A099553 Number of rooted 2-connected loopless 4-regular planar maps with n inner faces.
1, 2, 10, 42, 209, 1066, 5726, 31688, 180234, 1047356, 6198500, 37253790, 226891665, 1397880330, 8699804598, 54629525808, 345778883678, 2204263514460, 14142192816908, 91263177339092, 592069697914170, 3859674384409668, 25272938482712044
Offset: 3
Keywords
Examples
A(x) = x^3 + 2*x^4 + 10*x^5 + 42*x^6 + 209*x^7 + 1066*x^8 + 5726*x^9 + ...
Links
- Gheorghe Coserea, Table of n, a(n) for n = 3..305
- Han Ren, Yanpei Liu and Zhaoxiang Li, Enumeration of 2-connected Loopless 4-regular Maps on the Plane, European J. Combin., 23 (2002), 93-111.
- T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. III: Nonseparable maps, J. Combinatorial Theory Ser. B 18 (1975), 222-259, eq (7).
Programs
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Maple
A099553 := proc(n) local e; e := n-1 ; add(binomial(2*e-r,e-2-2*r)*2^r*binomial(2*e,r),r=0..floor(e/2-1)) ; %-3*add(binomial(2*e-r,e-3-2*r)*2^r*binomial(2*e,r),r=0..floor((e-3)/2)) ; %*2/e ; end proc: seq(A099553(n),n=3..30) ; # R. J. Mathar, Aug 28 2018
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Mathematica
a[n_] := Module[{e, s}, e = n-1; s = Sum[Binomial[2e-r, e-2-2r]*2^r*Binomial[2e, r], {r, 0, Floor[e/2-1]}]; s = s-3*Sum[Binomial[2e-r, e-3-2r]*2^r*Binomial[2e, r], {r, 0, Floor[(e-3)/2]}]; s=2s/e]; Table[a[n], {n, 3, 30}] (* Jean-François Alcover, Feb 14 2023, after R. J. Mathar *)
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PARI
F = (2*z^3*x^2 + (2*z^3 - 2*z)*x + (-z + 1))/(-2*z^3*x + 2*z); G = x*(4*x + 1)*z^4 + 4*x*z^3 - 5*x*z^2 - 2*z + 2; Z(N) = { my(z0=1+O('x^N), z1=0, n=1); while (n++, z1 = z0 - subst(G, 'z, z0)/subst(deriv(G,'z), 'z, z0); if (z1 == z0, break()); z0 = z1); z0; }; seq(N) = Vec(subst(F, 'z, Z(N+3))); seq(23) \\ test: y = Ser(seq(303))*'x^3; 0 == 8*y^4 + (32*x + 12)*y^3 + (48*x^2 + 23*x + 6)*y^2 + (32*x^3 + 10*x^2 - 10*x + 1)*y + x^3*(8*x - 1) \\ Gheorghe Coserea, Jul 13 2018
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PARI
seq(n)={my(g=1+x*sum(n=1,n,x^n*binomial(3*n, n)*2/((n+1)*(2*n+1))) + O(x*x^n)); Vec(-1 + sqrt(serreverse(x/g^2)/x))} \\ Andrew Howroyd, Apr 06 2021
Formula
From Gheorghe Coserea, Jul 12 2018: (Start)
G.f. A(x) = (2*z^3*x^2 + (2*z^3 - 2*z)*x + (-z + 1))/(-2*z^3*x + 2*z), where z = 1 + 2*x^2 + 6*x^3 + 34*x^4 + 176*x^5 + 1004*x^6 + ... satisfies 0 = x*(4*x + 1)*z^4 + 4*x*z^3 - 5*x*z^2 - 2*z + 2. (See Theorem D in reference.)
G.f. y=A(x) satisfies:
0 = 8*y^4 + (32*x + 12)*y^3 + (48*x^2 + 23*x + 6)*y^2 + (32*x^3 + 10*x^2 - 10*x + 1)*y + x^3*(8*x - 1).
0 = x^3*(2*x + 1)*(49*x - 18)*(196*x - 27)*y'''' + x^2*(96040*x^3 - 27587*x^2 - 9297*x + 972)*y''' + (72030*x^4 - 36309*x^3 + 2010*x^2 - 864*x)*y'' - 6*(8*x + 3)*(49*x + 12)*y' + (2352*x + 576)*y.
(End)
Conjecture: 3*n *(3*n-1) *(5*n-8) *(3*n-2)*a(n) -(n-2) *(2*n-3) *(355*n^2 -703*n +300)*a(n-1) -98*(n-2) *(5*n-3) *(2*n-3) *(2*n-5) *a(n-2)=0. - R. J. Mathar, Aug 28 2018
G.f.: x*(A(x) - 1) where A(x) satisfies A(x) = G(x*A(x)^2) and (G(x) + 2*x - 1)/x is the g.f. of A000139. - Andrew Howroyd, Apr 06 2021
Comments