Question

@dan815 Check this out, weird rule thingy

Answer 1

Ok so take a number, any number (preferably composite for the best results) and do this rule:\[ [3*2*2*3] = (3-1)(2*2*3)+(2-1)(2*3)+(2-1)(3)+(3-1)(1)\] So you see I used 36, then arbitrarily took a factor out each time and decreased it by one and multiplied it by what was left until there were no more factors.

So this will always decrease the number by 1, here I went from 36 to 35. Check it and test it out on any number. =P

Answer 2

I was using the [ ] brackets to denote a kind of "function" but if you put ( ) -1 then it will be equal, sorry if that is slightly confusing.

Answer 3

In a sense this is both a generalization of telescoping series and representing numbers in very weird bases where each individual digit can hold a different number of numbers.

For example:

1000 - 1 = 999 = 9*100 + 9*10 + 9*1
36 - 1 = 35 = 2*12 + 1*6 + 1*3 + 2*1

See how 9 is the highest digit in base 10? similarly, each of these numbers are the highest possible digit for their respective position in the basis, so it's sort of a hybrid of base 2 and base 3 I'm playing with and it's not even in a pattern. So in the basis I've shown, we could write 35 as 2112 and when you add 1 to it you get 10000 in the basis which represents 36.

Answer 4

has like this telescoping attribute

Answer 5

x y z
(x-1)*yz + (y-1)*z+z-1
xyz -yz +yz -z+z -1

Answer 6

An interesting rule thing is prime numbers will only have a single decomposition, [p]=(p-1)*1

Answer 7

But a weird thing about it is that it's completely arbitrary which order you decide to decompose it, so you can probably set up some interesting equations with it or something.

Answer 8

right

Answer 9

What do you think @wio? Kinda fun huh? haha =P

Answer 10

But you could have factored to \(6\times 6\)

Answer 11

Sure you can do that as well. \[\Large [6*6]=(6-1)*6+(6-1)*1=30+5\]

Answer 12

i know mathematicians wont like this proof but
for x y z
decomposition:
(x-1)*yz + (y-1)*z+z-1
xyz -yz +yz -z+z -1
-------------------
pattern must follow
a*b*c....*z
and since the first one is
a*b*c*...*z -b*c*...*z + .....-1
it wont matter what order u decompose

Answer 13

\[
p^n = n(p-1)p^{n-1} = np^n-np^{n-1}
\]

Answer 14

In my mind it has to work because you can always represent a number as 100...000 in any base, and when you subtract 1 you will get all the previous places maxed out. It's not a proof, but my intuition led me to it and it is right lol. XD

Answer 15

Oh wait, you remove a term each time

Answer 16

its a cool pattern to find out of the blue like that lol

Answer 17

Yeah it sort of makes "homogeneous" bases seem like a special case of something more interesting. Have you seen a factorial base? It basically counts like this:

1
10
11
20
21
100
101
110
121
200
...
1000

so each new digit 1, 10, 100, and 1000 represent 1, 2, 6, 24, etc... each a factorial. Then if you turn it around we can represent e from the power series as \[\Large e=10.\bar 1\] in this base. A repeating decimal, cool huh?

Answer 18

I hope you enjoy this kind of stuff as much as me, it's so totally weird and simple at the same time.

Answer 19

medal+fan if you find a basis for representing pi as a repeating decimal.=P

Answer 20

i was thinking about something similiar like

Answer 21

u know how 0.1 is an irrational number in binary

Answer 22

Perhaps we can find a basis to represent sqrt(2) as a repeating decimal as wll?

Answer 23

Binary doesn't make a number irrational

Answer 24

I don't know does irrational just mean infinite repeating decimals? I mix that up with transcendental

Answer 25

an irrational binary number

Answer 26

where theres no apparent pattern to the infinite sequence of numbers

Answer 27

like 3.141592632341251562462

Answer 28

wait irrational is infinite decimals and transcendental is no pattern? Does transcendental imply irrationality?

Answer 29

In other words can a number be transcendental but not irrational?

Answer 30

irrational is infinite decimal and no pattern

Answer 31

like 0.3333333333....... repeating is still rational since that is 1/3

Answer 32

irrational is all numbers that cannot be represented as a fraction of integers

Answer 33

i dunno what that means really interms of binary though

Answer 34

What do you mean by factorial bases?

Answer 35

Part of this began when trying to solve the goldbach conjecture, the other came from a question ganeshie had that we figured out, and that is that \[\Large 2^n-1\] is always composite when n is composite. The proof of it is quite fascinating and simple.

