[seqfan] Re: patterns continue for larger numbers? generalised question

[Ali Sada]

Thank you Neil.This is impressive. It could be a very interesting problem if we show that wewould get similar results for any power. I hope that we, as the OEIS community,would solve this. 

 

When I triedthe same algorithm with m as the  “smallestperfect power > n,” it seems that we will reach 1 only when n=2^k.

1.      When n=2^k+1 the algorithm goesup to infinity, since the next perfect power is obviously 2^k. (3,7,15,31,….)

2.      When n=(2^k)-(2^k1) , wherek1 is <k, the algorithm will also go to infinity since the continuous divisionby 2 will lead us to the numbers in point 1 above. Example, 32-4=28, 28/4=7,etc.

3.      When n=2^k-3, the algorithmwill go to infinity because the next perfect power is also 2^k (2^k-1 cannot bea perfect power because of the Catalan conjuncture; and 2^k-2 also cannot be a perfectpower because it’s 2 multiplied by an odd number,) and so on.

The other numbers I tried so far get “trapped”and will go up to infinity. Is this a correct assumption?

 

Best,

 

Ali