[seqfan] Re: mth squarefree = mth prime power

Edwin Clark eclark at math.usf.edu
Sat May 16 19:26:04 CEST 2009


Leroy,

I don't follow. For example, 2 is (by your definition) the mth-square free 
integer where m = 2. But 2 is not of the form p^2. In fact if m > 1 a 
square free number will never be an mth power of a prime.

What am I missing?

--Edwin



On Sat, 16 May 2009, Leroy Quet wrote:

>
> I just sent in this sequence. Is it complete?
>
> %I A160513
> %S A160513 1,2,3,7,17,19
> %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.
> %C A160513 Is this sequence complete?
> %C A160513 1 here is considered to be both the first squarefree integer and the first power of a prime.
> %Y A160513 A000961,A005117
> %K A160513 fini,more,nonn
> %O A160513 1,2
>
>
> 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.
> (Unless I am missing something here.)
>
> Thanks,
> Leroy Quet
>
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