# Binary symmetry

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Take the binary numbers of a cirtain length, for example.

00

01

10

11

Sort them into how many 1's they have.

00

01, 10

11

Write them in a more familiar form.

0

1, 2

3

And you have a pattern of numbers, which is more interesting with larger examples.

0

1

0

1, 2

3

0

1, 2, 4

3, 5, 6

7

0

1, 2, 4, 8

3, 5, 6, 9, 10, 12

7, 11, 13, 14

15

The first and last rows contain one element.

The 2nd row shows powers of 2.

More interestingly in all cases all but the last element of the 2nd to last row are prime.

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The number of elements in each row is according to the binomial coefficients. The 2nd row shows powers of two due to construction: It contains all values $\sum_n a_n 2^n$ for which exactly one $a_n$ is one and the others are zero. I must admit that I do not find it particularly interesting that you see an approximate pattern in the distribution of the primes. I think you could see an approximate prime pattern in any random arrangement of the numbers from 1 to 15.

Edited by timo
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The 2nd row shows powers of two due to construction: It contains all values $\sum_n a_n 2^n$

The 2nd row is an ordered set where $s_{n}$ = $2^{n}$.

How is it the sum of something times by $2^{n}$ where n is the sum counter? Isn't that going to return a number?

for which exactly one $a_n$ is one and the others are zero. I must admit that I do not find it particularly interesting that you see an approximate pattern in the distribution of the primes. I think you could see an approximate prime pattern in any random arrangement of the numbers from 1 to 15.

Are you saying that for subsequent elements in the sequence this observation of primes wouldn't hold true?

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The 2nd row is an ordered set where $s_{n}$ = $2^{n}$.

How is it the sum of something times by $2^{n}$ where n is the sum counter? Isn't that going to return a number?

I don't understand what you are asking. Any (natural) number x can be written as $x = a_0 2^0 + a_1 2^1 + a_2 2^2 + a_3 2^3 + \dots = \sum_n a_n 2^n$. If you demand that exactly one of the $a_n$ is one and the others are zero then it is obvious that the resulting x is a power of two. Iow: notice that there is another half to the half-sentence you quoted which contains important information.

Are you saying that for subsequent elements in the sequence this observation of primes wouldn't hold true?

I can probably see a pattern in every result coming out of rolling two dice if I want to. You would probably not agree that this pattern I see is an interesting find. But if you find the pattern you see interesting (I don't think I fully understand what you are saying), brute-force your arrangement on a computer to however far you get there. If you still think you see a pattern, then things might become interesting.

Edited by timo
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(from the thread title I thought you might be talking about digitally balanced numbers, which you might find interesting)

I'll try to do some brute forcing, should that help you in your quest to find something interesting.

Merged post follows:

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The first and last rows contain one element.
This was fairly obviously as a result of you choosing numbers of the form [imath]\sum_n 1\cdot 2^n[/imath], so as that the largest number in your list would be a series of all ones.

Anyway, brute forced results up to the very, very contrived figure of 511, (or, the sum of powers of two from 0 to 8)

[size="1"]{0}
{1, 2, 4, 8, 16, 32, 64, 128, 256}
{3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48, 65, 66, 68, 72, 80, 96, 129, 130, 132, 136, 144, 160, 192, 257, 258, 260, 264, 272, 288, 320, 384}
{7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 67, 69, 70, 73, 74, 76, 81, 82, 84, 88, 97, 98, 100, 104, 112, 131, 133, 134, 137, 138, 140, 145, 146, 148, 152, 161, 162, 164, 168, 176, 193, 194, 196, 200, 208, 224, 259, 261, 262, 265, 266, 268, 273, 274, 276, 280, 289, 290, 292, 296, 304, 321, 322, 324, 328, 336, 352, 385, 386, 388, 392, 400, 416, 448}
{15, 23, 27, 29, 30, 39, 43, 45, 46, 51, 53, 54, 57, 58, 60, 71, 75, 77, 78, 83, 85, 86, 89, 90, 92, 99, 101, 102, 105, 106, 108, 113, 114, 116, 120, 135, 139, 141, 142, 147, 149, 150, 153, 154, 156, 163, 165, 166, 169, 170, 172, 177, 178, 180, 184, 195, 197, 198, 201, 202, 204, 209, 210, 212, 216, 225, 226, 228, 232, 240, 263, 267, 269, 270, 275, 277, 278, 281, 282, 284, 291, 293, 294, 297, 298, 300, 305, 306, 308, 312, 323, 325, 326, 329, 330, 332, 337, 338, 340, 344, 353, 354, 356, 360, 368, 387, 389, 390, 393, 394, 396, 401, 402, 404, 408, 417, 418, 420, 424, 432, 449, 450, 452, 456, 464, 480}
{31, 47, 55, 59, 61, 62, 79, 87, 91, 93, 94, 103, 107, 109, 110, 115, 117, 118, 121, 122, 124, 143, 151, 155, 157, 158, 167, 171, 173, 174, 179, 181, 182, 185, 186, 188, 199, 203, 205, 206, 211, 213, 214, 217, 218, 220, 227, 229, 230, 233, 234, 236, 241, 242, 244, 248, 271, 279, 283, 285, 286, 295, 299, 301, 302, 307, 309, 310, 313, 314, 316, 327, 331, 333, 334, 339, 341, 342, 345, 346, 348, 355, 357, 358, 361, 362, 364, 369, 370, 372, 376, 391, 395, 397, 398, 403, 405, 406, 409, 410, 412, 419, 421, 422, 425, 426, 428, 433, 434, 436, 440, 451, 453, 454, 457, 458, 460, 465, 466, 468, 472, 481, 482, 484, 488, 496}
{63, 95, 111, 119, 123, 125, 126, 159, 175, 183, 187, 189, 190, 207, 215, 219, 221, 222, 231, 235, 237, 238, 243, 245, 246, 249, 250, 252, 287, 303, 311, 315, 317, 318, 335, 343, 347, 349, 350, 359, 363, 365, 366, 371, 373, 374, 377, 378, 380, 399, 407, 411, 413, 414, 423, 427, 429, 430, 435, 437, 438, 441, 442, 444, 455, 459, 461, 462, 467, 469, 470, 473, 474, 476, 483, 485, 486, 489, 490, 492, 497, 498, 500, 504}
{127, 191, 223, 239, 247, 251, 253, 254, 319, 351, 367, 375, 379, 381, 382, 415, 431, 439, 443, 445, 446, 463, 471, 475, 477, 478, 487, 491, 493, 494, 499, 501, 502, 505, 506, 508}
{255, 383, 447, 479, 495, 503, 507, 509, 510}
{511}[/size]

