[seqfan] Re: Questions on A230492
Jack Brennen
jfb at brennen.net
Fri Dec 20 19:52:08 CET 2013
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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