Answer 36

Represent a number in base 2 raised to a composite number and subtract 1 and you get this, here is an example: \[\Large 2^6-1=111111_2= \\ \Large 1*111111 \\ \Large 11*10101 \\ \Large 111*1001 \\ \Large 111111*1\] So here it's the same kind of idea, you will always end up with a composite number of 1's that can be factored out , here it's 1,2,3, and 6. Fun huh?

Answer 37

Factorial base I mean each place holds that many digits @wio I'l make an example one sec

Answer 38

The thing is \[
\frac{1}{2!}=\frac{3}{3!}
\]Also, since you start with \(1!\), that would mean that \[
1.0=0.1
\]

Answer 39

So here I am comparing a factorial base to base 10 for clarity, c's are the constants and follow this pattern. \[\Large x=c_1 *1! + c_2 * 2! + c_3 * 3! + ... + c_n * n! \\ \Large y= c_1 *10^0 + c_2 * 10^1 + c_3 * 10^2 + ... + c_n *10^{n-1}\]

Note that just like the highest number in base 10 for a digit is 9, we have a similar case for the factorial base, except it's slightly skewed since each digit holds from 0 to itself. So just like 21 in base 10 represents 2*10 + 1*1 we have 21 in the factorial base as being 311 which represents 3*3! + 1*2! + 1*1!

Another example is 999 in base 10 is 1000-1 and 321 in factorial base is 1000-1 which is in base 10 4!-1.

Sure enough, we can check, 3*3! + 2*2! + 1*1! = 3*6+2*2+1*1=18+4+1=23=24-1=4!-1

Answer 40

You would want \(6\) to be written as \(100\), because this basis is redundant

Answer 41

6 is written as 100 and it's not redundant

Answer 42

Counting up to 24 in factorial base:

1
10
11
20
21
100
101
110
111
120
121
200
201
210
211
220
221
300
301
310
311
320
321
100

Answer 43

Think of the "ones" place as being binary, "tens" place as being base 3, "hundreds" place as being base 4, etc...

Answer 44

Of course this is just a special case of the original statement I began this off with where we are specifically choosing to decompose \[[24]=[4*3*2*1]=(4-1)(3*2*1)+(3-1)(2*1)+(2-1)*1\] But we could have chosen any base to do this, like a messed up factorial base where the ones place was base 3, then the next one was base 2, etc...

Answer 45

Sorry it's kind of confusing since I just realized this myself recently but I can reexplain or make more examples of anything if you are curious. I have been thinking about this for a while and had a lot of difficulties so I tried to explain it best I could based on what was hard to me.

Answer 46

For a basis of odd numbers, you count like:
1
2
10
11
100
101
1000
1001
And express:\[
\pi = 11(1.\overline{01}-0.\overline{10})
\]

Answer 47

Non-homogeneous bases can't be multiplied or added easily because per-digit arithmetic rules don't work

Answer 48

Hmm I don't think that's exactly a legitimate basis since you aren't allowing each digit place to cycle through a predictable set of numbers.

Answer 49

is that pi from the power series

Answer 50

Like, for instance, 2 only shows up once and there's no predictability to it and it seems like there is not a unique representation for each number. But I'm still sort of curious how you got there since I don't understand it but seems cool.

Answer 51

How exactly are you defining a legitimate basis then?

Answer 52

Each digit has to cycle consistently through the same digits. So like in the factorial basis the only values it takes in the ones place are 0 and 1. The only values it takes in the tens place are 0,1,2. The only values in the hundreds place is 0,1,2,3. This pattern continues, but they are always cycling, kind of like a clock or combination lock.

Just like in base 10 you can't get to 30 from 20 without going through 0,1,2,3,4,5,6,7,8,9, these bases work the same in that sense, except now each place can have a different number of digits. A digital clock is a good example of this in that if we look at say, 11:31 AM as really a 4 digit number 1131 then we see that the ones place can hold 0,1,2,3,4,5,6,7,8,9 the tens place can hold 1,2,3,4,5, the hundreds place can hold 1,2,3,4,5,6,7,8,9 and the thousands place can hold 0,1 (or maybe 2 if you're on a 24 hour clock, but after that it doesn't go any higher.)

Answer 53

To extend the digital clock, we can see that the next digit after 1131 on a 24 hour clock will represent number of days. The next digit up can represent weeks, then years, decades, centuries, etc... if we so desire. But it's all an arbitrary method of counting that lines up with a nonhomogeneous basis.

