[seqfan] Re: mth squarefree = mth prime power
Edwin Clark
eclark at math.usf.edu
Sat May 16 19:43:39 CEST 2009
Nevermind, I now see what you meant. Namely find the values A000961(m)
such that A006117(m) = A0000961(m).
On Sat, 16 May 2009, Edwin Clark wrote:
>
> 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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