## Archive for the ‘**History**’ Category

## The b-numbers below 256

The numbers k with k == 5 (mod 8), and s(k-2) = k with s(.) the hypothetical sequence generated by Turing Machine #4, are what I’ve called “b-numbers” for basic numbers.

I haven’t succeeded in finding any rule that determines the whole sequence. Below, I copy the 26 b-numbers below 256. I’ve found this to be a hard problem.

b-numbers below 256:

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

## b-numbers with up to 10 bits

I refer to an earlier post a few days ago as an introduction to the problem:

https://meditationatae.wordpress.com/2017/09/23/the-b-numbers-again/

b-numbers are positive integers k with k == 5 (mod 8) such that s(k-2) = k; here, s(1), s(2), s(3), … is the hypothetical integer sequence computed by TM #4 (chaotic) of Heiner Marxen and Buntrock (discovered around 1990).

There are 42 b-numbers in the interval [1, 511], that is b-numbers with 1 through 9 bits. There are 22 b-numbers in the interval [512, 1023], that is numbers with 10 bits.

If y is a 10-bit b-number, then y > 512 in fact, and y-512 is a number in the range [1, 511] with 1 to 9 bits.

It turns out that y-512 is a b-number with 1 to 9 bits, when y is a 10-bit b-number. There are 22 possible 10-bit

b-numbers, and each of them gives us a different 1 to 9-bit b-number when we subtract 512 from it.

There are 42 b-numbers in the interval [1, 511], so if x is one of them, x+512 is a 10-bit number satisfying x == 5 (mod 8). For 22 of those 42, x+512 is also a b-number, and in the remaining 20 cases, x+512 isn’t a b-number.

For the 42 values of x, I’m trying to figure out when

x and x+512 are both b-numbers, contrasted with x being a b-number and x+512 not being a b-number.

The table below has 42 rows, one per b-number in the interval [1,511]. First comes the binary expansion of

x+512, x being a b-number with 1 to 9 bits. Then comes x in decimal, x being a 1 to 9-bit b-number. Then comes x+512, a 10-bit number. Next comes ‘1’ if x+512 is not a b-number, and ‘2’ if x+512 is indeed a 10-bit b-number, which happens in 22 cases.

I looked for a relationship between the ‘2’/’1′ which tells us whether x+512 is a b-number, and x+512 ‘s binary expansion.

Up to now, for this and the 13-bit case, I’ve found no

logical relationship.

The table with 42 rows is copied below:

