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%I #13 Jul 14 2019 08:34:32
%S 0,1,1,1,1,2,1,2,1,2,2,1,2,2,1,2,2,3,1,2,2,2,3,1,2,2,2,3,1,2,2,3,2,3,
%T 2,1,2,2,3,2,3,2,1,2,2,3,3,2,3,2,3,1,2,2,3,4,2,3,2,3,2,3,1,2,2,3,4,2,
%U 3,2,3,2,3,1,2,3,2,3,4,2,3,3,2,3,2,3,1,4
%N Number of distinct prime factors of A181800(n) (n-th powerful number that is the first integer of its prime signature).
%C Since each prime factor of A181800(n) divides A181800(n) at least twice, this is also the number of exponents > 2 in prime factorization of A181800(n).
%C Length of row A181800(n) of table A212171 equals a(n) for n > 1. Row A181800(n) of table A212172 has the same length when n > 1 (length = 1 if n = 1).
%H Amiram Eldar, <a href="/A212179/b212179.txt">Table of n, a(n) for n = 1..10000</a>
%H Will Nicholes, <a href="http://willnicholes.com/math/primesiglist.htm">Prime Signatures</a>
%F a(n) = A001221(A181800(n)) = A056170(A181800(n)).
%e 72 (2^3*3^2) has 2 distinct prime factors. Since 72 = A181800(8), a(8) = 2.
%Y Cf. A181800, A001694, A025487, A212171, A212172, A212176.
%K nonn
%O 1,6
%A _Matthew Vandermast_, Jun 04 2012
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