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Post by fcg710647 on Aug 18, 2015 22:33:20 GMT -8
Let's say you're using some base system and you want to know how quickly the powers of the base become more composite. In other words, the numbers 10, 100, 1000, 10000, 100000, and so on. Let's say you did this in binary. 1000...000 with n zeros would just be 2^n. Because 2 is prime, the number 2^n has n+1 factors. This means that adding another zero (doubling a given power of two) only adds one factor, so the number of factors grows linearly with the size of the number. What about decimal, base ten? 10 itself is 2*5; two prime factors contained. Thus, 100...000 with n zeros is not just 10^n, it's 2^n * 5^n. How many factors does this have? (n+1)^2. Unlike in a prime base, the number of factors rises quadratically with the length of the number. A thousand (1000 in decimal) has 4*4 = 16 factors, and ten thousand (10000) has 5*5 = 25 factors. Even though each power of ten is equivalent to more than three powers of two, the increased divisibility from two prime factors rapidly surpasses the binary powers. For instance, the decimal number 65536 (which is 2^16) has only 17 factors, well below the 25 of 10000, even though it is smaller. Continuing on, 2^40 is 1099511627776, which has 41 factors. 2^40 is just above 10^12; 10^12, or 1000000000000, has 13 * 13 = 169 factors, quite an upset! Uuummmm... can we have three prime factors?! That's where Trigesimal (base 2*3*5 or thirty) and Duoquadragesimal (base 2*3*7 or forty-two) come in handy. Even the powers of 42 can surpass the famous dozenal system in divisibility (twelve only has two prime factors). Initially, twelve looks pretty good. 12^3 is very close to 42^2. 12^3 is 1728 and 42^2 is 1764. 12^3 is 2^6 * 3^3, which has 7 * 4 = 28 factors. 42^2 = 2^2 * 3^2 * 7^2, which has 3^3 = 27 factors. Dozenal's slight lead can't last. 12^6 is 2985984 and 42^4 is 3111696, a bit larger. 2985984 is 2^12 * 3^6 -> 13 * 7 = 91 factors. 3111696 is 2^4 * 3^4 * 7^4 -> 5^3 = 125 factors O.o Let's keep goin'... and try letting dozenal have the larger number. 12^11 is 743008370688, or 2^22 * 3^11 -> 23 * 12 = 276 divisors 42^7 = 230539333248 = 2^7 * 3^7 * 7^7, which has 8^3 or 512 factors 12^17 = 2218611106740436992 (19 digits), 18 * 35 = 630 factors 42^11 = 717368321110468608 (18 digits), 12^3 = 1728 factors Simple enough, 100...000 with n zeros in base forty-two is 42^n (42 in decimal) and has (n+1)^3 factors, cubic growth A base like 42 allows the easy creation of nice, divisible numbers without the need for factorials (which use the multiplication of all those integers from 1 to n). On the other hand, what if we go for a base with four prime factors, quartic (4th order divisor growth)? Crud. 2*3*5*7 = 210, five times higher than even 42. How much do those last two comparison numbers lag behind the factorials in divisibility? 19! (19 factorial) is 121645100408832000 (18 digits)... its prime factorization is 2^16 * 3^8 * 5^3 * 7^2 * 11 * 13 * 17 * 19; that yields 17 * 9 * 4 * 3 * 2^4 = 17 * 9 * 192 = 29376 divisors, holy crap!!! I have tried using the factorial counting system, where 100...00 with n zeros is (n+1)! and fractions have a similar pattern... it's a nuisance. What a shame.
