- Empty

Number with only two factors.

- jamiebookeater

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- Empty

If you're given \(n= 2^a 3^b\) then can you know what the next largest number is: \(m=2^x3^y\)?
Basically, given a,b can we find x,y?

- Empty

Here is an example:
\(2 = 2^1 3^0\) so the next number is \(3=2^03^1\).
Here's the start of the trend: \(2 < 3 < 4 < 6 < 8 < 9 < \cdots\)

- UsukiDoll

that's when n = 2 a = 1 and b = 0 for n =2^a3^b
I guess the same trend will be for m = 2^x3^y too?

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## More answers

- Empty

Well it's simple to see the trend within numbers that have the same number of prime factors, for instance:
\(2*2<2*3<3*3\) Which is the same as saying \(4 < 6 <9 \)
However there is overlap sometimes, when does this occur? Afterall, we have:
\(2*3 < 2*2*2 < 3*3\) This is a simple example \(6<8<9\)
So is this kind of behavior predictable?

- UsukiDoll

can n and m be a number besides 2 and 3?

- Empty

Yeah, they are just consecutive numbers that have only factors of 2 and 3.

- UsukiDoll

\[4= 2^1 3^0\]
\[5= 2^0 3^1\]
like this?

- UsukiDoll

NUGH!!!!!!!!!!!! I don't know how to start this afterwards

- Empty

Oh ok well first off, wait I'm confused, these are just exponents:
\(4 =2^2 3^0 \ne 2^1 3^0 = 2\)
\(5 \ne 2^03^1 = 3\)
So n and m depends on the exponents:
\(2^13^0 < 2^0 3^1 < 2^23^0 < 2^1 3^1\)
So this is the same as
\(2 <3 <4 <6 \)

- UsukiDoll

wow. 5 got skipped.

- UsukiDoll

maybe
\[2^03^2 =1(9) =9 \]
\[2^23^1 =4(3) =12 \]
\[2^13^2 =2(9) =18 \]
\[2^23^2 =4(9) =36 \]

- Empty

There is no 5, only factors of 2 and 3, and you can't make 5 by multiplying 2 and 3 together so it's not included. :P

- UsukiDoll

yeah I know... 2 x 3 or 2^13^1 = 2(3) = 6

- UsukiDoll

so with the exponent numbers we can only produce certain results.
like
\[2^13^1 = (2)(3) = 6\]
\[2^13^2 = 2(9) = 18 \]
but I'm getting big numbers XD

- UsukiDoll

like let 2 be fixed with that exponent 1 only and let 3 have the b = 1,2,3,4,5,...............
\[2^13^3 = 2(27) = 54\]
\[2^13^4 = 2(81) = 162\]
\[2^13^5 = 2(243) = 486\]

- UsukiDoll

oh I see what I did wrong earlier XD
I just assigned m and n without thinking about a and b. That's why
\[4 \neq 2 , 5 \neq 3 \]

- Empty

Yeah! Now you understand where I'm stuck now haha. I just sorta came up with this problem on my own earlier and I forget why but it's not the first time I've seen this before so it must be important to me.

- UsukiDoll

if all of the exponents for 2 were 0
\[2^03^3 = 1(27) =27......................2^03^4 =(1)(81) = 81..........2^03^5=(1)(243) = 243\]

- UsukiDoll

it's like if the exponent number is bigger for 3, then we have more threes if we write it out... it dominates or something like that.
like
\[2^13^3 = 2(3)(3)(3) =54\]

- ganeshie8

\[\large n= 2^a 3^b = 2^{a+b\log_2 3}\]

- UsukiDoll

-_- now the problem *at least for me* looks a bit complicated... in that log form

- UsukiDoll

unless a = 1 and b = 1

- ganeshie8

Not really, notice that we no longer have to worry about two different bases

- UsukiDoll

\[\large n= 2^a 3^b = 2^{a+b\log_2 3}\]
if a = 1 and b = 1
\[\large n= 2^1 3^1 = 2^{1+1\log_2 3}\]

- UsukiDoll

nah I prefer the different bases.

