|
%I #44 Jun 10 2019 00:56:44
%S 1,1,2,5,14,42,132,428,1421,4807,16510,57421,201824,715768,2558167,
%T 9204651,33315919,121218195,443107245,1626546453,5993256280,
%U 22158739970,82182744284,305670888560,1139892935454,4261095044346,15964169665031,59933390160322
%N Number of ordered rooted trees with n non-root nodes and all outdegrees <= six.
%H Alois P. Heinz, <a href="/A036768/b036768.txt">Table of n, a(n) for n = 0..1000</a>
%H Vladimir Kruchinin and D. V. Kruchinin, <a href="http://arxiv.org/abs/1103.2582">Composita and their properties</a>, arXiv:1103.2582 [math.CO], 2011-2013.
%H Nickolas Hein and Jia Huang, <a href="http://arxiv.org/abs/1508.01688">Modular Catalan Numbers</a>, arXiv:1508.01688 [math.CO], 2015.
%H Nickolas Hein and Jia Huang, <a href="https://doi.org/10.1016/j.ejc.2016.11.004">Modular Catalan Numbers</a>, European Journal of Combinatorics 61 (2017), 197-218.
%H Lajos Takacs, <a href="http://www.appliedprobability.org/data/files/TMS%20articles/18_1_1.pdf">Enumeration of rooted trees and forests</a>, Math. Scientist 18 (1993), 1-10, esp. Eq. (6).
%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>
%F G.f. A(x) satisfies A(x)=1+sum(n=1..6, (x*A(x))^n). - _Vladimir Kruchinin_, Feb 22 2011
%p r := 6; [ 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) ];
%p # second Maple program:
%p b:= proc(u, o) option remember; `if`(u+o=0, 1,
%p add(b(u-j, o+j-1), j=1..min(1, u))+
%p add(b(u+j-1, o-j), j=1..min(6, o)))
%p end:
%p a:= n-> b(0, n):
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Aug 28 2017
%t 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]}]];
%t a[n_] := b[0, n, 6];
%t Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Nov 07 2017, after _Alois P. Heinz_ *)
%o (PARI) a(n)=if(n<0,0,polcoeff(serreverse(x/polcyclo(7)+O(x^(n+2))),n+1)) /* _Ralf Stephan_ */
%Y Column k=6 of A288942.
%K nonn
%O 0,3
%A _N. J. A. Sloane_
%E Name clarified by _Andrew Howroyd_, Dec 04 2017
|
