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Empty
 one year ago
Number with only two factors.
Empty
 one year ago
Number with only two factors.

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Empty
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0Here 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
 one year ago
Best ResponseYou've already chosen the best response.3that'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?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Well 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
 one year ago
Best ResponseYou've already chosen the best response.3can n and m be a number besides 2 and 3?

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, they are just consecutive numbers that have only factors of 2 and 3.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[4= 2^1 3^0\] \[5= 2^0 3^1\] like this?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3NUGH!!!!!!!!!!!! I don't know how to start this afterwards

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.3maybe \[2^03^2 =1(9) =9 \] \[2^23^1 =4(3) =12 \] \[2^13^2 =2(9) =18 \] \[2^23^2 =4(9) =36 \]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0There 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
 one year ago
Best ResponseYou've already chosen the best response.3yeah I know... 2 x 3 or 2^13^1 = 2(3) = 6

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.3like 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
 one year ago
Best ResponseYou've already chosen the best response.3oh 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah! 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
 one year ago
Best ResponseYou've already chosen the best response.3if 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
 one year ago
Best ResponseYou've already chosen the best response.3it'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
 one year ago
Best ResponseYou've already chosen the best response.2\[\large n= 2^a 3^b = 2^{a+b\log_2 3}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3_ now the problem *at least for me* looks a bit complicated... in that log form

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3unless a = 1 and b = 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Not really, notice that we no longer have to worry about two different bases

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\large n= 2^a 3^b = 2^{a+b\log_2 3}\] if a = 1 and b = 1 \[\large n= 2^1 3^1 = 2^{1+1\log_2 3}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3nah I prefer the different bases.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.2at least the logs give us some alternative but concrete thing to mess with..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2multiplying \(\log 2\) through out that becomes \( a\log 2 + b\log 3\le x\log 2+y\log 3 \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2let \(n = 2^a3^b\) 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
 one year ago
Best ResponseYou've already chosen the best response.2that simplifies to \[\large (\left\lfloor \log_2 n\right\rfloor+1)(b+1)+\sum\limits_{i = 0}^ {\left\lfloor \log_2 n\right\rfloor}\, \left\lfloor (ai)\log_3 2\right\rfloor \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2in 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
 one year ago
Best ResponseYou've already chosen the best response.2Here 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
 one year ago
Best ResponseYou've already chosen the best response.0this is called the (generalized) hamming problem

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I 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 = 61=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 121 =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
 one year ago
Best ResponseYou've already chosen the best response.0@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 oneandahalf times as big; etc.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we 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
 one year ago
Best ResponseYou've already chosen the best response.0essentially, we're looking for simple rational numbers close to \(\log_2(3)\approx1.5850\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2from 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
 one year ago
Best ResponseYou've already chosen the best response.2i 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
 one year ago
Best ResponseYou've already chosen the best response.2Here is a fiddle if you want to play with it more.. https://jsfiddle.net/ganeshie8/uzj8prf6/ @UsukiDoll
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