An interesting discovery (nothing to do with Collatz.co.uk)

[mugbuff]

Counting as a step a complete calculation of the next odd intetger,
e.g. a step = (3x+1)/2^n, I was investigating certain start values and
counting the steps taken to converge to 1.
I started with 2^k+c and for various values of c listed all the counts
for k = 1 to 100.
I thought the counts would look random, but certain counts seem to
predominate, e,g 32,44,56,162, 191and 210 over ranges of k (and
possibly 309 for high k).
Could anyone confirm this pattern ?
Putting c = n and 4n+1 produces a sequence of counts that eventually
match.

[mensa...@aol.com]

On Dec 25, 3:43�pm, mugbuff <at...@waitrose.com> wrote:
> Counting as a step a complete calculation of the next odd intetger,
> e.g. a step = (3x+1)/2^n, I was investigating certain start values and
> counting the steps taken to converge to 1.
> I started with 2^k+c and for various values of c listed all the counts
> for k = 1 to 100.
> I thought the counts would look random, but certain counts seem to
> predominate, e,g 32,44,56,162, 191and 210 over ranges of k (and
> possibly 309 for high k).
> Could anyone confirm this pattern ?

Sure.

## c = 1
## count: 1 occurences: 1
## count: 2 occurences: 1
## count: 3 occurences: 1
## count: 6 occurences: 1
## count: 8 occurences: 2
## count: 10 occurences: 2
## count: 15 occurences: 2
## count: 20 occurences: 1
## count: 32 occurences: 5
## count: 39 occurences: 1
## count: 44 occurences: 8
## count: 56 occurences: 10
## count: 61 occurences: 4
## count: 80 occurences: 1
## count: 85 occurences: 3
## count: 109 occurences: 8
## count: 138 occurences: 4
## count: 162 occurences: 6
## count: 174 occurences: 2
## count: 191 occurences: 22
## count: 210 occurences: 12
## count: 309 occurences: 3

Don't expct randomness because the sequences usually
converge long before they reach 1. For example,
the 1st and 2nd of the two 309 step cases above have
259 steps in common. The 2nd and 3rd have 243 steps
in common.

And it's usually not a coincidence that the uncommon
parts have the same step count. Often, highly patterened
numbers like 2**k+c exhibit binary resonnance until the
pattern is consumed. Since I chose c=1, every starting
value is 1000...0001 in binary which is a known resonator,
so numbers with close k values will have similar periods
of resonnance at the start of the sequence which can
easily be balanced by the residue of the sequence before
reaching the convergence point.

Also, starting values close in value are often found
in close proximity on the Collatz graph

v_w
a_b x_y
c z
d_e e
f f

Here, f is common to both sequences, a & v are similar
values but if we only count steps (the "_"), they are
the same up to the common point.

Here's a study I did on Mersenne numbers (c=-1) that
exhibits this behavior as plataeus of identical step
counts that cluster about the mean calculated value
of 4.819 steps/bit for Mersenne numbers:

## <http://members.aol.com/mensanator666/Page.htm>

> Putting c = n and 4n+1 produces a sequence of counts that eventually
> match.

Not sure what you mean by this.

[mugbuff]

put c = 7 then 29 then 117 for example.
I aslo put in huge random values for c, like 456738457 ( not that I
tried
that particular value) and the same counts kept occurring.

[mensa...@aol.com]

On Dec 26, 12:21 am, mugbuff <at...@waitrose.com> wrote:
> put c = 7 then 29 then 117 for example.

Which are 111, 11101 and 1110101 in binary, respectively,
which make them relatively insignificant as k goes to 100.

c=1 c=7 c=29 c=117
count: 1 1
count: 2 1
count: 3 1
count: 4 2 2 2
count: 5 1 1 1
count: 6 1 1 1 1
count: 8 2 1 1
count: 10 2 2 2 3
count: 11 1 1 1
count: 13 1 1 1
count: 15 2 2 2 2
count: 20 1 2 2 2
count: 27 1 1 1
count: 32 5 5 5 5
count: 34 1
count: 37 1 1 1
count: 39 1 1 2 2
count: 44 8 6 6 6
count: 56 10 10 10 10
count: 61 4 4 4 4
count: 80 1 1 1 1
count: 85 3 3 3 3
count: 109 8 8 8 8
count: 138 4 4 4 4
count: 162 6 6 6 6
count: 174 2 2 2 2
count: 191 22 22 22 22
count: 210 12 11 10 9
count: 309 3 3 2 1

And while they are similar, they're not identical.

