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Number of prime factors of A181800(n) (n-th powerful number that is the first integer of its prime signature), counted with multiplicity.
2

%I #12 Jul 14 2019 09:49:23

%S 0,2,3,4,5,4,6,5,7,6,6,8,7,7,9,8,8,6,10,9,8,9,7,11,10,9,10,8,12,11,10,

%T 8,11,9,10,13,12,11,9,12,10,11,14,13,12,10,9,13,11,12,10,15,14,13,11,

%U 8,12,10,14,12,13,11,16,15,14,12,9,13,11,15,13,14,12

%N Number of prime factors of A181800(n) (n-th powerful number that is the first integer of its prime signature), counted with multiplicity.

%C Every nonnegative integer n appears A002865(n) times.

%H Amiram Eldar, <a href="/A212639/b212639.txt">Table of n, a(n) for n = 1..10000</a>

%H Will Nicholes, <a href="http://willnicholes.com/math/primesiglist.htm">List of prime signatures</a>

%F a(n) = A001222(A181800(n)).

%e 72 (2^3*3^2, or 2*2*2*3*3) has a total of 5 prime factors when repetitions are counted. Since 72 = A181800(8), a(8) = 5.

%Y Cf. A181800, A001222, A212172, A212176, A212179, A212645.

%K nonn

%O 1,2

%A _Matthew Vandermast_, Jun 09 2012