Re: Re: how to solve the integer equation Abs[3^x-2^y]=1
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- Subject: [mg103156] Re: [mg103076] Re: how to solve the integer equation Abs[3^x-2^y]=1
- From: a boy <a.dozy.boy at gmail.com>
- Date: Wed, 9 Sep 2009 04:45:37 -0400 (EDT)
- References: <200909031110.HAA24198@smc.vnet.net> <h7tbeb$fs6$1@smc.vnet.net>
On Wed, Sep 9, 2009 at 12:18 AM, Daniel Lichtblau <danl at wolfram.com> wrote:
>
> [...]
>> For any integer k and 3^k, suppose 2^j is the closest to 3^k, Gap[k]=|
>> 3^k-2^j| is the subtraction .
>>
>> Gap = Function[k, x = k*Log[2, 3]; Min[3^k - 2^Floor[x], 2^Ceiling
>> [x] - 3^k]];
>> Table[{i, Gap[i]}, {i, 1, 100}]
>>
>> Out[24]:={{1, 1}, {2, 1}, {3, 5}, {4, 17}, {5, 13}, {6, 217}, {7,
>> 139}, {8,
>> 1631}, {9, 3299}, {10, 6487}, {11, 46075}, {12, 7153},.....
>> I find {Gap[i]} is not a increasing sequence. Suppose D is a strict
>> decreasing sub sequence of {Gap[i]} .
>> Q1: is the length of D always less than 3?
>>
>
> I suspect it is straightforward to show that you cannot have three
> consecutive decreases.
>
> As for getting any such subsequence, let's first define, for given
> nonnegative integers mj and nj, the real value tj by
>
> | mj*log(2) / (nj*log(3)) | = 1 + tj
>
> The idea being, we want to find pairs {mj,nj} with corresponding tj very
> small. In this setting, we have
>
> 2^mj - 3^nj = 3^nj * (3^(tj*nj)-1)
>
> So what we require is an increasing set m1, m2, m3 and corresponding n1,
> n2, n3 such that the sequence 3^nj * (3^(tj*nj)-1) decreases. To first order
> approximation, this value is tj*nj*3^nj.
>
> Can we have such trios? Perhaps naively, I think this would depend on
> having "large" convergents somewhere in the continued fraction
> representation of log(2)/log(3). But regardless, the answer is yes, we do
> have such trios. Here is one such.
>
> In[48]:= {Gap[666], Gap[661], Gap[660]} // N
>
> 317 314
> Out[48]= {1.930005508972960 10 , 6.328896257794369 10 ,
>
> 313
> > 4.037250828437273 10 }
>
~~~~~~~~~~~~~~~~~~~~
666>661>660
317>314>313
In[48] do not show a trio which we need. Up to now, I have not find a trio
(i,j,k) that
i<j<k and Gap[i]>Gap[j]>Gap[k]
>
> I found this using the code below.
>
> Gap[k_] := With[{x=k*Log[2, 3]}, Min[3^k-2^Floor[x], 2^Ceiling[x]-3^k]]
> orderedlogs = Ordering[Table[Log[N[Gap[k]]], {k, 1, 5000}]]
> orderdiffs = ListConvolve[{1,-1}, orderedlogs]
>
> Now just look for two consecutive negative signs:
>
> In[61]:= conseqs = Position[Partition[orderdiffs,2,1],
> {a_,b_} /; a<0&&b<0]
> Out[61]= {{659}, {1324}, {1989}, {2654}}
>
> It is reasonable to conjecture that there is an upper bound on these
> decreasing subsequence lengths. If I up the size from 5000 to 10000
> elements, I do not get further trios, which indicates it might be reasonable
> to conjecture that the maximum length of decreasing gap subsequences is in
> fact 3. But when I increas again to 20000, I get a sizeably larger set:
>
> Out[67]= {{659}, {1324}, {1989}, {2654}, {12935}, {13600}, {14265},
> {14930}, {16925}, {17590}, {18255}, {18920}}
>
> Does this mean we might expect decreasing subsequences of length 4 or
> larger? I do not know. One sign that would make me suspect a negative answer
> is that the pattern near such trios is always the same.
>
> In[78]:= Map[orderdiffs[[#[[1]]-2;;#[[1]]+3]]&, conseqs]
>
> Out[78]= {
> {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1},
> {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1},
> {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1},
> {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1}, {1, 7, -5, -1, 2, 1}}
>
> So we have recurring gap undulations, in a manner of speaking.
>
>
> -------------
>> I have another question.
>>
>> Table[Abs[s2 * 2^m + s3 *3^n], {s2, {-1, 1}}, {s3, {-1, 1}}, {m, 0,
>> 100}, {n, 0, 100}];
>> Tally[Sort[Flatten[%]]]
>>
>> The result shows that 21 != 2^i+3^j or |2^i-3^j| and 53 can not be
>> expressed as these form also.
>> But 53= 2 * 3^3 - 1
>> My another question is:
>> Q2: Is any odd prime number p can be expressed as one of these forms:
>> 1. 2^i + 3^j
>> 2. 2^i - 3^j or 3^i - 2^j
>> 3. 2^i * 3^j +1
>> 4. 2^i * 3^j -1
>>
>> The answer to Q2 is true of false? How to prove or disprove it?
>>
>
> Again I do not know the answer but my guess is it is false. Call the values
> you cannot attain in your table (extended to infinity...) non-gaps. I would
> expect the density of such nongaps to be far too large to recover them all
> as numbers in the form 3 or 4 above.
>
> Again, this might be tied to behavior of the continued fraction of
> log(2)/log(3).
>
> Daniel Lichtblau
> Wolfram Research
>
- References:
- how to solve the integer equation Abs[3^x-2^y]=1
- From: a boy <a.dozy.boy@gmail.com>
- how to solve the integer equation Abs[3^x-2^y]=1