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Number of nonseparable planar tree-rooted maps with n edges.
(Formerly M0364)
9

%I M0364 #30 Mar 28 2020 10:13:01

%S 1,2,2,6,28,160,1036,7294,54548,426960,3463304,28910816,247104976,

%T 2154192248,19097610480,171769942086,1564484503044,14407366963440,

%U 133978878618904,1256799271555872,11881860129979440

%N Number of nonseparable planar tree-rooted maps with n edges.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Gheorghe Coserea, <a href="/A004304/b004304.txt">Table of n, a(n) for n = 0..200</a>

%H Dov Tamari, <a href="https://doi.org/10.24033/bsmf.1446">Monoïdes préordonnés et chaînes de Malcev</a>, Bulletin de la Société Mathématique de France, Volume 82 (1954), 53-96. See end of Appendix II.

%H T. R. S. Walsh, A. B. Lehman, <a href="http://dx.doi.org/10.1016/0095-8956(75)90050-7">Counting rooted maps by genus. III: Nonseparable maps</a>, J. Combinatorial Theory Ser. B 18 (1975), 222-259. See Table IVc.

%F From _Paul D. Hanna_, Nov 26 2009: (Start)

%F G.f.: A(x) = [x/Series_Reversion(x*F(x)^2)]^(1/2) where F(x) = g.f. of A005568, where A005568(n) is the product of two successive Catalan numbers C(n)*C(n+1).

%F G.f.: A(x) = F(x/A(x)^2) where A(x*F(x)^2) = F(x) where F(x) = g.f. of A005568.

%F G.f.: A(x) = G(x/A(x)) where A(x*G(x)) = G(x) where F(x) = g.f. of A168450.

%F G.f.: A(x) = x/Series_Reversion(x*G(x)) where G(x) = g.f. of A168450.

%F Self-convolution yields A168451.

%F (End)

%p A004304 := proc(n) local N,x,ode ; Order := n+1 ; ode := x^2*diff(N(x),x,x)*(N(x)^3-16*x*N(x)) ; ode := ode + (x*diff(N(x),x))^3*(16-6*N(x)) ; ode := ode + (x*diff(N(x),x))^2*(12*N(x)^2-16*x-24*N(x)) ; ode := ode + x*diff(N(x),x)*(-8*N(x)^3+24*x*N(x)+12*N(x)^2) ; ode := ode + 2*N(x)^2*(N(x)^2-N(x)-6*x) ; dsolve({ode=0,N(0)=1,D(N)(0)=2},N(x),type=series) ; convert(%,polynom) ; rhs(%) ; RETURN( coeftayl(%,x=0,n)) ; end; for n from 0 to 20 do printf("%d,",A004304(n)) ; od ; # _R. J. Mathar_, Aug 18 2006

%t m = 22;

%t F[x_] = Sum[2 (2n+1) Binomial[2n, n]^2 x^n/((n+2)(n+1)^2), {n, 0, m}];

%t A[x_] = (x/InverseSeries[x F[x]^2 + O[x]^m, x])^(1/2);

%t CoefficientList[A[x], x] (* _Jean-François Alcover_, Mar 28 2020 *)

%o (PARI) {a(n)=local(C_2=vector(n+1,m,(binomial(2*m-2,m-1)/m)*(binomial(2*m,m)/(m+1))));polcoeff((x/serreverse(x*Ser(C_2)^2))^(1/2),n)} \\ _Paul D. Hanna_, Nov 26 2009

%o (PARI)

%o seq(N) = {

%o my(c(n)=binomial(2*n,n)/(n+1), s=Ser(apply(n->c(n)*c(n+1), [0..N])));

%o Vec(subst(s, 'x, serreverse('x*s^2)));

%o };

%o seq(20)

%o \\ test: y=Ser(seq(200)); 0 == x^2*y''*(y^3 - 16*x*y) + (x*y')^3*(16-6*y) + (x*y')^2*(12*y^2-16*x-24*y) + x*y'*(-8*y^3 + 24*x*y + 12*y^2) + 2*y^2*(y^2-y-6*x)

%o \\ _Gheorghe Coserea_, Jun 13 2018

%Y Cf. A000264.

%Y Cf. A005568, A168450, A168451, A168452. - _Paul D. Hanna_, Nov 26 2009

%K nonn

%O 0,2

%A _N. J. A. Sloane_

%E More terms from _R. J. Mathar_, Aug 18 2006