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Post by fcg710647 on Aug 18, 2015 22:39:29 GMT -8
Dozenal and octodecimal looked pretty good but the lack of a third prime has become annoying. I can live with ignoring the multiplication table for now and using mental shortcuts hahahaha
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Post by fcg710647 on Aug 19, 2015 11:26:05 GMT -8
Using the symbols for duoquadragesimal... try some trigesimal fractions. 1/2 = 0.a 1/3 = 0.t 1/4 = 0.7A 1/5 = 0.6 1/6 = 0.5 1/7 = 0.48N48N48N48N48N48N48N.... 1/8 = 0.3]A 1/9 = 0.3t 1/t = 0.3 1/r = 0.2aEJte85HG2aEJte85HG2aEJte85... // Cyclic -.- 1/U = 0.2A 1/H = 0.296ePj296ePj296ePj296ePj296... 1/C = 0.248N48N48N48N48N48N48... 1/A = 0.2 1/J = 0.1?7A
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Post by fcg710647 on Sept 8, 2015 13:17:16 GMT -8
Ha ha I'm in the habit of forty-two and made an error in that fraction list. The little a is twenty-one; 0.a is 1/2 in base forty-two but 0.A is 1/2 in base thirty because I'm using A for fifteen. 0.a in base thirty is seven tenths lol
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Post by Buffoonery on Sept 10, 2015 15:37:13 GMT -8
What use does the third prime give us? I'm curious as to what Base 36 would give us, 36 is a highly composite number. It's the main reason we have 360 degrees in a circle. With all the numbers we have: [1,2,3,4,5,6,7,8,9,A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,10] The factors would be: 1,2,3,4,6,9,C,I,10 so fractions would be: 1/2 = 0.I 1/3 = 0.C 1/4 = 0.9 1/5 = 0.777-- 1/6 = 0.6 1/7 = 0.555-- 1/8 = 0.4I 1/9 = 0.4 Very interesting. Since 36 is on a 6x6 grid, it also has the advantage in symmetry. (4=2x2; 9=3x3; 16=4x4; 25=5x5; 36=6x6) It has the advantage of thirds and quarters just like base 12, and an easy integration with 5's and 7's with the repeating decimals. However I spot one disadvantage with eighths. It has 2 decimal placements instead of 1 (0.4I), and since 8 is a power of 2, cutting a pie into eighths is gonna give you the same deal with our quarters in decimal (0.25). But that's not the end of the world. Oh, back on my original question, what advantage does a third prime give us?
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Post by fcg710647 on Sept 13, 2015 17:56:19 GMT -8
Dozenal gives two floating places with eigths as well, such as 1/8 being 0.16 and 5/8 is 0.76, so if you want super short expressions for eigths (and of course divisibility by three), you're stuck with quadrivigesimal (base twenty-four), for which I already wrote fractions in another post. 1/8 is 0.3 and 5/8 is 0.A in base twenty-four, since I have A for fifteen. More: 1/3 = 0.8 1/4 = 0.6 1/6 = 0.4 1/U = 0.2 (one twelfth) // Sounds like 2^3 * 3 satisfies you if you're getting pickier with those powers of two.
Ok, the number of prime factors in the base can be imagined as the number of dimensions a base has. The denominators that give terminating expressions are built out of the primes the base has. If the base is a power of two, *only* 1/2, 1/4, 1/8, 1/16, 1/32, and so on have finite representations. In other words, you can just list the powers of two along a line or row (one dimension). In bases six and twelve (two prime factors), you can make a multiplication table where the top row is the powers of two and the left column is the powers of three (or vice versa), which might look like this:
1____2____4____8____.... 3____6____12___24___.... 9____18___36___72___.... 27___54___108__216__.... .... Now, if you use trigesimal or duoquadragesimal (three primes), you can imagine growing a cube with each axis having a different prime factor and being able to multiply all those different possibilities together. It literally is an increase in terminating power equivalent to going from a Flatland to the 3D world we're used to. In a Flatland, a view from inside the plane is only an edge, as opposed to the 2D images we see.
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Post by fcg710647 on Sept 13, 2015 18:27:01 GMT -8
So you mentioned the versatility of hexadecimal when dealing with binary data, but encoding such data in hex isn't that efficient, since characters themselves are stored in at least eight bits. The perfect efficiency would be base 2^8 (256 in decimal), but you know how hellish that many symbols would be. Anyway, two bytes allow 65536(decimal) values. Base 256 would encode this in two digits and hexadecimal would require four. Can two bytes be encoded in three digits as a compromise? Take the cube root of 65536 to find out: 40.31747..... the smallest base that could do it is 41, but who in hell wants a giant prime base?! Sounds like another advantage for forty-two The cube of 42 is 74088 in decimal, so it loses a little efficiency (larger than 65536), however, 42 is a better deal than the currently applied Base64 encoding scheme, found in the Standard Mail Transfer Protocol. Base64 encodes three bytes in four digits, which is slightly more efficient, but the number of bytes in the data package must be a multiple of three; if it's not, padding or other methods must be used when (number of bytes) mod 3 is 1 or 2. When using 42, the number of bytes only needs to be even, and only one possible slip-up can arise if it's not, as opposed to two other possibilities for Base64.
