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

**References**:**Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p***From:*Tangerine Luo <tangerine.luo@gmail.com>

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