Before you think this too spooky, consider one of the
ones with 309 steps. Or 309 odd numbers. There is a mean
of 2 even numbers for every odd, so those 309 odd numbers
imply 618 even numbers. Now, how many ways can 618 balls
be placed in 309 urns such that each urn contains at least
1 ball? C(618-1,309-1) or

174492780558417293865771989230596657905904181694817273530206
483126338444928373706902419659908515704518416030264817014875
903949713246162796835099230825178927660791905310899573348586
77360

so getting a lot of dulpicate step counts is not surprising.

> I aslo put in huge random values for c, like 456738457 ( not that I
> tried
> that particular value)

I did:

## c = 456738457
## count: 29 occurences: 1
## count: 36 occurences: 1
## count: 38 occurences: 1
## count: 41 occurences: 2
## count: 46 occurences: 1
## count: 53 occurences: 1
## count: 58 occurences: 3
## count: 61 occurences: 2
## count: 63 occurences: 1
## count: 66 occurences: 1
## count: 68 occurences: 1
## count: 70 occurences: 2
## count: 74 occurences: 1
## count: 75 occurences: 1
## count: 77 occurences: 4
## count: 78 occurences: 1
## count: 80 occurences: 3
## count: 82 occurences: 1
## count: 85 occurences: 1
## count: 94 occurences: 2
## count: 97 occurences: 3
## count: 103 occurences: 1
## count: 106 occurences: 3
## count: 109 occurences: 3
## count: 116 occurences: 2
## count: 119 occurences: 1
## count: 133 occurences: 3
## count: 138 occurences: 1
## count: 147 occurences: 2
## count: 162 occurences: 1
## count: 174 occurences: 8
## count: 191 occurences: 15
## count: 200 occurences: 1

## count: 210 occurences: 12

## count: 215 occurences: 6
## count: 220 occurences: 2
## count: 256 occurences: 2
## count: 306 occurences: 2
## count: 374 occurences: 1

> and the same counts kept occurring.

Sure, because we frequently pass throught the same
part of the graph. But note that even though old favorites
like 191 & 210 still show up, we get a lot of step counts
not seen before. And not just big ones like 374. In the
first table there were only 3 step counts between 60
and 109, here there are 16. But also note that 456738457
(11011001110010100011010011001 binary) is much more
signifcant when k is large.

Also, look at the difference in step counts when c=-1.

## c = -1

## count: 1 occurences: 1
## count: 2 occurences: 1

## count: 5 occurences: 2

## count: 15 occurences: 2

## count: 20 occurences: 2
## count: 39 occurences: 2
## count: 44 occurences: 2
## count: 56 occurences: 4
## count: 61 occurences: 2
## count: 80 occurences: 2
## count: 109 occurences: 2
## count: 138 occurences: 2

## count: 162 occurences: 6
## count: 174 occurences: 2

## count: 191 occurences: 10

## count: 210 occurences: 12

## count: 256 occurences: 2
## count: 306 occurences: 2
## count: 309 occurences: 10
## count: 333 occurences: 10
## count: 386 occurences: 2
## count: 492 occurences: 2
## count: 528 occurences: 18

Here, the magnitude of c is less important than the fact that,
in binary, our starting number becomes mostly 1's instead of
mostly 0's.

There are two major factors involved in step counts:
- number of bits
- patterning of bits

The general rule of thumb is: larger values (having more bits)
tend to have larger step counts.

Of course, there are numerous exceptions to that rule, often
dependent on bit patterns.

For example, the Floating Zero exercise.

Here, the formula is (2**100-1)-(2**(100-k)) for k=1 to 99.
That makes every starting numberr 100 bits making the number
of bits criteria constant. So step count differences will be
due to patterning. Here, there are always 99 1-bits with a
single 0-bit at the kth position. Thus, at each starting value,
the 0-bit "floats" from the MSB to the LSB. Of course, that
means the magnitude of each starting number is increasing.

