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%I #13 Feb 02 2019 08:20:20
%S 1,1,1,4,1,7,1,16,4,7,1,40,1,7,7,64,1,40,1,40,7,7,1,208,4,7,16,40,1,
%T 73,1,256,7,7,7,292,1,7,7,208,1,73,1,40,40,7,1,1024,4,40,7,40,1,208,7,
%U 208,7,7,1,544,1,7,40,1024,7,73,1,40,7,73,1,1840,1,7,40,40,7,73,1
%N Number of ordered factorizations of n with 2 levels 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).
%H Antti Karttunen, <a href="/A050356/b050356.txt">Table of n, a(n) for n = 1..10000</a>
%F Dirichlet g.f.: (3-2*zeta(s))/(4-3*zeta(s)).
%F a(p^k) = 4^(k-1).
%F a(A002110(n)) = A050352(n).
%F Sum_{k=1..n} a(k) ~ -n^r / (9*r*Zeta'(r)), where r = 2.52138975790328306967497455387140053675965539610041801606891036... is the root of the equation Zeta(r) = 4/3. - _Vaclav Kotesovec_, Feb 02 2019
%e For n=6 we have ((6)) = ((3*2)) = ((2*3)) = ((3)*(2)) = ((2)*(3)) = ((3))*((2)) = ((2))*((3)), thus a(6) = 7.
%o (PARI)
%o A050356aux(n) = if(1==n,1/3, 3*sumdiv(n,d, if(d<n, A050356aux(d), 0)));
%o A050356(n) = if(1==n,n,A050356aux(n)); \\ _Antti Karttunen_, May 19 2017, after the general recurrence given by _Vladeta Jovovic_ May 25 2005 in A050354.
%Y Cf. A002033, A050351, A050352, A050353, A050354, A050355, A050357, A050358, A050359.
%K nonn
%O 1,4
%A _Christian G. Bower_, Oct 15 1999
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