A131178 Non-plane increasing unary binary (0-1-2) trees where the nodes of outdegree 1 come in 2 colors.
1, 2, 5, 16, 64, 308, 1730, 11104, 80176, 643232, 5676560, 54650176, 569980384, 6401959328, 77042282000, 988949446144, 13488013248256, 194780492544512, 2969094574403840, 47640794742439936, 802644553810683904, 14166772337295285248, 261410917571703825920
Offset: 1
Examples
G.f. = x + 2*x^2 + 5*x^3 + 16*x^4 + 64*x^5 + 308*x^6 + 1730*x^7 + 11104*x^8 + ... a(3) = 5: Denoting the two types of node of outdegree 1 by the letters a or b, the 5 possible trees are . . 1a 1b 1a 1b 1 . | | | | / \ . 2a 2b 2b 2a 2 3 . | | | | . 3 3 3 3 - _Peter Bala_, Sep 01 2011
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..100
- Jean-Luc Baril, Sergey Kirgizov, Vincent Vajnovszki, Patterns in treeshelves, arXiv:1611.07793 [cs.DM], 2016.
- 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.
- Lapo Cioni, Luca Ferrari, and Corentin Henriet. A direct bijection between two-stack sortable permutations and fighting fish, Euro. Conf. Comb., Graph Theory Appl. (2023) No. 12, 283-289.
- D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA], 2005.
Programs
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Maple
E:= (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)): S:= map(simplify,series(E,x,101)): seq(coeff(S,x,j)*j!, j=1..100); # Robert Israel, Nov 23 2016
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Mathematica
max = 25; f[x_] := (2*(Exp[Sqrt[2]*x] - 1))/((2 + Sqrt[2]) - (2 - Sqrt[2])*Exp[Sqrt[2]*x]); Drop[ Simplify[ CoefficientList[ Series[f[x], {x, 0, max}], x]*Range[0, max]!], 1] (* Jean-François Alcover, Oct 05 2011 *) -
PARI
x='x+O('x^66); /* that many terms */ default(realprecision,1000); /* working with floats here */ egf=(2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)); round(Vec(serlaplace(egf))) /* show terms */ /* Joerg Arndt, Sep 01 2011 */ -
PARI
/* the following program should be preferred. */ Vec( serlaplace( serreverse( intformal( 1/(1+2*x+1/2*x^2) + O(x^66) ) ) ) ) \\ Joerg Arndt, Mar 01 2014
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PARI
{a(n) = if( n<1, 0, n! * polcoeff( 2 / (-2 + quadgen(8) * (-1 + 2 / (1 - exp(-quadgen(8) * x + x * O(x^n))))), n))};
Formula
E.g.f.: A(x) = (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)) = x+2*x^2/2!+5*x^3/3!+16*x^4/4!+64*x^5/5!+....
From Peter Bala, Sep 01 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A' = 1+2*A+1/2*A^2 with A(0) = 0. It follows that the inverse function A(x)^-1 may be expressed as an integral A(x)^-1 = int {t = 0..x} 1/(1+2*t+1/2*t^2).
Applying [Dominici, Theorem 4.1] to invert the integral gives the following method for calculating the terms of the sequence: let f(x) = 1+2*x+1/2*x^2. Let D be the operator f(x)*d/dx. Then a(n) = D^n(f(x)) evaluated at x = 0. Compare with A000111(n+1) = D^n(1+x+x^2/2!) evaluated at x = 0.
(End)
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ) and m=1; (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - 1/2*x^2*(k+1)*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2013
a(n) ~ n! * 2^((n+3)/2) / log(3+2*sqrt(2))^(n+1). - Vaclav Kotesovec, Oct 08 2013
G.f.: conjecture: T(0)/(1-2*x) -1, where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-2*x*(k+1))*(1-2*x*(k+2))/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2013
E.g.f.: x/(T(0)-x), where T(k) = 4*k + 1 + x^2/(8*k+6 + x^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2013
Extensions
Terms >= 80176 from Peter Bala, Sep 01 2011
Changed offset to 1 to agree with name and example. - Michael Somos, Nov 23 2016
Comments