Max. Power of 2 and number of Odd integers in the loop

[roupam]

I have found, that if there exists a loop, then if
n = number of odd integers in the loop
k = maximum power of 2 that divides a number in the loop

then, the following conclusions hold...

for positive integers
k <= n+1
for negative integers
k < n log3//log2 - n + 1

Consider the negative loop starting with
-17 -50 -25 -74 -37 -110 -55 -164 -82 -41 -122 -61 -182 -91 -272 -136
-68 -34 and then back to -17

Here...
n = 7
k = 4

4 < 7 * log3/log2 -7 + 1 = 5.09...

[roupam]

it doesnt mean that opposite is true...

[jillbones]

Works for 1, 4, 2, 1 anyway.

[Mensanator]

On Jun 4, 5:03 pm, jillbones <b92...@yahoo.com> wrote:
> Works for 1, 4, 2, 1 anyway.

But *WHY* does it work?

> > 4 < 7 * log3/log2 -7 + 1 = 5.09...- Hide quoted text -
>
> - Show quoted text -

[roupam ghosh]

> Works for 1, 4, 2, 1 anyway.

But *WHY* does it work?

 

I have a proof... but i'll recheck it and then post it here...

 

[jillbones]

On Jun 3, 2:40 am, roupam <roupam.gh...@gmail.com> wrote:

Suppose that the first (n-1)= i numbers

1) O_(k+1) = (3*O_k + 1)/2. Then,

2) O_i ~ O_1 * 1.5^(i).

To complete the loop;

3) O_n = O_1 = (3*O_i + 1)/(2^k)

IMMHO, this will give the maximum possible
value for 'k'.

Bill J