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  • one year ago

Number with only two factors.

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  1. Empty
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    can n and m be a number besides 2 and 3?

  6. Empty
    • one year ago
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    Yeah, they are just consecutive numbers that have only factors of 2 and 3.

  7. UsukiDoll
    • one year ago
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    \[4= 2^1 3^0\] \[5= 2^0 3^1\] like this?

  8. UsukiDoll
    • one year ago
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    NUGH!!!!!!!!!!!! I don't know how to start this afterwards

  9. Empty
    • one year ago
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    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
    • one year ago
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    wow. 5 got skipped.

  11. UsukiDoll
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yeah I know... 2 x 3 or 2^13^1 = 2(3) = 6

  14. UsukiDoll
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    \[\large n= 2^a 3^b = 2^{a+b\log_2 3}\]

  21. UsukiDoll
    • one year ago
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    -_- now the problem *at least for me* looks a bit complicated... in that log form

  22. UsukiDoll
    • one year ago
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    unless a = 1 and b = 1

  23. ganeshie8
    • one year ago
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    Not really, notice that we no longer have to worry about two different bases

  24. UsukiDoll
    • one year ago
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    \[\large n= 2^a 3^b = 2^{a+b\log_2 3}\] if a = 1 and b = 1 \[\large n= 2^1 3^1 = 2^{1+1\log_2 3}\]

  25. UsukiDoll
    • one year ago
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    nah I prefer the different bases.

  26. ganeshie8
    • one year ago
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    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
    • one year ago
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    at least the logs give us some alternative but concrete thing to mess with..

  28. ganeshie8
    • one year ago
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    multiplying \(\log 2\) through out that becomes \( a\log 2 + b\log 3\le x\log 2+y\log 3 \)

  29. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    this is called the (generalized) hamming problem

  34. UsukiDoll
    • one year ago
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    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
    • one year ago
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    @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
    • one year ago
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    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
    • one year ago
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    essentially, we're looking for simple rational numbers close to \(\log_2(3)\approx1.5850\)

  38. anonymous
    • one year ago
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    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
    • one year ago
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    at least I tried :(

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

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