Note how there is a large frequency of primes in the second to last row, but it's not all but one.

Edited by the tree
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Any (natural) number x can be written as $x = a_0 2^0 + a_1 2^1 + a_2 2^2 + a_3 2^3 + \dots = \sum_n a_n 2^n$

I see, binomial coefficients? I've never come across this before. How would the number 4 be represented in this notation?

Merged post follows:

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(from the thread title I thought you might be talking about digitally balanced numbers, which you might find interesting)

Thanks. This means that the central row of each one shows digitally balanced numbers, with varying degrees of imbalance towards the top and bottom.

This was fairly obviously as a result of you choosing numbers of the form [imath]\sum_n 1\cdot 2^n[/imath]

I think the [imath]\cdot[/imath] operator means multiply?

Note how there is a large frequency of primes in the second to last row, but it's not all but one.

It's every other one instead. Seems I was wrong, it does appear as if there is some pattern of prime distributions though. Seems to much of a coincidence.

Wolfram alpha shows that several rows including the middle one (digitally balanced numbers) produce a fractal like graph.

Edited by alan2here
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I see, binomial coefficients? I've never come across this before. How would the number 4 be represented in this notation?

Four would be 100, or $a_2=1, a_{n \neq 2}=0$.

The $a_n$ are not the binomial coefficients, the number of entries in each row are. That is because the binomial coefficients $\left( \begin{array}{c} N \\ k \end{array} \right)$ Tell you how many possible combinations there are to distribute k equal elements on N places (in your case ones on the available digits).

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Just in case the OP still isn't clear on this, the expression of any natural number in any base [imath]b[/imath] expression in the form [imath]\sum_{n=0}^{\infty} a_n b^n \;| a_n \in [0\dots b-1][/imath].

For instance, 5427 = 5x103 + 4x102 + 2x101 + 7x100.

And the binomial co-efficients [imath]\binom{n}{k}= \,^{n}C_k \,=\frac{n!}{k!\,(n-k)!}[/imath] is the count of the amount of ways that, for instance, [imath]k[/imath] out of [imath]n[/imath] boxes can be ticked.

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Four would be 100, or $a_2=1, a_{n \neq 2}=0$.

Thanks. It's a differnt way to describe binary numbers.

Merged post follows:

Consecutive posts merged
Just in case the OP still isn't clear on this, the expression of any natural number in any base [imath]b[/imath] expression in the form [imath]\sum_{n=0}^{\infty} a_n b^n \;| a_n \in [0\dots b-1][/imath].

Thats a verry confusing looking equation.

For instance, 5427 = 5x103 + 4x102 + 2x101 + 7x100.

Yes, this makes sence.

Thanks. I see now how the equation works now (b = base)

A = 542 = A2*b2 + A1*b1 + A0*b0.

542 = $\sum$Anbn

And the binomial co-efficients [imath]\binom{n}{k}= \,^{n}C_k \,=\frac{n!}{k!\,(n-k)!}[/imath] is the count of the amount of ways that, for instance, [imath]k[/imath] out of [imath]n[/imath] boxes can be ticked.

So an equation calculates how many differnt K of N combinations exist is. [imath]\frac{n!}{k!\,(n-k)!}[/imath]

edit: seems to work.

Edited by alan2here
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So an equation calculates how many differnt K of N combinations exist is. [imath]\frac{n!}{k!\,(n-k)!}[/imath]

edit: seems to work.

Definitely works.

It can be proved by induction, but fairly awkwardly since you have to prove it for varying n and varying k separately.

They are called binomial coefficients because of the binomial theorem:

$(x+y)^n = \sum_{i=0}^n x^{n-i}y^{i}\binom{n}{i}$

Small ones can be calculated by hand using Pascal's triangle.

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