Answer 54

Although I suppose it sort of breaks down after a while since the way months and years are defined is arbitrary to the point of leap years and each month having different numbers of days etc, which isn't regular so can't be accounted for in this system.

Answer 55

Okay, here is an idea, make your bases from the sequence \(a_n=16n^2-16+3\).
Except, also have our primary base be \(1\).
So we get bases like: \[
1,3,15,35,63
\]We count like: \[
1\\2\\10\\11\\12\\20\\21\\22\\30\\31\\32\\40\\41\\42\\100
\]

Answer 56

Ahhhh that's what I'm talking about awesome! =O

Answer 57

Well, I got \(\pi=2.1\bar 2\) but I need to go over it a bit.

Answer 58

Ok this is a good idea I now feel like we're on the same page, I'm trying to check this somehow.

Answer 59

I started with \[
\pi=4\sum_{k=1}^{\infty}\frac{(-1)^k}{2k-1}
\]I tried to get rid of alternation, first I said: \[
\sum_{k=1}^{\infty}\frac{(-1)^k}{2k-1}=\sum_{k=1}^{\infty}(-1)^ka_k = (a_1-a_{2})+(a_3-a_4)+\ldots
\]Then I said: \[
b_k = a_{2k-1}-a_{2k} \implies \sum_{k=1}^{\infty}(-1)^ka_k=\sum_{k=1}^{\infty}b_k
\]Which is not an alternating series.
To be explicit:\[
b_k = a_{2k-1}-a_{2k}=\frac{1}{2(2k-1)-1}-\frac{1}{2(2k)-1} = \frac{1}{4k-3}-\frac{1}{4k-1}\\
~~~~=\frac{2}{16k^2-16k+3}
\]This ends up giving us:\[
\pi = 4\sum_{k=1}^{\infty}b_k=\sum_{k=1}^{\infty}\frac{8}{16k^2-16k+3}
\]The digit \(8\) fits for any base such that \(k\geq2\).
We only have a problem with \(8/3\). This ends up as \(2+1/3\).
So we have: \[
\pi=2+\frac{1}{3}+\sum_{k=2}^{\infty}\frac{8}{16k^2-16k+3} = 2.1\overline{8}
\]

Answer 60

Prime numbers don't form a legitimate basis if we are using addition.
When using addition, it seems that legitimate basis will be formed by any increasing sequence.

I hypothesize that sequences won't form legitimate basis.

Answer 61

Ok I'm working it out, fascinating. I was trying to bootstrap my way in through these formulas, which you might have pretty much nearly derived I'm not sure: http://en.wikipedia.org/wiki/Approximations_of_%CF%80#Digit_extraction_methods

Answer 62

arithmetic sequences wont form legitimate bases

Answer 63

I know what bases do work, ones where the next base up is a multiple of the previous one. So 1 is first, 10*1 is next, 10*10*1 is the one after, etc... and in factorial that ended up being 1, 1*2, 1*2*3, etc...

That's part of the reason why the original interesting thing works. Although the original thing also works for nonintegers as well, for example: \[[5]=[\frac{1}{3}3*5]=(\frac{1}{3}-1)(3*5)+(3-1)(5)+(5-1)(1)=4\]

Answer 64

Geometric sequences definitely work

Answer 65

Homogeneous basis are just geometric sequences

Answer 66

Binary is\[
a_n=a_1r^{n-1}=2^n
\]

Answer 67

But I seemed to have gotten a quadratic to work, or maybe we just didn't really explore it far enough to know. I assumed that \(n^2\) would be a legit basis.

Answer 68

Proof that arithmetic sequences won't work for basis: \[
a_n=a_1+(n-1)d\implies a_{n+1}-a_n=d
\]Once you reach \(n\) such that \(a_n>d\), you'll find that the \(a_{n-1}\) digit no longer becomes necessary to reach the next number.

Answer 69

Actually, I might be off by a bit, but the point is \(d\), can be express with digits \(1\ldots n\). So if you have \(a_{n+3}\), then you can get high enough to reach \(a_{n+4}\) without needing to increment \(a_{n+2}\).

Answer 70

In the case of \[
a_n=n^2\implies a_{n+1}-a_n=(n+1)^2-n^2 = 2n+1
\]The space between each digit is increasing.

Answer 71

But it might not be increasing fast enough? I am not sure.

Answer 72

I just worked out that if there is a common pattern you can always represent it as a geometric sequence. So there are also "semi" homogeneous bases like this for instance: \[\large \frac{1}{2^03^0}+ \frac{1}{2^13^0}+ \frac{1}{2^13^1}+ \frac{1}{2^23^1}+ \frac{1}{2^23^2}+...\] Here the basis steps between incrementing 2 and 3 back and forth, we can factor out the first step and turn it into a geometric sequence again. Interesting, good observation on the geometric sequence thing!