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Post by fcg710647 on Sept 13, 2015 18:48:36 GMT -8
Have you read my first description of the divisibility of the powers of the base? Meaning 100....000 in whatever base.
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Post by fcg710647 on Sept 17, 2015 16:29:56 GMT -8
On a side note, given that the default account image is a symmetrical arrangement of twelve red circles, how might 36 or 42 objects be arranged? Perhaps using six-fold symmetry...
___O_O_O_O___(6^2) __O_O_O_O_O__ _O_O_O_O_O_O_ O_O_O___O_O_O _O_O_O_O_O_O_ __O_O_O_O_O__ ___O_O_O_O___
______O______(2*3*7 = 6*7) ___O_O_O_O___ O_O_O_O_O_O_O _O_O_O_O_O_O_ O_O_O___O_O_O _O_O_O_O_O_O_ O_O_O_O_O_O_O ___O_O_O_O___ ______O______
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Post by Buffoonery on Sept 17, 2015 18:17:37 GMT -8
Ohh man, let me get back to this page in a bit, that's a lot of information.
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Post by Buffoonery on Sept 17, 2015 20:00:52 GMT -8
10 = [2] *[5] ---2 Dimensions 12 = [2^2] *[3] ---2 Dimensions 18 = [2] *[3^2] ---2 Dimensions 24 = [2^3] *[3] ---2 Dimensions 30 = [2] *[3] *[5] ---3 Dimensions 32 = [2^5] ---1 Dimension 36 = [2^2] *[3^2] ---2 Dimensions 42 = [2] *[3] *[7] ---3 Dimensions So are you saying the advantage would be that adding a 'dimension' would allow more routes to arriving at a number? Having 2, 3, and 5 as your friends, why is this necessary, I'm a bit confused.
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Post by fcg710647 on Sept 17, 2015 20:32:54 GMT -8
Denominators that give terminating fractions in hexadecimal and octal: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, ... (powers of two only) Terminating denoms. in decimal: 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128, 160, 200, 250, 320, 400, 500, 625, 640, ... (lots more) Terminators for dozenal, octodecimal, alphadecimal: 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, ... (a few more) How about for 42? 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 27, 28, 32, 36, 42, 48, 49, 54, 56, 63, 64, 72, 81, 84, 96, 98, 108, 112, 126, 128, ... now that's impressive. Octal has six such numbers below 100, decimal has thirteen, dozenal has nineteen, and forty-two has twenty-nine below a hundred, despite skipping the prime 5 in its factorization!
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Post by Buffoonery on Sept 17, 2015 21:01:07 GMT -8
About that Hex topic and the computers, I wasn't talking about efficiency, since the greatest efficiency would be Binary since the initial computer signals are binary. I was talking about human understanding, We understand what 0-9 means, adding A-F is pretty simple to add on top. - "Have you read my first description of the divisibility of the powers of the base? Meaning 100....000 in whatever base." Where is this? Could you link me the page? I'll be sure to check it out. :) - Ooooooooooo, I like those 6-fold symmetries! Imagine designing a city around that, Imagine how efficient it would be if each circle represented a house location, or at least part of a house. Funny how that's the first thing I think of when I see hexagons. Bees do it, but building furniture would be a bit difficult for us, many problems would arise I would think.
