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%I #29 Jan 03 2021 15:56:41
%S 1,1,1,2,1,2,1,4,2,2,1,5,1,2,2,8,1,5,1,5,2,2,1,12,2,2,4,5,1,6,1,16,2,
%T 2,2,14,1,2,2,12,1,6,1,5,5,2,1,28,2,5,2,5,1,12,2,12,2,2,1,18,1,2,5,32,
%U 2,6,1,5,2,6,1,37,1,2,5,5,2,6,1,28,8,2,1,18,2,2,2,12,1,18,2,5,2,2,2,64
%N Number of ordered factorizations into prime powers greater than 1.
%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 The Dirichlet inverse is in A010055, turning all but the first element in A010055 negative. - _R. J. Mathar_, Jul 15 2010
%C Not multiplicative: a(6) =2 <> a(2)*a(3) = 1*1. - _R. J. Mathar_, May 25 2017
%H R. J. Mathar, <a href="/A050363/b050363.txt">Table of n, a(n) for n = 1..5000</a>
%F Dirichlet g.f.: 1/(1-B(s)) where B(s) is D.g.f. of characteristic function of prime powers >1.
%F a(p^k) = 2^(k-1).
%F a(A002110(k)) = k!.
%F a(n) = A050364(A101296(n)). - _R. J. Mathar_, May 26 2017
%F G.f. A(x) satisfies: A(x) = x + Sum_{p prime, k>=1} A(x^(p^k)). - _Ilya Gutkovskiy_, May 11 2019
%e From _R. J. Mathar_, May 25 2017: (Start)
%e a(p^2) = 2: factorizations p^2, p*p.
%e a(p^3) = 4: factorizations p^3, p^2*p, p*p^2, p*p*p.
%e a(p*q) = 2: factorizations p*q, q*p.
%e a(p*q^2)= 5: factorizations p*q^2, q^2*p, p*q*q, q*p*q, q*q*p. (End)
%p read(transforms) ;
%p [1,seq(-A010055(n),n=2..100)] ;
%p DIRICHLETi(%) ; # _R. J. Mathar_, May 25 2017
%Y Cf. A000961, A050360, A050361, A050362, A050363, A050364, A076277.
%K nonn
%O 1,4
%A _Christian G. Bower_, Oct 15 1999
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