Answer 73

I am beginning to think that power sequence basis won't work either.

Answer 74

One way that works but I don't think follows a pattern (just a random thought) is to increment by whichever number is smaller of the two. So for instance when you have 2^3 and 3^2 in the denominator, then because 8<9 we increase to 2^4*3^2 and then 16>9 so we increment the next basis by multiplying by 3, etc...

Answer 75

Your semi homogeneous seems like a product of two homogeneous basis

Answer 76

A problem I would like to tackle with this new discovery is an old problem I've had. I want to be able to convert between different bases in a linear algeba kind of way. So we can do something like matrix multiply the coefficients on base 2 (the digits) and turn them into base 5, or whatevr.

Actually the semihomogeneous is just a linear combination of homogeneous bases it looks like.

Answer 77

To finish the example and represent it in its own basis: \[\large \frac{1}{2^03^0}+ \frac{1}{2^13^0}+ \frac{1}{2^13^1}+ \frac{1}{2^23^1}+ \frac{1}{2^23^2}+...\]\[\large \frac{1}{6^0}+\frac{1}{2} \frac{1}{6^0}+ \frac{1}{6^1}+\frac{1}{2} \frac{1}{6^1}+...\] \[\large \frac{3}{2} \left( \frac{1}{6^0}+ \frac{1}{6^1}+ \frac{1}{6^2}+... \right)\]\[\Large \frac{3}{2} \frac{1}{1-(1/6)}=\frac{9}{5}\] which we can then turn around and represent by the original basis (loook above, all 1s) is really just \[\large 1.\bar1\] interesting.

Answer 78

If you make a basis out of \(n^2\).
You have \(1,4,9,16,25,36\)

Answer 79

I don't know if that works since 9 isn't an integer multiple of 4, etc... but I see where you're going with this haha.

Answer 80

I think any sequence breaks the rules if it is the case that: \[
a_{n+1}-a_{n}<a_{n-1}
\]Because it means \(a_{n-1}\) is not necessary to go from \(a_n\) to \(a_{n+1}\).

Answer 81

So in our case: \[
(n+1)^2-n^2<(n-1)^2\implies 2n+1<n^2-n+1
\]

Answer 82

Well maybe we could make something like this work if we allowed a "digit" to take on noninteger values. I'm sort of trying to lean into the idea that every number has this form in base 10

This we already know:
\[\Large \sum_{n=-\infty}^\infty c_n 10^n\]

I'm trying to strive for: \[\Large \int\limits_{-\infty}^\infty c(n)10^ndn\] maybe we can do something like this and have an infinitesimal basis or something of the sort... hmmm...

Answer 83

I think that allowing rational values results in basis becoming redundant.

Answer 84

Yeah, I don't think there's much to gain from representing as a rational number anyways even if it isn't redundant. Suppose we allow only half integers as being allowed, that would be the same as doubling the basis in the first place I think since 0,1/2,1,3/2,4, etc... are really just the numbers from 0 to 4 renamed in a sense.

Answer 85

That isn't the relationship between integrals and sums. Also a sum needs to have a infinitesimal factor for it to relate to an integral

Answer 86

It needs a \(\Delta x\to 0\) to relate to \(dx\).

Answer 87

Yeah I know, it was just sort of like a shot in the dark to give the idea of what I was striving for

Answer 88

For any polynomial sequence \(a_n\) degree \(k\), we have: \[
a_{n+1}-a_{n} = O(n^{k-1}) < O(n^k) =a_{n-1}
\]

Answer 89

Which means that polynomial sequences can't be used to form basis...

Answer 90

Meaning that we'd need to find another way to express \(\pi\).

Answer 91

I'm going to try my test on factorials...

Answer 92

\[
a_{n+1}-a_n = (n+1)!-n! = (n+1)n!-n! = nn! = n^2(n-1)! <(n-1)!
\]So factorials will check out.

Answer 93

Hmm interesting. I think this will be something to sleep on and come back to, kinda need a break from this one I think. Thanks, maybe we can pick this back up tomorrow or something.

Answer 94

For geometric series: \[
r^{n-1}-r^{n} = rr^n-r^n=(r-1)r^n =(r^2-r)r^{n-1}< r^{n-1}
\]

Answer 95

Hmm, so \(r>1\). You could have a homogenous base of \(1.5\)?