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%I #5 Jun 23 2019 11:51:47
%S 0,1,2,9,64,625,7776,117649,2097144,43046136,999970020,25936053990,
%T 742947675624,23295384644532,793591829158128,29187143427692250,
%U 1152639088016576160,48646833059722978080,2185150741063924810176,104085328898784937079376,5240461483486301616704160
%N a(n) is the number of labeled rooted trees on a set of size n where each node has at most 6 neighbors that are further away from the root than the node itself.
%C A preimage constraint on a function is a set of nonnegative integers such that the size of the inverse image of any element is one of the values in that set. View a labeled rooted tree as an endofunction on the set {1,2,...,n} by sending every non-root node to its neighbor that is closer to the root and sending the root to itself. Thus, a(n) is the number of endofunctions on a set of size n with exactly one cyclic point and such that each preimage has at most 6 entries.
%H B. Otto, <a href="https://arxiv.org/abs/1903.00542">Coalescence under Preimage Constraints</a>, arXiv:1903.00542 [math.CO], 2019, Corollaries 5.3 and 7.8.
%F a(n) = (n-1)! * [x^(n-1)] e_6(x)^n, where e_k(x) is the truncated exponential 1 + x + x^2/2! + ... + x^k/k!. The link above yields explicit constants c_k, r_k so that the columns are asymptotically c_6 * n^(-3/2) * r_6^-n.
%o (Python)
%o # print first num_entries entries in the sequence
%o import math, sympy; x=sympy.symbols('x')
%o k=6; num_entries = 64
%o P=range(k+1); eP=sum([x**d/math.factorial(d) for d in P]); r = [0,1]; curr_pow = eP
%o for term in range(1,num_entries-1):
%o ...curr_pow=(curr_pow*eP).expand()
%o ...r.append(curr_pow.coeff(x**term)*math.factorial(term))
%o print(r)
%Y Column k=6 of A325201; see that entry for sequences related to other preimage constraints constructions.
%K easy,nonn
%O 0,3
%A _Benjamin Otto_, Apr 11 2019
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