|
| |
|
|
A325205
|
|
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.
|
|
1
|
|
|
|
0, 1, 2, 9, 64, 625, 7776, 117649, 2097144, 43046136, 999970020, 25936053990, 742947675624, 23295384644532, 793591829158128, 29187143427692250, 1152639088016576160, 48646833059722978080, 2185150741063924810176, 104085328898784937079376, 5240461483486301616704160
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
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.
|
|
|
LINKS
|
|
|
|
FORMULA
|
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.
|
|
|
PROG
|
(Python)
# print first num_entries entries in the sequence
import math, sympy; x=sympy.symbols('x')
k=6; num_entries = 64
P=range(k+1); eP=sum([x**d/math.factorial(d) for d in P]); r = [0, 1]; curr_pow = eP
for term in range(1, num_entries-1):
...curr_pow=(curr_pow*eP).expand()
...r.append(curr_pow.coeff(x**term)*math.factorial(term))
print(r)
|
|
|
CROSSREFS
|
Column k=6 of A325201; see that entry for sequences related to other preimage constraints constructions.
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
