[seqfan] Re: Questions on A230492
Jack Brennen
jfb at brennen.net
Sat Dec 21 20:05:56 CET 2013
I assumed the fact that all non-trivial multiples of a perfect number
are abundant was so well-known it could be stated without proof.
See Corollary 1 on the OEIS wiki:
http://oeis.org/wiki/Abundant_numbers
On 12/21/2013 9:19 AM, Vladimir Shevelev wrote:
>
> One moment, in my opinion, need an exactness.
> It is clear that terms of A230492 could not be abundant. Thus they are either
> perfect or deficient. In particular, if there is no an odd perfect, then all odd terms
> of the sequence are deficient. At the end you write: "If N is divisible by
> 2^x*(2^(x+1)-1), for x>0, N is either a perfect number or an abundant number."
> It is not quite clear.
> I think that the completion of proof should be as the following.
> Since N is divisible by 2^(x+1)-1, then the smallest prime divisor p<= 2^(x+1)-1;
> but you proved that p>= 2^(x+1)-1. Thus p= 2^(x+1)-1 is a Mersenne prime.
> From this it follows that x+1(=r) is prime. So, N is divisible by 2^(r-1)*(2^r-1).
> Thus, by Euclid-Euler theorem (here it is sufficient to use Euclid theorem) N is
> divisible by an even perfect number. But why N is also perfect? - you need to prove.
> For example, why all deficient numbers could not be multiple of perfect 6?
> (most likely, it is true). Conversly, perfect 28 is not a multiple of another perfect. Thus, you should prove the EQUALITY N=2^(r-1)*(2^r-1).
>
> Vladimir
> ________________________________________
> From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Jack Brennen [jfb at brennen.net]
> Sent: 20 December 2013 20:52
> To: seqfan at list.seqfan.eu
> Subject: [seqfan] Re: Questions on A230492
>
> Proof that all even terms of the sequence are perfect numbers:
>
> A number N is in A230492 if each divisor of N is >= the sum of all
> of the smaller divisors of N, with equality in at least one case.
>
> No power of 2 is in the sequence; it's easy to see you never get
> equality in that case. So we have some set of odd divisors of N:
> (1,p,q,r,...).
>
> It's also easy to see that if N is divisible by 2^x, then the
> smallest odd divisor p >= 2^(x+1)-1. (If smaller than that,
> you would have a divisor which is less than the sum of the
> smaller divisors).
>
> Finally, it's easy to see that the divisors (p,p*2,...,p*2^x)
> must occur with no intervening divisors -- any intervening divisors
> would also lead to a divisor less than the sum of the smaller divisors.
>
> So the set of divisors of N can be grouped like this, when sorted
> from smallest to largest:
>
> (1,2,...,2^x)
> (p,p*2,...,p*2^x)
> (q,q*2,...,q*2^x)
> ...
>
> In order to get equality in at least one case -- where a divisor
> is equal to the sum of the smaller divisors -- you will need to
> introduce it at the beginning of one of the rows as laid out
> above. (For instance, if p*2 is the sum of the smaller divisors,
> so is p.)
>
> But each row is divisible by 2^(x+1)-1. So if the first term of
> any row is equal to the sum of the previous rows, it must be
> divisible by 2^(x+1)-1, meaning than N must be divisible by 2^(x+1)-1.
>
> If N is divisible by 2^x*(2^(x+1)-1), for x>0, N is either a perfect
> number or an abundant number. Abundant numbers can't be in the
> sequence because the largest divisor is less than the sum of
> the smaller divisors. So N must be a perfect number.
>
>
>
> On 12/20/2013 9:42 AM, Jack Brennen wrote:
>> No need to check the even perfect numbers; yes, they are all in
>> A230492. The proof is fairly easy to see because each even
>> perfect number satisfies the property that any divisor N is
>> at least equal to the sum of all of the smaller divisors.
>>
>>
>>
>>
>>
>> On 12/19/2013 6:21 PM, Donovan Johnson wrote:
>>> Michel,
>>>
>>> The 7th perfect number is also a term in A230492. I checked all 2^38-1
>>> positive sums. Each one is >= the previous sum. 524288 times they are
>>> equal.
>>>
>>> Donovan
>>>
>>>
>>>
>>>
>>>
>>> On Thursday, December 19, 2013 3:52 PM, Donovan Johnson
>>> <donovan.johnson at yahoo.com> wrote:
>>>
>>> Michel,
>>>
>>> The 6th perfect number is a term in A230492. I checked all 2^34-1
>>> positive sums. Each one is >= the previous sum. 131072 times they are
>>> equal.
>>>
>>> I am checking the 7th perfect number right now.
>>>
>>> Regards,
>>> Donovan
>>>
>>>
>>>
>>>
>>>
>>>
>>> On Thursday, December 19, 2013 12:33 PM, "michel.marcus at free.fr"
>>> <michel.marcus at free.fr> wrote:
>>>
>>>
>>> Hi SeqFans,
>>>
>>> I have recently extended A230492 to 250 terms. Unless mistaken, it
>>> still verify the property that even terms are perfect.
>>> I have also checked that the 5 first perfect numbers (A000396) belong
>>> to this sequence (the 6th one is taking some time).
>>>
>>> Is it possible to prove that perfect numbers are in A230492, but no
>>> other even numbers ?
>>> And what could be said about the odd terms ?
>>>
>>> I have searched the OEIS and found only 2 other sequences with the
>>> same property:
>>> A034897 Hyperperfect numbers.
>>> A225417 Composite numbers which contain their sum of
>>> aliquot parts as a substring.
>>> Are there others ?
>>>
>>> Thank you for your help.
>>> Michel
>>>
>>>
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>>>
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>>>
>>
>>
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