A007853 Number of maximal antichains in rooted plane trees on n nodes.
1, 2, 5, 15, 50, 178, 663, 2553, 10086, 40669, 166752, 693331, 2917088, 12398545, 53164201, 229729439, 999460624, 4374546305, 19250233408, 85120272755, 378021050306, 1685406494673, 7541226435054, 33852474532769, 152415463629568, 688099122024944
Offset: 1
Keywords
Examples
G.f. = x + 2*x^2 + 5*x^3 + 15*x^4 + 50*x^5 + 178*x^6 + 663*x^7 + 2553*x^8 + ... - _Michael Somos_, Nov 07 2019
Links
- R. Bacher, On generating series of complementary plane trees arXiv:math/0409050 [math.CO], 2004.
- M. Klazar, Twelve countings with rooted plane trees, European Journal of Combinatorics 18 (1997), 195-210; Addendum, 18 (1997), 739-740.
- Index entries for sequences related to rooted trees
Crossrefs
Programs
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Mathematica
ie[t_]:=If[Length[t]==0,1,1+Product[ie[b],{b,t}]]; allplane[n_]:=If[n==1,{{}},Join@@Function[c,Tuples[allplane/@c]]/@Join@@Permutations/@IntegerPartitions[n-1]]; Table[Sum[ie[t],{t,allplane[n]}],{n,9}] (* Gus Wiseman, Aug 13 2018 *)
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Maxima
a(n):=1/(n+1)*binomial(2*n,n)+sum((k+2)/(n+1)*binomial(2*n-k-1,n-k-1)*(sum(((binomial(2*i,i))*(binomial(k+i,3*i)))/(i+1),i,0,floor(k/2))),k,0,n-1); /* Vladimir Kruchinin, Apr 05 2019 */
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PARI
{a(n) = my(A); if( n<0, 0, A = sqrt(1 - 4*x + x * O(x^n)); polcoeff( (3 - 2*x - A - sqrt(2 - 16*x + 4*x^2 + (2 + 4*x) * A)) / 4, n))}; /* Michael Somos, Nov 07 2019 */
Formula
G.f.: (1/4) * (3 - 2*x - sqrt(1-4*x) - sqrt(2) * sqrt((1+2*x) * sqrt(1-4*x) + 1 - 8*x + 2*x^2)) [from Klazar]. - Sean A. Irvine, Feb 06 2018
a(n) = (1/(n+1))*C(2*n,n) + Sum_{k=0..n-1} ((k+2)/(n+1))*C(2*n-k-1,n-k-1)*Sum_{i=0..floor(k/2)} C(2*i,i)*C(k+i,3*i)/(i+1). - Vladimir Kruchinin, Apr 05 2019
Given the g.f. A(x) and the g.f. of A213705 B(x), then -x = A(-B(x)). - Michael Somos, Nov 07 2019
Extensions
More terms from Sean A. Irvine, Feb 06 2018
Comments