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a(n) = 2*T(n) - pi(n), where T(n) (A208251) is the number of refactorable/tau numbers (A033950) <= n and pi(n) (A000720) is the number of primes <= n.
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%I #13 Jan 11 2019 07:57:48

%S 2,3,2,2,1,1,0,2,4,4,3,5,4,4,4,4,3,5,4,4,4,4,3,5,5,5,5,5,4,4,3,3,3,3,

%T 3,5,4,4,4,6,5,5,4,4,4,4,3,3,3,3,3,3,2,2,2,4,4,4,3,5,4,4,4,4,4,4,3,3,

%U 3,3,2,4,3,3,3,3,3,3,2,4,4,4,3,5,5,5,5,7,6,6,6,6,6,6,6,8,7,7,7,7,6,6,5,7,7,7,6,8,7,7,7,7,6,6,6,6,6,6,6,6,6,6,6

%N a(n) = 2*T(n) - pi(n), where T(n) (A208251) is the number of refactorable/tau numbers (A033950) <= n and pi(n) (A000720) is the number of primes <= n.

%C Colton conjectured that T(n) >= pi(n)/2 for all n, i.e., this sequence is nonnegative. Zelinsky proved it for n > 7.42*10^13 (see the Zelinsky reference). This calculation went to 7.44*10^13, proving the conjecture.

%H Simon Colton, <a href="https://cs.uwaterloo.ca/journals/JIS/colton/joisol.html">Refactorable Numbers - A Machine Invention</a>, J. Integer Sequences, Vol. 2, 1999.

%H Joshua Zelinsky, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL5/Zelinsky/zelinsky9.html">Tau Numbers: A Partial Proof of a Conjecture and Other Results </a>, Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.8.

%e For n=6, pi(6)=3, T(6)=2, so a(6) = 2*2 - 3 = 1.

%Y Cf. A033950, A208251, A000720.

%K nonn

%O 1,1

%A _Jud McCranie_, Jan 11 2019