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A325208
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a(n) is the number of labeled rooted trees on a set of size n where each node has at most 9 neighbors that are further away from the root than the node itself.
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1
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0, 1, 2, 9, 64, 625, 7776, 117649, 2097152, 43046721, 1000000000, 25937424590, 743008369224, 23298084997044, 793714764270428, 29192925433321650, 1152921466989795360, 48661189511753527280, 2185911410555033096364, 104127340753401006230046, 5242879377215160617336400
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OFFSET
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0,3
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COMMENTS
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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 9 entries.
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LINKS
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FORMULA
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a(n) = (n-1)! * [x^(n-1)] e_9(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_9 * n^(-3/2) * r_9^-n.
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MATHEMATICA
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e[k_][x_] := Sum[x^j/j!, {j, 0, k}];
a[0] = 0; a[n_] := (n - 1)! Coefficient[e[9][x]^n, x, n - 1];
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PROG
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(Python) # print first num_entries entries in the sequence
import math, sympy; x=sympy.symbols('x')
k=9; 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)
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CROSSREFS
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Column k=9 of A325201; see that entry for sequences related to other preimage constraints constructions.
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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