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a(n) = 100 for all n > N. What is the least such N? - Charles R Greathouse IV, May 12 2013

(PARI) a(n)=my(s, i, N=2^n); forprime(p=2, sqrtint(N), s+=primepi(N\p); i++); s-=i*(i-1)/2; i=primepi(sqrtint(N))+primepi(N/2)-1; round(100*(s-i)/s) \\ Charles R Greathouse IV, May 12 2013

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W. Bomfim (webonfim(AT)bol.com.br), Oct 27 2008

Washington Bomfim, Oct 27 2008

Percentage (rounded) of semiprimes <= 2^n which are odd and square-freesquarefree.

More than 84% of the semiprimes in the interval [4, 2^32] are odd and square-freesquarefree. This percentage appears to rise indefinitely as n grows.

a(5)= 20 since the interval [4, 2^5] contains 10 semiprimes, namely 4,6,9,10,14,15,21,22,25 and 26; and two of those semiprimes, (15 and 21), are odd and square-freesquarefree.

Cf. A001358(semiprimes), A125527(Number of semiprimes <= 2^n), A146168(Number of odd square-free squarefree semiprimes < 2^n).

nonn,new

nonn

a(5)= 20 since the interval [4, 2^5] contains 10 semiprimes, namely 4,6,9,10,14,15,21,22,25, and 26; and two of those semiprimes, (15, and 21), are odd and square-free.

nonn,new

nonn

Percentage (rounded) of semiprimes <= 2^n which are odd and square-free.

0, 0, 17, 20, 36, 48, 56, 61, 65, 69, 71, 73, 75, 76, 77, 78, 79, 80, 80, 81, 81, 82, 82, 82, 83, 83, 83, 83, 84, 84, 84, 84

2,3

More than 84% of the semiprimes in the interval [4, 2^32] are odd and square-free. This percentage appears to rise indefinitely as n grows.

a(n) = round(A146168(n)/A125527(n)*100)

a(5)= 20 since the interval [4, 2^5] contains 10 semiprimes, namely 4,6,9,10,14,15,21,22,25, and 26; and two of those semiprimes, (15, and 21), are odd and square-free.

Cf. A001358(semiprimes), A125527(Number of semiprimes <= 2^n), A146168(Number of odd square-free semiprimes < 2^n).

nonn

W. Bomfim (webonfim(AT)bol.com.br), Oct 27 2008

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