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 A050354 Number of ordered factorizations of n with one level of parentheses. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS 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). Dirichlet inverse of (A074206*A153881). - Mats Granvik, Jan 12 2009 LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 FORMULA 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, d1, a(1)= 1/k. - Vladeta Jovovic, May 25 2005 a(p^k) = 3^(k-1). a(A002110(n)) = A050351(n). 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 EXAMPLE For n=6, we have (6) = (3*2) = (2*3) = (3)*(2) = (2)*(3), thus a(6) = 5. MATHEMATICA A[n_]:=If[n==1, n/2, 2*Sum[If[d

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Last modified February 20 21:44 EST 2019. Contains 320362 sequences. (Running on oeis4.)