[seqfan] Re: A process connected with proper divisors of positive integers

[Vladimir Shevelev]

A property of positive integers.
 
If instead of proper divisors of n+1 to begin the same process with the proper divisors of n (including d=1), then it seems that we always obtain n-1. For example, if n=8, then d={1,2,4}; n+1=9,n+2=10,n+4=12; d={3,5,6}; n+3=11,n+5=13,n+6=14; d={7}; n+7=15. Thus we have the set of d's: {1,2,4,3,5,6,7} which contains n-1=7 elements. Firstly I thought that it could be a variant of "law of  small numbers" but Peter Moses verified it for much larger numbers. How to prove this very plausible conjecture?
 
Best regards,
Vladimir


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Monday, June 17, 2013 8:18
Subject: [seqfan] A process connected with proper divisors of positive integers
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Dear SeqFans,
>  
> Let n be a positive integer. We begin our process with n+1. Let 
> n+1 have proper divisors >1: d_1, ..., d_k. Consider proper 
> divisors of  numbers n+d_1,...,n+d_k which not earlier 
> appeared. Let they d^(1)_1,...,d^(1)_t. Further, consider proper
> divisors of  numbers n+d^(1)_1,...,n+d^(1)_t which  
> not earlier appeared, etc. Let a(n) be the total number of different
> divisors which appeared in the considered process. Then sequence 
> {a(n)} begins 0,0,1,0,3,0,5,1,5,0,9,0,11,2,3,0,15,...(A226770).
> For example, for n=9,  the proper divisors >1 of n+1 are 
> 2,5; consider n+2=11 and n+5=14. These numbers give only one 
> "new" proper divisor >1 7; the "new" proper divisors >1 of 
> n+7=16 are 4,8 and n+4=13, n+8=17 do not have proper divisors 
> >1. The set of   
> proper divisors of all considered sums is {2,5,7,4,8}. It 
> contains 5 elements. Thus 
> a(9)=5. 
> It is clear that a(n)=0 iff n=p-1, where p is prime. 
> Furthermore, I believe that a(p)=p-2. What one can say about 
> other n's?
>  
> Best regards,
> Vladimir
> 
>  Shevelev Vladimir‎
> 
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 Shevelev Vladimir‎