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A050354
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Number of ordered factorizations of n with one level of parentheses.
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8
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1, 1, 1, 3, 1, 5, 1, 9, 3, 5, 1, 21, 1, 5, 5, 27, 1, 21, 1, 21, 5, 5, 1, 81, 3, 5, 9, 21, 1, 37, 1, 81, 5, 5, 5, 111, 1, 5, 5, 81, 1, 37, 1, 21, 21, 5, 1, 297, 3, 21, 5, 21, 1, 81, 5, 81, 5, 5, 1, 201, 1, 5, 21, 243, 5, 37, 1, 21, 5, 37, 1, 513, 1, 5, 21, 21, 5, 37, 1, 297, 27, 5, 1, 201
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OFFSET
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1,4
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COMMENTS
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a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3,1).
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LINKS
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FORMULA
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Dirichlet g.f.: (2-zeta(s))/(3-2*zeta(s)).
Recurrence for number of ordered factorizations of n with k-1 levels of parentheses is a(n) = k*Sum_{d|n, d<n} a(d), n>1, a(1)= 1/k. - Vladeta Jovovic, May 25 2005
a(p^k) = 3^(k-1).
Sum_{k=1..n} a(k) ~ -n^r / (4*r*Zeta'(r)), where r = 2.185285451787482231198145140899733642292971552057774261555354324536... is the root of the equation Zeta(r) = 3/2. - Vaclav Kotesovec, Feb 02 2019
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EXAMPLE
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For n=6, we have (6) = (3*2) = (2*3) = (3)*(2) = (2)*(3), thus a(6) = 5.
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MATHEMATICA
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A[n_]:=If[n==1, n/2, 2*Sum[If[d<n, A[d], 0], {d, Divisors[n]}]]; Table[If[n==1, n, A[n]], {n, 1, 100}] (* Indranil Ghosh, May 19 2017 *)
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PROG
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(PARI)
A050354aux(n) = if(1==n, n/2, 2*sumdiv(n, d, if(d<n, A050354aux(d), 0)));
(Sage)
def A(n): return 1/2 if n==1 else 2*sum(A(d) for d in divisors(n) if d<n)
def a(n): return 1 if n==1 else A(n)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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