[seqfan] Re: Questions on A230492
Vladimir Shevelev
shevelev at bgu.ac.il
Fri Dec 20 16:09:01 CET 2013
It is easy to show that, if N is an even perfect number, then it is in the sequence.
Indeed, by Euclid-Euler theorem, N is an even perfect number iff it has form 2^(p-1)*(2^p-1), where p and 2^p-1 are primes. Thus N has 2*p divisors:
N, 2^(p-2)*(2^p-1),...,2*(2^p-1), 2^p-1, 2^(p-1), 2^(p-2),...,2,1 (1)
Note that from them the divisors divisiblle by 2^p-1 are equal sum of all previous.
For example, 2^p-1=2^(p-1)+2^(p-2)+...+1; 2*(2^p-1)=2*p-1+
2^(p-1)+...+1, etc.
Using your definition, we see that the sums being ordered by their characteristic functions, seen as binary number are 1,2,3,...,2^p-1,2^p-1, 2^p, 2^p+1,2^p+2,...,
2^(p+1)-2, 2^(p+1)-2 , 2^(p+1)-1,..., which are increasing, except for
repetition two times of several sums. Let us describe them. Let s be a number
from the interval [0, 2^p-2]. Then consecutive numbers of the form s*2^p+2^p-1,
(s+1)*2^p and only they have the same sums of divisors of N corresponding of their
binary 1's. So such sums repeat two times. So every even perfect N is in your sequence.
Regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of michel.marcus at free.fr [michel.marcus at free.fr]
Sent: 19 December 2013 18:16
To: seqfan at list.seqfan.eu
Subject: [seqfan] Questions on A230492
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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