recursive numbering and symmetric numbers

[phillip1882]

this is a cool idea i found over at xkcd forums.

for a number n, if n is composite, break it into its prime factors. if n is prime, place '<' '>' around the number and determine the n'th prime its is.

 

then the first 20 numbers are...

 

1

<> 2

<<>> 3

<><> 4

<<<>>> 5

<><<>> 6

<<><>> 7

<><><> 8

<<>><<>> 9

<><<<>>> 10

<<<<>>>> 11

<><><<>> 12

<<><<>>> 13

<><<><>> 14

<<>><<<>>> 15

<><><><> 16

<<<><>>> 17

<><<>><<>> 18

<<><><>> 19

<><><<<>>> 20

 

a symmetric number is a number that can be written symmetric about the middle. the first few are

1

<> 2

<<>> 3

<><> 4

<<<>>> 5

<<><>> 7

<><><> 8

<<>><<>> 9

<<<<>>>> 11

<><<>><> 12

<><><><> 16

<<<><>>> 17

<<>><><<>> 18

<<><><>> 19

<><<<>>><> 20

<<<>><<>>> 23

<<<>>><<<>>> 25

<<>><<>><<>> 27

<><<><>><> 28

<<<<<>>>>> 31

<><><><><> 32

<><<>><<>><> 36

<<><<>><>> 37

 

hypothesis:

1) numbers that differ by true value of 1 won't differ by number of backets by more than 2.

2) symmetric numbers follow a logrithmic growth rate similar to the primes.

3) there is no general algorithm for adding recursive numbers.

[imatfaal]

that's very intriguing and delightfully good to look at on the page. Thanks for mentioning it

[Keen]

Very interesting idea, but unless I did not get something, I believe your first hypothesis to be false. You claim that numbers which differ by one have no more than two brackets of difference between them in their bracket notation. I tried among the Fermat's primes as they grow rather quickly and it's easy to decompose the preceeding number.

I have found [math]257=2^8+1[/math]. 257 is 65th prime, so according to your notiation it should have 65 opening brackets and 65 closing brackets.

 

The preceeding number [math]256=2^8[/math] has eight twos in it's decomposition therefore if I get it correctly it should have 8 opening and 8 closing brackets, which is quite a difference with the next number.

 

I'd personally speculate that there is no bound for the difference between brackets of consecutive numbers simply because when you multiply small prime numbers among them the result tends to grow rather quickly and if the consecutive number is a prime then it would be a rather large prime with lots of brackets, while the preceding number won't have that many brackets, since it's composed of a relatively small number of small primes.

[imatfaal]

Very interesting idea, but unless I did not get something, I believe your first hypothesis to be false. You claim that numbers which differ by one have no more than two brackets of difference between them in their bracket notation. I tried among the Fermat's primes as they grow rather quickly and it's easy to decompose the preceeding number.

I have found [math]257=2^8+1[/math]. 257 is 65th prime, so according to your notiation it should have 65 opening brackets and 65 closing brackets.

 

The preceeding number [math]256=2^8[/math] has eight twos in it's decomposition therefore if I get it correctly it should have 8 opening and 8 closing brackets, which is quite a difference with the next number.

 

I'd personally speculate that there is no bound for the difference between brackets of consecutive numbers simply because when you multiply small prime numbers among them the result tends to grow rather quickly and if the consecutive number is a prime then it would be a rather large prime with lots of brackets, while the preceding number won't have that many brackets, since it's composed of a relatively small number of small primes.

 

Great thinking Keen.

 

I haven't thought this through deeply - but following your line of reasoning wouldn't most Mersenne Primes also invalidate the proposed theorem 1.

 

For the 4th Mersenne it no longer holds

 

127 is the 31st prime ie 31+31 brackets

128 is 2^7 so 14 brackets

 

intuition would tell me that no higher Mersenne would hold as the magnitude of the ordinal of hte prime grows much quicker than the exponent of 2 - eg the next mersenne is 8191 (2^13 -1) in which the 2^n would have only 12 more bracket than the previous example but the prime would have about a thousand brackets

---

 

But having re-read the OP

 

the 65th prime is not just 65 opening followed by 65 closing

 

it is <5*13> which in turn is < <<<>>> <<><<>>> >

 

which is 16 brackets

 

as is 2^8

 

oh well - it was a good thought