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Post by fcg710647 on Sept 17, 2015 21:01:29 GMT -8
Duoquadragesimal is one of the best bases for a prime sieve. Sexagesimal and above are too high to handle. Except for 2, 3, and 7, any prime ends in one of duoquadragesimal's totative digits. We make a list of possible primes using this feature. Composite numbers are labeled with underscores. Every single prime number *must* appear in this list - it's rather useful when compared with other prime number strategies, such as spirals or Mill primes. Note that duoquadragesimal has one composite totative - it has to be included as a possible ending digit as some of its results are prime, and having every prime in the list is its best characteristic. A prime sieve normally begins with all the integers before removing multiples of primes, or with 2 plus all the odd numbers above 2. Duoquadragesimal creates a much more efficient list, which can still be sieved further. Using digit summation to check for multiples of b-1 or b+1 for a base b is possible, but is an extra step and a huge complication if included in a programming implementation, since one must derive the digits of the number in base b if b is not 2; only bitwise operators are immediately available. Something like if (c == 109) with 109 written in decimal does not imply that the digits of c are already calculated; the 109 is translated into binary, not the other way around.
Duoquadragesimal totatives: 1, 5, r, H, N, G, j, _T, m, F, L, b
2, 3, 5, 7, r, H, N, G, j, _T, m, F, L, b, 11, 15, 1r, _1H, 1N, 1G, _1j, 1T, 1m, 1F, 1L, 1b, _21, 25, _2r, 2H, 2N, 2G, 2j, 2T, 2m, _2F, _2L, _2b, 31, 35, 3r, 3H, _3N, _3G, 3j, 3T, _3m, 3F, 3L, 3b, _41, 45, 4r, 4H, _4N, _4G, 4j, 4T, 4m, 4F, _4L, _4b, 51, _55, _5r, 5H, 5N, 5G, 5j, _5T, 5m, 5F, _5L, 5b, _61, 65, 6r, _6H, 6N, 6G, _6j, 6T, 6m, 6F, _6L, 6b, _71, _75, _7r, 7H, 7N, 7G, 7j, _7T, _7m, _7F, 7L, .....
Duoquadragesimal removes the following composite numbers, which can't be sieved out by senary, dozenal, octodecimal, quadrivigesimal, alphadecimal, etc. etc.:
D, 17, 1D, 27, 2D, 37, 3D, 47, 4D, 57, 5D, 67, 6D, 77, 7D, .... in other words, any multiple of 42(decimal) +- 7.
Time to make even more fun of octal and hexadecimal, which list 2 and then all odd numbers above 2; they have weak sieves, so write the numbers they produce in duoquadragesimal and notice how many are clearly composite from their ending digit: in other words, the last digit must be a totative (except the specific numbers 2, 3, and 7) for the number to be prime.
2, 3, 5, 7, _9, r, H, _A, N, G, _a, j, T, _e, m, F, _n, _D, L, _Y, b, 11, _13, 15, _17, _19, 1r, 1H, _1A, 1N, 1G, _1a, 1j, 1T, _1e, 1m, 1F, _1n, _1D, 1L, _1Y, 1b, 21, _23, 25, _27, 29, 2r, 2H, _2A, 2N, 2G, _2a, 2j, 2T, _2e, 2m, 2F, _2n, _2D, 2L, _2Y, 2b, 31, _33, 35, _37, _39, 3r, 3H, _3A, 3N, 3G, ... (Now the numbers are written in decimal) Next time someone gives you a number and wonders if it's prime, just take its mod 42 (unless the number is 2, 3, or 7), and the result is the ending digit it would have in base forty-two (the remainder). For instance, mod(49, 42) = 7 -> 49 is divisible by 7.
Mod(151, 42) = 25, oh, 151 isn't divisible by 2, 3, or 7. Turns out it's prime lol Mod(161, 42) = 35, O.o 35 is a multiple of 7, not a totative, 161 is divisible by 7. Mod(111, 42) = 27, 27 is 3^3 so 111 is divisible by 3. Mod(1722, 42) = 0, oh ha ha 1722 is divisible by 42. Mod(666, 42) = 36, 36 and 42 have 6 in common so 666 is divisible by 6, as expected. Mod(196, 42) = 28, 28 and 42 have 14 in common so 196 is divisible by 14 (2*7) Mod(421, 42) = 1 (totative), turns out 421 is prime. Mod(441, 42) = 21, 21 is half of 42 so 441 is divisible by 21 (3*7) Mod(501, 42) = 39, 39 and 42 are both multiples of 3 so 501 is divisible by 3.
