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

1. 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?

2. 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: $$2 < 3 < 4 < 6 < 8 < 9 < \cdots$$

3. 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?

4. 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?

5. UsukiDoll

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

6. Empty

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

7. UsukiDoll

$4= 2^1 3^0$ $5= 2^0 3^1$ like this?

8. UsukiDoll

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

9. 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 $$2 <3 <4 <6$$

10. UsukiDoll

wow. 5 got skipped.

11. UsukiDoll

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

12. 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

13. UsukiDoll

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

14. 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

15. UsukiDoll

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

16. 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 $4 \neq 2 , 5 \neq 3$

17. 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.

18. UsukiDoll

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

19. 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 $2^13^3 = 2(3)(3)(3) =54$

20. ganeshie8

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

21. UsukiDoll

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

22. UsukiDoll

unless a = 1 and b = 1

23. ganeshie8

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

24. UsukiDoll

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

25. UsukiDoll

nah I prefer the different bases.

26. ganeshie8

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

27. ganeshie8

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

28. ganeshie8

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

29. 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 $\large \sum\limits_{i = 0}^ {\left\lfloor \log_2 n\right\rfloor}\, \left(\left\lfloor\log_3 \frac{n}{2^i}\right\rfloor + 1\right)$

30. ganeshie8

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

31. 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

32. 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..

33. anonymous

this is called the (generalized) hamming problem

34. 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

35. 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.

36. 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

37. anonymous

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

38. 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

39. UsukiDoll

at least I tried :(

40. 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...

41. 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.. )

42. ganeshie8

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