And, yet, the general trend (barring some hailstoning) is for
step counts to DECREASE as the 0-bit floats to the LSB.

## floating 0 test
## k: 1 steps: 528
## k: 2 steps: 527
## k: 3 steps: 458
## k: 4 steps: 528
## k: 5 steps: 492
## k: 6 steps: 456
## k: 7 steps: 475
## k: 8 steps: 427
## k: 9 steps: 521
## k: 10 steps: 391
## k: 11 steps: 468
## k: 12 steps: 492
## k: 13 steps: 516
## k: 14 steps: 506
## k: 15 steps: 456
## k: 16 steps: 456
## k: 17 steps: 415
## k: 18 steps: 350
## k: 19 steps: 456
## k: 20 steps: 468
## k: 21 steps: 415
## k: 22 steps: 439
## k: 23 steps: 386
## k: 24 steps: 386
## k: 25 steps: 453
## k: 26 steps: 456
## k: 27 steps: 456
## k: 28 steps: 439
## k: 29 steps: 439
## k: 30 steps: 403
## k: 31 steps: 316
## k: 32 steps: 439
## k: 33 steps: 516
## k: 34 steps: 333
## k: 35 steps: 350
## k: 36 steps: 350
## k: 37 steps: 316
## k: 38 steps: 456
## k: 39 steps: 350
## k: 40 steps: 453
## k: 41 steps: 516
## k: 42 steps: 316
## k: 43 steps: 306
## k: 44 steps: 386
## k: 45 steps: 350
## k: 46 steps: 350
## k: 47 steps: 516
## k: 48 steps: 350
## k: 49 steps: 306
## k: 50 steps: 350
## k: 51 steps: 297
## k: 52 steps: 492
## k: 53 steps: 453
## k: 54 steps: 374
## k: 55 steps: 350
## k: 56 steps: 297
## k: 57 steps: 350
## k: 58 steps: 333
## k: 59 steps: 386
## k: 60 steps: 333
## k: 61 steps: 333
## k: 62 steps: 326
## k: 63 steps: 309
## k: 64 steps: 256
## k: 65 steps: 386
## k: 66 steps: 326
## k: 67 steps: 297
## k: 68 steps: 386
## k: 69 steps: 306
## k: 70 steps: 326
## k: 71 steps: 306
## k: 72 steps: 306
## k: 73 steps: 210
## k: 74 steps: 386
## k: 75 steps: 210
## k: 76 steps: 306
## k: 77 steps: 306
## k: 78 steps: 306
## k: 79 steps: 306
## k: 80 steps: 386
## k: 81 steps: 256
## k: 82 steps: 386
## k: 83 steps: 309
## k: 84 steps: 256
## k: 85 steps: 306
## k: 86 steps: 210
## k: 87 steps: 256
## k: 88 steps: 256
## k: 89 steps: 309
## k: 90 steps: 309
## k: 91 steps: 256
## k: 92 steps: 256
## k: 93 steps: 528
## k: 94 steps: 492
## k: 95 steps: 306
## k: 96 steps: 256
## k: 97 steps: 256
## k: 98 steps: 256
## k: 99 steps: 528

What's happening is that the lone 0-bit acts as a firebreak
to the binary resonnace induced by a pattern of contiguous
1-bits at the LSB. As the 0 floats towards the LSB, the
resonating pattern (which greatly influences the step count)
keeps getting smaller and the step count decreases accordingly.
But note that at k=93 & 94 & 99 the step counts zoom back up.

We hit a different kind of resonnance at those points. Elswhere,
there is simple ressonance with the 0-bit acting as a firebreak.
But there is also complex resonnace (bi-stable, tri-stable,
quadra-stable, etc.) and these can involve not only multi-bit
rep-unit patterns, but also a resonator (a bit pattern distinct
from the rep-units at the LSB). So, instead of acting like a
firebreak with a small pattern at the LSB, it suddenly appears
to be a resonator attached to a 96-bit pattern and a whole new
set of rules determining step counts is invoked.