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%I #30 Mar 27 2021 22:47:16
%S 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,
%T 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,
%U 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
%N Number of ordered factorizations of n with one level of parentheses.
%C 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).
%C Dirichlet inverse of (A074206*A153881). - _Mats Granvik_, Jan 12 2009
%H Antti Karttunen, <a href="/A050354/b050354.txt">Table of n, a(n) for n = 1..10000</a>
%F Dirichlet g.f.: (2-zeta(s))/(3-2*zeta(s)).
%F 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
%F a(p^k) = 3^(k-1).
%F a(A002110(n)) = A050351(n).
%F 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
%e For n=6, we have (6) = (3*2) = (2*3) = (3)*(2) = (2)*(3), thus a(6) = 5.
%t 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 *)
%o (PARI)
%o A050354aux(n) = if(1==n,n/2, 2*sumdiv(n,d, if(d<n, A050354aux(d), 0)));
%o A050354(n) = if(1==n,n,A050354aux(n)); \\ _Antti Karttunen_, May 19 2017, after Jovovic's general recurrence.
%o (Sage)
%o def A(n): return 1/2 if n==1 else 2*sum(A(d) for d in divisors(n) if d<n)
%o def a(n): return 1 if n==1 else A(n)
%o print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, May 19 2017, after _Antti Karttunen_'s PARI program
%Y Cf. A002033, A050351, A050352, A050353, A050355, A050356, A050357, A050358, A050359.
%K nonn
%O 1,4
%A _Christian G. Bower_, Oct 15 1999
%E Duplicate comment removed by _R. J. Mathar_, Jul 15 2010
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