Indexes' Difference Divides Sum Of 2 Terms

[f.firoozbakht at sci.ui.ac.ir]

On Fri, 13 Feb 2004, Leroy Quet wrote:

> Let b(1) = 1;
> Let b(m) = lowest positive unpicked integers such that:
> (m-k) divides evenly into (b(m) + b(k))
> for EACH k, 1 <= k <= m-1.

> I get (again, figured by hand, so not completely believable): 
> b(m) : 1, 2, 3, 8, 7, 54,...
> (That last IS 54, not "5, 4".)
> 
> Is this sequence a permutation of the integers >= 1?
> 
> (I am not even sure it is infinite, without looking harder at the 
> sequence's mathematics.)


> Is this sequence a permutation of the integers >= 1?


No,your sequence is finite,in fact we can show b(7) doesn't exist.
So for n > 6 ,b(n) doesn't exist.

b(7) doesn't exist because:

we have 7-1 divides b(7)+ b(1) ,since b(1)=1 we get  b(7) == -1  mod 6 
,so  b(7) == -1  mod 3    (I) 

we also have 7-4 divides b(7)+ b(4),since b(4)=8 we get  b(7) == 1 mod 3 (II)

 From (I) ,(II)  we conclude that b(7) doesn't exist.

Regards,

Farideh



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