Sequence: (m-k) divides a(m)*a(k)

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

On Mon, 16 Feb 2004, Leroy Quet wrote:

> Let a(1)  = 1; 
> Let a(m) = lowest positive unpicked integers such that:
> (m-k) divides evenly into (a(m) * a(k))
> for EACH k, 1 <= k <= m-1.
> I get (again, figured by hand, so not completely believable):
> a(m) : 1, 2, 4, 3, 12, 30,...
> Now this looks less like it might be a permutation of the positive 
> integers, less likely a permutation than the sequence in the link above, 
> in any case.
> 
> Is it a permutation of the positive integers?

No ,it isn't a permutation of the positive integers.
In fact I prove that for m > 5 , 10 divides a(m).

Proof:

We know m > 5 and  a(1)=1, a(2)=2, a(3)=4, a(4)=3 & a(5)=12  .
There exist 10 cases;

Case 1. m == 0  mod 10 ,
 since  m-5 divides   a(5)*a(m) ,we deduce that  5 divides a(m)  (I-1)
 since  m-4 divides   a(4)*a(m) ,we deduce that  2 divides a(m)  (I-2)
from (I-1) and (I-2) we conclude that 10 divides a(m).

Case 2. m == 1  mod 10 ,
 since  m-1 divides   a(1)*a(m) ,we deduce that 10 divides a(m).  

Case 3. m == 2  mod 10 ,
 since  m - 2 divides   a(2)*a(m) ,we deduce that  5 divides a(m)  (III-1)
 since  m - 4 divides   a(4)*a(m) ,we deduce that  2 divides a(m)  (III-2)
so from (III-1) and (III-2) we conclude that 10 divides a(m).

Case 4. m == 3  mod 10 ,
  since  m - 1 divides   a(1)*a(m) ,we deduce that  2 divides a(m) (IV-1)
  since  m - 3 divides   a(3)*a(m) ,we deduce that  5 divides a(m) (IV-2)
so from (IV-1) and (IV-2) we conclude that 10 divides a(m).

Case 5. m == 4  mod 10 ,
  since  m - 4 divides   a(4)*a(m) ,we deduce that  10 divides a(m) .

Case 6. m == 5  mod 10 ,
  since  m - 1 divides   a(1)*a(m) ,we deduce that  2 divides a(m) (VI-1)
  since  m - 5 divides   a(5)*a(m) ,we deduce that  5 divides a(m) (VI-2)
so from (VI-1) and (VI-2) we conclude that 10 divides a(m).

Case 7. m == 6  mod 10 ,
  since  m - 1 divides   a(1)*a(m) ,we deduce that  5 divides a(m) (VII-1)
  since  m - 4 divides   a(4)*a(m) ,we deduce that  2 divides a(m) (VII-2)
so from (VII-1) and (VII-2) we conclude that 10 divides a(m).

Case 8. m == 7  mod 10 ,
  since  m - 1 divides   a(1)*a(m) ,we deduce that  2 divides a(m) (VIII-1)
  since  m - 2 divides   a(2)*a(m) ,we deduce that  5 divides a(m) (VIII-2)
so from (VIII-1) and (VIII-2) we conclude that 10 divides a(m).

Case 9. m == 8  mod 10 ,
  since  m - 3 divides   a(3)*a(m) ,we deduce that  5 divides a(m) (IX-1)
  since  m - 4 divides   a(4)*a(m) ,we deduce that  2 divides a(m) (IX-2)
so from (IX-1) and (IX-2) we conclude that 10 divides a(m).

Case 10. m == 9  mod 10 ,
  since  m - 1 divides   a(1)*a(m) ,we deduce that  2 divides a(m) (X-1)
  since  m - 4 divides   a(4)*a(m) ,we deduce that  5 divides a(m) (X-2)
so from (X-1) and (X-2) we conclude that 10 divides a(m).

Hence in all cases we see 10 divides a(m) and the proof is complete.

a(m) for m=1,2,...,20 are:

           1,2,4,3,12,30,60,420,840,1260,630,27720,4620,60060,
           30030,120120,240240,2042040,3063060,232792560

Regards,

Farideh






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