Re: Frequency of relative primes.
- --- In email@example.com, Yann Guidon <whygee@...> wrote:
>Yann, I see what you mean. In this case we can indeed simplify and still get an exact result. More on that later.
> why simplify at the cost of precision ?
> losing precision creates the risk of uncertainty,
> not the best thing to keep around in this situation.
> > If we filter out factors 2,3,5 and 7 the formula for theHere's more then. In the example above, it is seen that for every 210 units, there are exactly 15 'twins' (two numbers two apart) without factors of 2,3,5 and 7.
> > exact number of twins is getting out of hand:
> > floor((n-1)/210) + floor((n+11)/210) + floor((n+17)/210) +
> > floor((n+29)/210) + floor((n+41)/210) + floor((n+59)/210) +
> > floor((n+71)/210) + floor((n+101)/210) + floor((n+107)/210) +
> > floor((n+137)/210) + floor((n+149)/210) + floor((n+167)/210) +
> > floor((n+179)/210) + floor((n+191)/210) + floor((n+197)/210)
> this looks nice but I fear that it's inefficient,
> otherwise the TPC would be proven for a long time, right ?
> thank you for this refreshing post, and hoping for more
This is an example where we have simplified a result without losing exactness. We're just focusing on whole intervals of 210 rather than exactly what goes on in within these intervals.
(On a similarly note, as we saw previously, there are exactly 3 twins every 30 units for numbers without factors of 2,3 and 5.)
Chris D. sent me an email that was intended for the list (how many times has *that* happened!) and mentioned that he wanted the numbers with factors of two to remain, not to be filtered out.
Well as it turns out, when we don't filter out the factor of two (and only filter out factors of 3,5 and 7), we have exactly the same number of twins (15) in an interval exactly half as large (105). Now to some this may be a self evident and familiar result, but I have trouble getting my head around this observation.
And there's more. When we filter out factors of 2,3,5,7 and 11, we get exactly 135 twins in every interval of 2*3*5*7*11.
When we filter out factors of 2,3,5,7,11 and 13 we get exactly 1485 twins in every interval of 2*3*5*7*11*13.
The formula is pretty easy to see from this data (and which I have little doubt is already known): If we filter out all the numbers from 1 to n that have prime factors p1,p2,p3...pk, with p1>2, then in every interval of p1*p2*p3*...*pk numbers there will be exactly (p1-2)*(p2-2)*(p3-2)*...*(pk-2) twins. If we want to filter out numbers with factors of two as well then we just double the interval length to 2*p1*p2*p3*...*pk and arrive at the same number of twins: (p1-2)*(p2-2)*p(p3-2)*...*(pk-2). Note that p1,p2,p3... do not have to be consecutive primes.