNCTION

The e.g.f. is

(1)... F(x, t) = E(t)^x = Sum_{n >= 0} P(n, x) * t^n/n!,

where

E(t) = 1/2+sqrt(3)/2*tan[sqrt(3)/2*t+Pi/6] = 1 + t + t^2/2! + 3*t^3/3! + 9*t^4/4! + ... is the e.g.f. for A080635.

ROW POLYNOMIALS

One easily checks that

... d/dt(F(x,t)) = x*(F(x-1,t)-F(x,t)+F(x+1,t))

and hence the row generating polynomials P(n,x) satisfy the recurrence relation

(2)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}.

RELATIONS WITH OTHER SEQUENCES

A080635(n) = P(n,1).

A185422(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).

A185423(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).