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%I #12 Nov 11 2017 17:44:23
%S 0,0,0,1,0,1,0,1,1,1,0,2,0,1,1,1,0,2,0,2,1,1,0,2,1,1,1,2,0,2,0,1,1,1,
%T 1,3,0,1,1,2,0,2,0,2,2,1,0,2,1,2,1,2,0,2,1,2,1,1,0,3,0,1,2,1,1,2,0,2,
%U 1,2,0,3,0,1,2,2,1,2,0,2,1,1,0,3,1,1,1,2,0,3,1,2,1,1,1,2,0,2,2,3,0,2,0,2,2
%N Number of symbols '*' and '^' to write the canonical prime factorization of n.
%C a(n) = A001221(n) - 1 + A056170(n) for n > 1;
%C a(n) = 0 iff n=1 or n is prime: a(A008578(n)) = 0;
%C a(n) = 1 iff n is a semiprime or a prime power p^e with e > 1.
%C It is conjectured that 1,2,3,4,5,6,7,9,11 are the only positive integers which cannot be represented as the sum of two elements of indices n such that a(n) = 1. - _Jonathan Vos Post_, Sep 11 2005
%C a(n) = 2 iff n is a sphenic number (A007304) or n is a prime p times a prime power q^e with e > 1 and q not equal to p. a(n) = 3 iff n has exactly four distinct prime factors (A046386); or n is the product of two prime powers (p^e)*(q^f) with e > 1, f > 1 and p not equal to q; or n is a semiprime s times a prime power r^g with g > 1 and r relatively prime to s. For a(n) > 3, _Reinhard Zumkeller_'s description is a simpler description than the above compound descriptions. - _Jonathan Vos Post_, Sep 11 2005
%H Reinhard Zumkeller, <a href="/A110475/b110475.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeFactorization.html">Prime Factorization</a>
%e a(208029250) = a(2*5^3*11^2*13*23^2) = 4 '*' + 3 '^' = 7.
%o (Haskell)
%o a110475 1 = 0
%o a110475 n = length us - 1 + 2 * length vs where
%o (us, vs) = span (== 1) $ a118914_row n
%o -- _Reinhard Zumkeller_, Mar 23 2014
%Y Cf. A050252, A001358, A025475, A000040.
%Y Cf. A007304, A046386.
%Y Cf. A118914.
%K nonn
%O 1,12
%A _Reinhard Zumkeller_, Sep 08 2005
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