[seqfan] Re: (kein Betreff)

Hagen von EItzen math at von-eitzen.de
Tue May 19 20:57:55 CEST 2009

    lq> Leroy Quet q1qq2qqq3qqqq at yahoo.com
    lq> Sat May 16 18:45:48 CEST 2009
    lq> I just sent in this sequence. Is it complete?
    lq> %I A160513
    lq> %S A160513 1,2,3,7,17,19
    lq> %N A160513 Positive squarefree integers n such that, if n is the mth squarefree integer, then n is the mth power of a prime as well. 
    lq> %C A160513 Is this sequence complete? 
    lq> %C A160513 1 here is considered to be both the first squarefree integer and the first power of a prime. 
    lq> %Y A160513 A000961,A005117 
    lq> %K A160513 fini,more,nonn
    lq> %O A160513 1,2
    lq> Since the squarefree integers have a density of 6/pi^2; and since the density, as n approaches infinity, of the number of prime powers <= n approaches 0 (right?), then it is certain that this sequence is finite.
    lq> (Unless I am missing something here.)

    rm> An exhaustive search shows that in the index range m <= 100000, these 6 numbers
    rm> are the only A000961(m) such that A000961(m)=A005117(m); that is, the
    rm> next entry would have to be larger than 1295947. So it is very likely
    rm> that the sequence contains only these 6 numbers.


    One just has to make a few limits explicit.
    For example, the number of square-free numbers below  x  is  sqf(x)
    = 6x/pi^2 + O(sqrt(x)).
    More precisely one can show
    | sqf(x) - 6x/pi^2 | < sqrt(x) + 1
    If the m'th squarefree number is  n,  this implies 

    (1)   m > 6n/pi^2  - sqrt(n) - 1.

    And if the m'th prime power is  n,  then there are at least 
    m-pi(n)  numbers below  n  that are not square-free, i.e.  n-sqf(n)
     >= m-pi(n)  or

    (2)   m <= pi(n) + n - 6n/pi^2 +sqrt(n)+1.

    (One should use a better estimate for prime powers, but we'll see
    that it is indeed sufficient to merely note that higher prime powers
    are not squarefree).
     From (1) and (2), we see

    (3)   pi(n) > (12/pi^2 - 1)*n - 2*sqrt(n) -2

    For  n>55  we have the estimate  pi(n) < n/(ln(n) - 4) , hence with (3)

    (4)   1/(ln(n) - 4) > 12/pi^2 -1 -2/sqrt(n) -2/n.

    For big enough  n ,  say  n > 4*10^4 > e^10 , this implies
    1/6 > 1/(ln(n) - 4) > 12/pi^2 -1 -2/sqrt(n) -2/n > 12/pi^2 - 1 -
    2/200 - 2/40000 = 0.2058...
     From this contradiction, it follows that all entries in the
    sequence are  <= 4*10^4.
    Since Richard Mathar's exhaustive search covered an even bigger
    range, we are done.


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