[seqfan] Re: mth squarefree = mth prime power
Maximilian Hasler
maximilian.hasler at gmail.com
Thu May 21 16:51:41 CEST 2009
The second power of a prime does not mean a prime squared,
but the second NUMBER which is a power (including exponents 1 or 0) of
some prime.
Maximilian
On Thu, May 21, 2009 at 10:35 AM, Peter Pein <petsie at dordos.net> wrote:
> 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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>>>
>>
>>
>>
>>
>>
>>
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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)?
>
>
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