1 0 0 0 0 0 0 1 0 1 5 517 1

1 0 0 0 0 0 1 1 0 1 13 525 1

1 0 0 0 0 1 0 1 0 1 21 533 1

1 0 0 0 0 1 1 1 0 1 29 541 1

1 0 0 0 1 0 0 1 0 1 37 549 1

1 0 0 0 1 0 1 1 0 1 45 557 1

1 0 0 0 1 1 0 1 0 1 53 565 1

1 0 0 0 1 1 1 1 0 1 61 573 1

1 0 0 1 0 0 1 1 0 1 77 589 1

1 0 0 1 0 1 0 1 0 1 85 597 1

1 0 0 1 0 1 1 1 0 1 93 605 1

1 0 0 1 1 0 0 1 0 1 101 613 1

1 0 0 1 1 0 1 1 0 1 109 621 1

1 0 0 1 1 1 0 1 0 1 117 629 2

1 0 0 1 1 1 1 1 0 1 125 637 1

1 0 1 0 0 1 0 1 0 1 149 661 1

1 0 1 0 1 0 1 1 0 1 173 685 1

1 0 1 0 1 1 0 1 0 1 181 693 2

1 0 1 0 1 1 1 1 0 1 189 701 1

1 0 1 1 0 0 1 1 0 1 205 717 1

1 0 1 1 0 1 0 1 0 1 213 725 2

1 0 1 1 0 1 1 1 0 1 221 733 1

1 0 1 1 1 0 0 1 0 1 229 741 1

1 0 1 1 1 0 1 1 0 1 237 749 2

1 0 1 1 1 1 0 1 0 1 245 757 2

1 0 1 1 1 1 1 1 0 1 253 765 2

1 1 0 0 1 1 0 1 0 1 309 821 2

1 1 0 1 0 1 0 1 0 1 341 853 2

1 1 0 1 1 0 1 1 0 1 365 877 2

1 1 0 1 1 1 0 1 0 1 373 885 2

1 1 0 1 1 1 1 1 0 1 381 893 2

1 1 1 0 0 1 0 1 0 1 405 917 2

1 1 1 0 1 0 1 1 0 1 429 941 2

1 1 1 0 1 1 0 1 0 1 437 949 2

1 1 1 0 1 1 1 1 0 1 445 957 2

1 1 1 1 0 0 1 1 0 1 461 973 2

1 1 1 1 0 1 0 1 0 1 469 981 2

1 1 1 1 0 1 1 1 0 1 477 989 2

1 1 1 1 1 0 0 1 0 1 485 997 2

1 1 1 1 1 0 1 1 0 1 493 1005 2

1 1 1 1 1 1 0 1 0 1 501 1013 2

1 1 1 1 1 1 1 1 0 1 509 1021 2

## My GPG ciphertext test (on new key)

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Version: GnuPG v2.0.14 (GNU/Linux)

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lAVBy6ZprrWtB+5PDnflZsA3eRr9M5iRvHaDAVHfloyWRuuAKdu1GWeJeWTT6ffe

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LPwap7efgA7hVt6auuvljyxMRYDyob4/nvxIqsVJMDzl+bR/YgncGjRr8aSYNaDy

fsNERWT2gCCCI19AsF5ni9Zd91ZX6tzTnZJP9tQu1wEYHPCxsPNjGedNKSp6ZYWn

1PYbWdJTdrI7vMrRzuw4i1cASuK30gFlVxshcDidhxkWOIwvgu+ownfQ/tGeliKF

XjkVOkI74kxeIeMENfjx9/w0j7PtSGiAw49ZREmzaIY7rG+aaywkBhmqKNGs75rO

b4/Din+M2Gl/te5gPlCCAw35hMgis7e23iitkdJhIoerIdQygGStOK4VYW+gGZyD

loc87CFCYq56DM9D3ZrcjxkfQ/ka4VwQYF0=

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## cat test2 | sha256sum 4607a2d0c064c3de08b1537bff179b84b54e2a0fc8220e69f6f06a0938fec86d –

Hello World!

## $ cat topost | sha256sum 1e7c2e0e2352235a5f0f537a2c434b3f73307c91923da47762e873c14b187e80 –

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IDI4IDI5IDMwICAgMjkgMzAgMzEKCgo=

## More on b-numbers, part B

This is a continuation of the previous post on b-numbers.

===

For a b-number ‘x’ in the range 4096 to 8191 (13 bit length),

it appears at first glance that, modulo 4096,

x == y (mod 4096), where y is a b-number in

the range 1-4095;

Details:

k up to 4095:

5 // term 1

13

21

29

37

45

53

61

77

85 // term 10

93

101

109

117

125

149

173

181

189

205

213

221

229

237 x

245

253

309

341

365

373

381

405

429

437 x

445

461

469

477

485

493 x

501

509

629

693

725

749

757

765

821

853 // term 50

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 // term 100

2541

2557

2741

2773

2797

2813

2933

2997

3029

3053

3061

3069

3253

3285

3309

3325

3445

3509

3541

3565

3573

3581

3701

3765

3797

3821

3829

3837

3893

3925 // t. 130

3949

3957

3965

3989

4013

4021

4029

4045

4053

4061 // term 140

4069

4077

4085

4093 // term 144

k from 4096 to 8191, modulo 4096

237 // b-number term number 145, modulo 4096

437

493

725

765

885

1013

1205

1261

1461

1493

1517

1533

1717

1749

1773

1789

1909

1973

2005

2029

2037

2045

2229

2285

2485

2517

2541

2557

2741

2773

2797

2813

2933

2997

3029

3053

3061

3069

3253

3285

3309

3325

3445

3509

3541

3565

3573

3581

3701

3765

3797

3821

3829

3837

3893

3925

3949

3957

3965

3989

4013

4021

4029

4045

4053

4061

4069

4077

4085

4093

## 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