- ganeshie8

The problem translates to finding the smallest positive integers \(x,y\) given \(a, b\) that satisfy :
\( a + b\log_2 3\le x+y\log_2 3 \)

- ganeshie8

at least the logs give us some alternative but concrete thing to mess with..

- ganeshie8

multiplying \(\log 2\) through out that becomes
\( a\log 2 + b\log 3\le x\log 2+y\log 3 \)

- ganeshie8

let \(n = 2^a3^b\)
then the number of integers of form \(2^x2^y\) less than or equal to \(n\) is given by
\[\large \sum\limits_{i = 0}^ {\left\lfloor \log_2 n\right\rfloor}\, \left(\left\lfloor\log_3 \frac{n}{2^i}\right\rfloor + 1\right)\]

- ganeshie8

that simplifies to
\[\large (\left\lfloor \log_2 n\right\rfloor+1)(b+1)+\sum\limits_{i = 0}^ {\left\lfloor \log_2 n\right\rfloor}\, \left\lfloor (a-i)\log_3 2\right\rfloor \]

- ganeshie8

in other words, if we sort all the numbers of form \(2^x3^y\) in increasing order, the position of \(n=2^a3^b\) is given by above expression

- ganeshie8

Here are some other facts
https://oeis.org/A003586
I really want to figure out the solution but I need to go for now, wil try again over the night if it still remains unsolved..

- anonymous

this is called the (generalized) hamming problem

- UsukiDoll

I think the pattern could be predictable because of this theorem
Also numbers that are divisible by neither 6k - 1 nor 6k + 1, for all k > 0
like if we let k = 1
then 6(1)-1 = 6-1=5
6(1)+1 = 6+1=7
so maybe that would mean that numbers can't be divisible by 5 nor 7. We don't see 5 10 15..... on that list and similarly we don't see 7 14 21 on that list either.
so if k = 2, then 12-1 =11 or 12+1 =13 which means numbers can't be divisible by 11 nor 13. And again, we don't see 11,22,33 nor 13,26,39

- anonymous

@ganeshie8 has the right kind of idea; suppose we had \(2^4 3^7\) -- we could take a look at a few candidates that are bigger but only by somewhat: $$2^5 3^7,2^3 3^8,2^2 3^9,\dots$$ note that increasing the exponent of the 2 is twice as big; if we trade a 2 for a 3, we'll get something 3/2 or one-and-a-half times as big; etc.

- anonymous

we could also trade three 2s for a 3, which is only 9/8 = 1.125 as big; we're interested in making \(3^a,2^b\) as close as possible so that we can make their ratio a small number greater than \(1\) -- examples: $$\frac32=1.5\\\frac98=1.125\\\frac{32}{27}=1.\overline{185}\\\frac{256}{243}\approx1.0535$$and so on

- anonymous

essentially, we're looking for simple rational numbers close to \(\log_2(3)\approx1.5850\)

- anonymous

in terms of logs, we're looking for ways to exchange \(\log2,\log3\) terms in \(x\log 2+y\log 3\) to get something slightly bigger -- in other words, we want to find integers \(a,b\) such that \(a\log 2+b\log 3\) is small and positive

- UsukiDoll

at least I tried :(

- ganeshie8

from hamming problem, it seems they are generating the sequence by multiplying a prime to a term already existing in the sequence... it is essentially an algorithm...
i think we cannot find a "formula" for the next term...

- ganeshie8

i mean, if such a formula exists, then there is no need to use an algorithm to generate the sequence recursively.. (ofcourse cost could be a reason but im beginning to feel that a direct formula is impossible.. )

- ganeshie8

Here is a fiddle if you want to play with it more..
https://jsfiddle.net/ganeshie8/uzj8prf6/
@UsukiDoll

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