[seqfan] Re: Another Collatz-like algorithm

[David Seal]

> Please seethe algorithm below.
> 
> 1. Pick aninteger n>0
> 
> 2. If n iseven, divide by 2. If n is odd, find the least triangular number
> T greater thenn and add n+T.
> 
> 3. Repeatstep 2. with either n/2 or n+T
> 
> Other thanthe powers of 2, the numbers I tested take one of two routes:
>  
> a.      They go into loops that have5 at their lowest point.Ex. 31+36=67;
> 67+78=145; 145+153=298; 298/2=149; 149+153=302;302/2=151; 151+153=304;
> 304/16=19; 19+21=40; 40/8=5.5+6=11; 11+15=26; 26/2=13; 13+15=28; 28/4=7;
> 7+10=17; 17+21=38; 38/2=19; 19+21=40; 40/8=5.
> 
> b.      They go up to a point wherethe software doesn’t work anymore.
> Maybe they go up to infinity, but I cannotconfirm. Ex.173, 285, 331, etc. 

I notice that there are non-power-of-2 positive integers that end up in the 1+3=4->1 loop and so are in neither of those categories. Defining a 'step' between odd positive integers to be the process of calculating n+T and then dividing by the highest power of 2 that the result is divisible by, the smallest example that gets to a power of 2 in a single step is 1013: 1013+1035=2048->1. The next few are 16297, 65369, 67105823, 268434877, 1099511480047.

But there's a smaller example that does it in two steps: 991+1035=2026->1013; 1013+1035=2048->1. And 3923+4005=7928->991 then shows that there's a 3-step example; 7817+7875=15692->7846->3923 that there's a 4-step example; 7759+7875=15634->7817 that there's a 5-step example, etc. I think that given any odd positive integer, it has an infinite number of single-step predecessors, but haven't yet worked out all the details of a proof.

Best regards,

David