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%I #8 Mar 30 2012 18:40:27
%S 0,0,1,2,3,4,5,6,7,8,2,1,6,3,5,7,9,9,11,9,3,2,2,8,7,7,7,10,11,10,3,2,
%T 7,8,11,7,12,13,15,11,3,2,6,7,7,10,9,12,12,13,5,2,5,8,8,7,13,12,10,12,
%U 6,3,3,6,12,11,12,10,12,12,2,6,12,8,11,9,14,13,13,13,7,2
%N Number of prime factors, with multiplicity, of the nonzero 9-acci numbers.
%C Prime 9-acci numbers: b(3) = 2, b(12) = 1021, ... Semiprime 9-acci numbers: b(4) = 4 = 2^2, b(11) = 511 = 7 * 73, b(22) = 1035269 = 47 * 22027, b(23) = 2068498 = 2 * 1034249, b(32) = 1049716729 = 1051 * 998779 b(42) = 1064366053385 = 5 * 212873210677, b(52) = 1079219816432629 = 28669 * 37644138841, b(71) = 555323195719171835391 = 3 * 185107731906390611797, b(82) = 1125036467745713090813969 = 37 * 30406391020154407859837.
%F a(n) = A001222(A104144(n+7)).
%e a(1)=a(2)=0 because the first two nonzero 9-acci numbers are both 1, which has zero prime divisors.
%e a(3)=1 because the 3rd nonzero 9-acci number is 2, a prime, with only one prime divisor.
%e a(4)=2 because the 4th nonzero 9-acci number is 4 = 2^2 which has (with multiplicity) 2 prime divisors (which happen to be equal).
%e a(5)=3 because the 5th nonzero 9-acci number is 8 = 2^3.
%e a(13) = 6 because b(13) = 2040 = 2^3 * 3 * 5 * 17 so has 6 prime factors (2 with multiplicity 3 and 3, 5 and 17 once each).
%Y Cf. A001222, A104411, A104412, A104413, A104414, A104415.
%K easy,nonn
%O 1,4
%A _Jonathan Vos Post_, Mar 06 2005
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