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A048656 a(n) is the number of unitary (and also of squarefree) divisors of n!. 33
1, 2, 4, 4, 8, 8, 16, 16, 16, 16, 32, 32, 64, 64, 64, 64, 128, 128, 256, 256, 256, 256, 512, 512, 512, 512, 512, 512, 1024, 1024, 2048, 2048, 2048, 2048, 2048, 2048, 4096, 4096, 4096, 4096, 8192, 8192, 16384, 16384, 16384, 16384, 32768, 32768, 32768, 32768 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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
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
a(n) = number of subsets S of {1,2,...,n} such that every number in S is a prime. - Clark Kimberling, Sep 17 2022
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
International Mathematical Olympiad 2001, Hong Kong Preliminary Selection Contest, Problem 2.
FORMULA
A001221(n!) = A000720(n) so a(n) = A034444(n!) = 2^A000720(n).
EXAMPLE
For n=7, n! = 5040 = 16*9*5*7 with 4 distinct prime factors, so a(7) = A034444(7!) = 16.
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
MATHEMATICA
Table[2^PrimePi[n], {n, 1, 70}] (* Clark Kimberling, Sep 17 2022 *)
PROG
(PARI) a(n)=2^primepi(n) \\ Charles R Greathouse IV, Apr 07 2012
CROSSREFS
Sequence in context: A166632 A116596 A248692 * A107848 A285273 A353946
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified March 29 10:58 EDT 2024. Contains 371268 sequences. (Running on oeis4.)