[seqfan] Re: Questions on A230492

[Vladimir Shevelev]

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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