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Number of semiprimes s such that 2^n < s <= 2^(n+1).
15

%I #20 Mar 05 2019 01:56:04

%S 0,1,1,4,4,12,20,40,75,147,285,535,1062,2006,3918,7548,14595,28293,

%T 54761,106452,206421,401522,780966,1520543,2962226,5777162,11272279,

%U 22009839,43006972,84077384,164482781,321944211,630487562,1235382703

%N Number of semiprimes s such that 2^n < s <= 2^(n+1).

%C The partial sum equals the number of Pi_2(2^n) = A125527(n).

%H Dana Jacobsen, <a href="/A120033/b120033.txt">Table of n, a(n) for n = 0..62</a> (first 48 terms from Charles R Greathouse IV, corrected a(47)-a(48))

%e (2^2, 2^3] there is one semiprime, namely 6. 4 was counted in the previous entry.

%t SemiPrimePi[n_] := Sum[PrimePi[n/Prime[i]] - i + 1, {i, PrimePi[Sqrt[n]]}]; t = Table[SemiPrimePi[2^n], {n, 0, 35}]; Rest@t - Most@t

%o (PARI) pi2(n)=my(s,i); forprime(p=2, sqrt(n), s+=primepi(n\p); i++); s - i * (i-1)/2

%o a(n)=pi2(2^(n+1))-pi2(2^n) \\ _Charles R Greathouse IV_, May 15 2012

%o (Perl) use ntheory ":all"; print "$_ ",semiprime_count(1+(1<<$_), 1<<($_+1)),"\n" for 0..48; # _Dana Jacobsen_, Mar 04 2019

%o (Perl) use ntheory ":all"; my $l=0; for (0..48) { my $c=semiprime_count(1<<($_+1)); print "$_ ",$c-$l,"\n"; $l=$c; } # _Dana Jacobsen_, Mar 04 2019

%Y Cf. A001358, A072000, A066265, A036378, A120033-A120043.

%K nonn

%O 0,4

%A _Jonathan Vos Post_ and _Robert G. Wilson v_, Mar 20 2006