This is a similar idea to decimal, where numbers ending in 0, 2, 4, 6, and 8 are even, and those ending in 0 and 5 are multiples of five. The totatives in decimal are 1, 3, 7, and 9. 42 takes care of 3 and 7 instead of 5, when compared with decimal. 129(decimal) = 33(Dq.), ending digit is 3, mod(129, 42) = 3, 129 is a multiple of 3.
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Post by fcg710647 on Sept 17, 2015 21:05:09 GMT -8
I'll repost my Aug. 19th introduction... Let's say you're using some base system and you want to know how quickly the powers of the base become more composite. In other words, the numbers 10, 100, 1000, 10000, 100000, and so on. Let's say you did this in binary. 1000...000 with n zeros would just be 2^n. Because 2 is prime, the number 2^n has n+1 factors. This means that adding another zero (doubling a given power of two) only adds one factor, so the number of factors grows linearly with the size of the number. What about decimal, base ten? 10 itself is 2*5; two prime factors contained. Thus, 100...000 with n zeros is not just 10^n, it's 2^n * 5^n. How many factors does this have? (n+1)^2. Unlike in a prime base, the number of factors rises quadratically with the length of the number. A thousand (1000 in decimal) has 4*4 = 16 factors, and ten thousand (10000) has 5*5 = 25 factors. Even though each power of ten is equivalent to more than three powers of two, the increased divisibility from two prime factors rapidly surpasses the binary powers. For instance, the decimal number 65536 (which is 2^16) has only 17 factors, well below the 25 of 10000, even though it is smaller. Continuing on, 2^40 is 1099511627776, which has 41 factors. 2^40 is just above 10^12; 10^12, or 1000000000000, has 13 * 13 = 169 factors, quite an upset! Uuummmm... can we have three prime factors?! That's where Trigesimal (base 2*3*5 or thirty) and Duoquadragesimal (base 2*3*7 or forty-two) come in handy. Even the powers of 42 can surpass the famous dozenal system in divisibility (twelve only has two prime factors). Initially, twelve looks pretty good. 12^3 is very close to 42^2. 12^3 is 1728 and 42^2 is 1764. 12^3 is 2^6 * 3^3, which has 7 * 4 = 28 factors. 42^2 = 2^2 * 3^2 * 7^2, which has 3^3 = 27 factors. Dozenal's slight lead can't last. 12^6 is 2985984 and 42^4 is 3111696, a bit larger. 2985984 is 2^12 * 3^6 -> 13 * 7 = 91 factors. 3111696 is 2^4 * 3^4 * 7^4 -> 5^3 = 125 factors O.o Let's keep goin'... and try letting dozenal have the larger number. 12^11 is 743008370688, or 2^22 * 3^11 -> 23 * 12 = 276 divisors 42^7 = 230539333248 = 2^7 * 3^7 * 7^7, which has 8^3 or 512 factors 12^17 = 2218611106740436992 (19 digits), 18 * 35 = 630 factors 42^11 = 717368321110468608 (18 digits), 12^3 = 1728 factors Simple enough, 100...000 with n zeros in base forty-two is 42^n (42 in decimal) and has (n+1)^3 factors, cubic growth A base like 42 allows the easy creation of nice, divisible numbers without the need for factorials (which use the multiplication of all those integers from 1 to n). On the other hand, what if we go for a base with four prime factors, quartic (4th order divisor growth)? Crud. 2*3*5*7 = 210, five times higher than even 42. How much do those last two comparison numbers lag behind the factorials in divisibility? 19! (19 factorial) is 121645100408832000 (18 digits)... its prime factorization is 2^16 * 3^8 * 5^3 * 7^2 * 11 * 13 * 17 * 19; that yields 17 * 9 * 4 * 3 * 2^4 = 17 * 9 * 192 = 29376 divisors, holy crap!!! I have tried using the factorial counting system, where 100...00 with n zeros is (n+1)! and fractions have a similar pattern... it's a nuisance. What a shame.
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