[seqfan] Re: mth squarefree = mth prime power

[Leroy Quet]

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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