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%I #53 Oct 08 2022 08:04:30
%S 1,2,7,32,166,926,5419,32816,203902,1292612,8327254,54358280,
%T 358769152,2390130038,16051344307,108548774240,738563388214,
%U 5052324028508,34727816264050,239733805643552,1661351898336676,11553558997057772,80603609263563262,563972937201926432
%N Number of ordered rooted trees with n generators.
%C A generator is a leaf or a node with just one child.
%C The Hankel transform of this sequence is 3^C(n+1,2). The Hankel transform of this sequence with 1 prepended (1,1,2,7,...) is 3^C(n,2). - _Paul Barry_, Jan 26 2011
%C a(n) is the number of Schroder paths of semilength n-1 in which the (2,0)-steps that are not on the horizontal axis come in 2 colors. Example: a(3)=7 because we have HH, UDUD, UUDD, HUD, UDH, UBD, and URD, where U=(1,1), H=(2,0), D=(1,-1), while B and R are, respectively, blue and red (2,0)-steps. - _Emeric Deutsch_, May 02 2011
%C Also the compositional inverse of A327767. - _Peter Luschny_, Oct 08 2022
%H Vincenzo Librandi, <a href="/A108524/b108524.txt">Table of n, a(n) for n = 1..200</a>
%H Paul Barry, <a href="https://arxiv.org/abs/1803.10297">Generalized Eulerian Triangles and Some Special Production Matrices</a>, arXiv:1803.10297 [math.CO], 2018.
%H Shishuo Fu, Yaling Wang, <a href="https://arxiv.org/abs/1908.03912">Bijective recurrences concerning two Schröder triangles</a>, arXiv:1908.03912 [math.CO], 2019.
%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>
%F G.f.: (sqrt(4*x^2-8*x+1) - 1)/(2*x-4).
%F G.f.: 1/(1-x-x/(1-2x-x/(1-2x-x/(1-2x-x/(1-2x-x/(1-... (continued fraction). - _Paul Barry_, Feb 10 2009
%F a(n) = sum(i=1..n+1, (i*C(n+1,i)*sum(j=0..n-i, (-1)^j*2^(n-j)*C(n,j)*C(2*n-j-i-1,n-1)))/2^i)/(n*(n+1)). - _Vladimir Kruchinin_, May 10 2011
%F From _Gary W. Adamson_, Jul 11 2011: (Start)
%F a(n) is upper left term in the following infinite square production matrix:
%F 1, 1, 0, 0, 0, ...
%F 1, 1, 1, 0, 0, ...
%F 3, 3, 1, 1, 0, ...
%F 9, 9, 3, 1, 1, ...
%F ...
%F (where columns are (1, 1, 3, 9, 27, 81, ...) prefaced with (0,0,1,2,3,...) zeros. (End)
%F Conjecture: 2*n*a(n) +(24-17*n)*a(n-1) +4*(4*n-9)*a(n-2) +4*(3-n)*a(n-3)=0. - _R. J. Mathar_, Nov 14 2011
%F G.f.: A(x)=(sqrt(4*x^2-8*x+1) - 1)/x/(2*x-4) = 1/(G(0)-x); G(k) = 1 + 2*x - 3*x/G(k+1); (continued fraction, 1-step ). - Sergei N. Gladkovskii, Jan 05 2012
%F a(n) ~ 3^(1/4)*(3^(3/2)-5)*(4+2*sqrt(3))^n/(2*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 18 2012
%F From _Peter Bala_, Mar 13 2015: (Start)
%F The o.g.f. A(x) satisfies the differential equation (2 - 17*x + 16*x^2 - 4*x^3)A'(x) + (7 - 4*x)*A(x) = 2 - 2*x. Mathar's conjectural recurrence above follows from this.
%F The o.g.f. A(x) is the series reversion of the rational function x*(1 - 2*x)/(1 - x^2). - _Peter Bala_, Mar 13 2015
%p # Using function CompInv from A357588.
%p CompInv(24, n -> [1, -2][irem(n-1, 2) + 1]); # _Peter Luschny_, Oct 08 2022
%t Rest[CoefficientList[Series[(Sqrt[4*x^2-8*x+1]-1)/(2*x-4), {x, 0, 20}], x]] (* _Vaclav Kotesovec_, Oct 18 2012 *)
%o (Maxima)
%o a(n):=sum((i*binomial(n+1,i)*sum((-1)^j*2^(n-j)*binomial(n,j)*binomial(2*n-j-i-1,n-1),j,0,n-i))/2^i,i,1,n+1)/(n*(n+1)); // _Vladimir Kruchinin_, May 10 2011
%Y Cf. A108521-A108529, A108525, A000108, A001003, A327767.
%K nonn
%O 1,2
%A _Christian G. Bower_, Jun 07 2005
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