### 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.

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.