[seqfan] Re: mth squarefree = mth prime power

[Peter Pein]

Leroy Quet schrieb:
> Yes, the second squarefree is the second power of a prime. They both are 2.
> (Primes are both squarefree and powers of primes.)
> 
> All terms of my (short) sequence are either 1 or prime.
> 
> Thanks,
> Leroy Quet
> 
> 
> --- On Thu, 5/21/09, Peter Pein <petsie at dordos.net> wrote:
> 
>> From: Peter Pein <petsie at dordos.net>
>> Subject: [seqfan] Re: mth squarefree = mth prime power
>> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
>> Date: Thursday, May 21, 2009, 11:23 AM
>> Richard Mathar schrieb:
>>> lq> Leroy Quet q1qq2qqq3qqqq at yahoo.com
>>> lq> Sat May 16 18:45:48 CEST 2009
>>> lq> 
>>> lq> I just sent in this sequence. Is it complete?
>>> lq> 
>>> 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> 
>>> lq> 
>>> 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.)
>>>
>> Sorry, I don't get it...
>> Say m=2; then the second squarefree number shall be a
>> second power of a prime???
>> Enlighten me please!
>> Cheers,
>> Peter
>>
>>
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maybee I'm too dumb. 2 is the second squarefree integer. Therefore m=2. But if 
you raise any prime to the power of 2, you'll never get a squarefree number 
(by definition)?