## Re: [PrimeNumbers] prime +/- 2^n = primes: A tight sequence and a potential conj

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• In a message dated 28/06/03 07:47:47 GMT Daylight Time, ... An interesting and (to me) novel procedure, Mark. From a given number N, you are constructing
Message 1 of 1 , Jun 28 4:11 AM
In a message dated 28/06/03 07:47:47 GMT Daylight Time,
mark.underwood@... writes:

> I was looking at the first twenty or so primes in binary notation for
> any patterns and of course was made quite dizzy. Anyways, one thing
> that stood out was that if a binary 0 was replaced with a 1 or if a
> binary 1 was replaced with a 0, then another prime would result. In
> other words, if we start with a prime, then at least one other prime
> can be produced by adding or subtracting 2^n, where 2^n is less than
> the original prime.
>
>

An interesting and (to me) novel procedure, Mark.

From a given number N, you are constructing floor(log2(N))+1 new numbers by
"flipping" each of the bits in turn. For large N this is approximately log2(N)
new numbers.

Apart from the one got by flipping the top bit from 1 to 0, all these new
numbers are roughly the same size as N.

Provided the orginal number is odd (which it is, a fortiori, if N is prime),
all but one of these new numbers will be odd.
So the chance of each of them being prime is approximately 2/log(N).
So, as N -> oo, the expected no. of primes is 2*log2(N)/log(N) = 2/log(2) =
2.8854.

> The number of primes generated from the prime 151 to the prime 313
> goes like this:
> 3,3,4,4,3,4,5,4,3,4,5,4,3,4,5,3,3,4,4,3,4,4,2,2,3,4,3,4,4,3,4,3,3,5

which has average 3.618.

So my analysis seems to be on the right track, even for such relatively small
N.

If it is, and if the average number is indeed about 3, Poisson will ensure
that the number actually generated is 0 for infinitely many N.
Who can find the first such?

Mike Oakes

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