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%I #49 Sep 23 2022 03:26:52
%S 1,2,4,4,8,8,16,16,16,16,32,32,64,64,64,64,128,128,256,256,256,256,
%T 512,512,512,512,512,512,1024,1024,2048,2048,2048,2048,2048,2048,4096,
%U 4096,4096,4096,8192,8192,16384,16384,16384,16384,32768,32768,32768,32768
%N a(n) is the number of unitary (and also of squarefree) divisors of n!.
%C Let K(n) be the field that is generated over the rationals Q by adjoining the square roots of the numbers 1,2,3,...,n, i.e., K(n) = Q(sqrt(1),sqrt(2),...,sqrt(n)); a(n) is the degree of this field over Q. - Avi Peretz (njk(AT)netvision.net.il), Mar 20 2001
%C For n>1, a(n) is the number of ways n! can be expressed as the product of two coprime integers p and q such that 0 < p/q < 1, if negative integers are considered as well. This is the answer to the 2nd problem of the International Mathematical Olympiad 2001. Example, for n = 3, the a(3) = 4 products are 3! = (-2)*(-3) = (-1)*(-6) = 1*6 = 2*3. - _Bernard Schott_, Jan 21 2021
%C a(n) = number of subsets S of {1,2,...,n} such that every number in S is a prime. - _Clark Kimberling_, Sep 17 2022
%H Charles R Greathouse IV, <a href="/A048656/b048656.txt">Table of n, a(n) for n = 1..10000</a>
%H International Mathematical Olympiad 2001, <a href="/A048656/a048656.pdf">Hong Kong Preliminary Selection Contest</a>, Problem 2.
%H <a href="/index/Di#divseq">Index to divisibility sequences</a>.
%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F A001221(n!) = A000720(n) so a(n) = A034444(n!) = 2^A000720(n).
%e For n=7, n! = 5040 = 16*9*5*7 with 4 distinct prime factors, so a(7) = A034444(7!) = 16.
%e The subsets S of {1, 2, 3, 4} such that every number in S is a prime are these: {}, {2}, {3}, {2, 3}; thus, a(4) = 4. - _Clark Kimberling_, Sep 17 2022
%t Table[2^PrimePi[n], {n, 1, 70}] (* _Clark Kimberling_, Sep 17 2022 *)
%o (PARI) a(n)=2^primepi(n) \\ _Charles R Greathouse IV_, Apr 07 2012
%Y Cf. A000720, A001221, A034444, A357214, A357215.
%K nonn
%O 1,2
%A _Labos Elemer_
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