Re: Re: how to solve the integer equation Abs[3^x-2^y]=1
- To: mathgroup at smc.vnet.net
- Subject: [mg103143] Re: [mg103076] Re: how to solve the integer equation Abs[3^x-2^y]=1
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Wed, 9 Sep 2009 04:43:15 -0400 (EDT)
- References: <200909031110.HAA24198@smc.vnet.net> <h7tbeb$fs6$1@smc.vnet.net> <200909070635.CAA03927@smc.vnet.net>
> [...]
> 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 }
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>
- Re: how to solve the integer equation Abs[3^x-2^y]=1
- From: a boy <a.dozy.boy@gmail.com>
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