## The b-numbers again

If we write the (hypothetical, from now on understood) sequence from TM #4 (chaotic), 5-state, 2-symbol tape of Marxen and Buntrock in rows of 8, the 3rd column has a complex structure.

Suppose we use “index 1”-notation, which just means that the first term of the sequence, 3, is written s(1) instead of s(0) as would be the case with arrays in C, C++, and so on.

Then the third column is made up of s(3) = 5, s(11), s(19), and in general s(8j + 3) for some integer j with 0<= j < oo.

Empirically, s(j) <= j+2. When j is congruent to 3 modulo 8, so that s(j) is a third-column term, it happens that:

s(j) = j+2 exactly.

When this happens, in column 3, I call j+2 a b-number (basic number). Or in other words, if k is such that s(k-2) = k, and

k == 5 (mod 3), and of course k >2, then by definition k is a basic number or b-number.

When k == 5 (mod 3), but s(k-2) is not k, then I’ve called k a forbidden number. This is justified by studying the binary expansion of s(k-2) when k is a forbidden number.

Empirical study gives some credence to the belief that, when k is not a b-number, s(k-2) can be obtained from ‘k’ by omitting one or more of the most significant bits of k, all consecutive bits, while always leaving perhaps 3 bits of ‘k’ to yield 101_(Base 2) = 5_(Base 10).

Moreover, a first look gave me a clue that when k is forbidden, then s(k-2) can be obtained from the binary expansion of k by omitting the most significant ‘1’ bit from k in base 2, and going on striking-out mentally the most significant ‘1’ bits (starting at the leftmost bit) and stopping when and only when the modified number is a b-number. Thus, if k is forbidden, s(k-2) would still be a b-number or basic number.

The new feature I hadn’t looked at previously is the count of b-numbers by bit-length.

The following table has this , with a coming explanation:

n count(n)

======================

10 22

11 32

12 48

13 71

14 106

15 158

16 234

17 348

18 518

19 772

20 1152

n is a bit-length of a positive integer.

Examples: 6 has bit-length 3, 2^k has

bit length k+1, 100 has bit-length 7,

ans 2^k – 1 has bit-length k.

count(n) is the number of b-numbers with bit-length equal to n.

So there are 22 b-numbers with 10 bits, which means in the

range 512 to 1023 inclusive. And so on up to 20 bits.

You’ll see that c(n+1) is approximately (3/2)*c(n),

better than chance in my opinion.

In other words, while there are 48 b-numbers

in the range 2048 to 4095, there are only

23 in the range 4096 to 8191 (which is

twice as long).

To if we increase n by +1, the frequency of

b-numbers is reduced by a factor of 4 approximately.

Perhaps these correspond to two extra “pseudo-random”

boolean 0/1 conditions for a 13-bit b-number, as

compared to a 12-bit b-number. This line of

inquiry has not so far been pursued further.

If you look at the first few dozen b-numbers,

you might notice that 2^k – 3 is a b-number

for k = 3, 4, 5, 6, 7, … 15, …

There are other “magic numbers” like 3.

While consistent, these “magic numbers”

and their patterns still give complicated

sequences.

Perhaps I missed something that could be guessed

from the fact that c(n+1) is approximately (3/2)*c(n).

I find this problem of pattern recognition both

tantalizing and frustrating, because it is

produced by such a seemingly simple 5-state,

2-symbol Turing machine, starting with a

blank tape.

The first 100 b-numbers :

5

13

21

29

37

45

53

61

77

85

93

101

109

117

125

149

173

181

189

205

213

221

229

237

245

253

309

341

365

373

381

405

429

437

445

461

469

477

485

493

501

509

629

693

725

749

757

765

821

853

877

885

893

917

941

949

957

973

981

989

997

1005

1013

1021

1205

1237

1261

1277

1397

1461

1493

1517

1525

1533

1653

1717

1749

1773

1781

1789

1845

1877

1901

1909

1917

1941

1965

1973

1981

1997

2005

2013

2021

2029

2037

2045

2229

2285

2485

2517

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b-numbers with up to 10 bits | meditationataeSeptember 26, 2017 at 6:02 am