A036767 Number of ordered rooted trees with n non-root nodes and all outdegrees <= five.
1, 1, 2, 5, 14, 42, 131, 421, 1385, 4642, 15795, 54418, 189454, 665471, 2355510, 8393461, 30084695, 108394449, 392356788, 1426137550, 5203211200, 19048447855, 69951072700, 257609070810, 951172531880, 3520465229446, 13058843476526, 48540377627407
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- Colin Defant and Kai Zheng, Stack-Sorting with Consecutive-Pattern-Avoiding Stacks, arXiv:2008.12297 [math.CO], 2020.
- Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.
- Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, J. Ana. Num. Theor. Vol. 2, No. 2, 2014, 37-44.
- Nickolas Hein and Jia Huang, Modular Catalan Numbers, arXiv:1508.01688 [math.CO], 2015.
- Nickolas Hein and Jia Huang, Modular Catalan Numbers, European Journal of Combinatorics 61 (2017), 197-218.
- L. Takacs, Enumeration of rooted trees and forests, Math. Scientist 18 (1993), 1-10, esp. Eq. (6).
- Index entries for sequences related to rooted trees
Crossrefs
Column k=5 of A288942.
Programs
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Maple
r := 5; [ seq((1/n)*add( (-1)^j*binomial(n,j)*binomial(2*n-2-j*(r+1), n-1),j=0..floor((n-1)/(r+1))), n=1..30) ]; # second Maple program: b:= proc(u, o) option remember; `if`(u+o=0, 1, add(b(u-j, o+j-1), j=1..min(1, u))+ add(b(u+j-1, o-j), j=1..min(5, o))) end: a:= n-> b(0, n): seq(a(n), n=0..30); # Alois P. Heinz, Aug 28 2017
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Mathematica
nn=12;f[x_]:=Sum[a[n]x^n,{n,0,nn}];sol=SolveAlways[Series[0==f[x]-x -x f[x]-x f[x]^2-x f[x]^3-x f[x]^4- x f[x]^5,{x,0,nn}],x];Table[a[n],{n,0,nn}]/.sol (* Geoffrey Critzer, Jan 05 2013 *) b[u_, o_, k_] := b[u, o, k] = If[u + o == 0, 1, Sum[b[u - j, o + j - 1, k], {j, 1, Min[1, u]}] + Sum[b[u + j - 1, o - j, k], {j, 1, Min[k, o]}]]; a[n_] := b[0, n, 5]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 07 2017, after Alois P. Heinz *)
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PARI
a(n)=if(n<0,0,polcoeff(serreverse(x/sum(k=0,5,x^k)+O(x^(n+2))),n+1)) \\ Ralf Stephan
Formula
G.f. A(x) satisfies A(x) = 1 + sum(n=1..5, (x*A(x))^n). - Vladimir Kruchinin, Feb 22 2011
Extensions
Name clarified by Andrew Howroyd, Dec 04 2017
Comments