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Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
*To*: mathgroup at smc.vnet.net
*Subject*: [mg97091] Re: [mg97037] Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Thu, 5 Mar 2009 04:53:43 -0500 (EST)
*References*: <200903031056.FAA02950@smc.vnet.net>
Tangerine Luo wrote:
> I have a conjecture:
> Any odd positive number is the sum of 2 to an i-th power and a
> (negative) prime.
> 2n+1 = 2^i+p
>
> for example: 5 = 2+3 9=4+5 15=2^3+7 905=2^12-3191 ....
> as to 2293=2^i +p ，I don't know i , p . it is sure that i>30 000 if
> the conjecture is correct.
>
> More,
> n = 3^i+p, (if n=6k-2 or n=6k+2)
> for example:8 = 3+5 16=3^2+7 100=3+97, 562 = 3^6 -167
>
> I can't proof this. Do you have any idea?
You worded this in a way that is a bit confusing, and the typography
(strange symbols, in my email client) does not help.
What you conjecture: Given any positive odd integer n, there is an
i>Log[2,n] and a (positive) prime p such that n=2^i-p.
If primes are "random" (whatever that might mean), then this is true
with probability 1. Moreover we can formulate the expected number of
values i that work, for given n, from the distribution of primes as
provided by the Prime Number Theorem. It is
Sum[1/i,{i,Ceiling[Log[2,n]],Infinity}], and that is clearly infinity
(it's a tail of the harmonic series).
So what is happening with 2293? To put this into a sensible context, is
there reason to believe it might sometimes be difficult to find a pair
{i,p} for given odd n? We'll see what one might expect to happen when we
allow all i from Ceiling[Log[2,2293]]=12 to 100000.
In[7]:= Sum[1/i, {i,12,100000}] // N
Out[7]= 9.07027
So for n between 2^11 and 2^12, we'd expect on average around 9 values
of i between Log[2,n] and 100000 to work (that is, provide a difference
from n that is prime). From this, one might then expect that
occasionally some will really require going above 100000. We can test
this as follows. As ever assuming (perhaps incorrectly) "independent
random" locations for the primes, the probability of no i in the range
working is given as
In[11]:= InputForm[ Product[(i - 1)/i, {i, 12, 100000}]]
Out[11]//InputForm= 11/100000
That's a bit more than 1 in ten thousand. So it is not implausible that,
even if the conjecture is true, some odd n in the ballpark of a few
thousand will require i to be alrger than 10^5.
I do not know offhand whether 2293 will find a pair {i,p} with i<10^5.
All I can say is it has not done so as yet, and has been running overnight.
Daniel Lichtblau
Wolfram Research
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