Re: More number theory tidbits, paper?

From: No Way (Not_at_real.com)
Date: 08/01/04 

Date: Sun, 01 Aug 2004 17:39:41 +0200

On 30 Jul 2004 13:46:34 -0700, jstevh@msn.com (James Harris) wrote:

>My previous tidbit was that [(N-4)/6] for N>2 gives the count of odd
>numbers up to and including N that have 3 as a factor.
>
>So then it follows that
>
>[N/3] - [N/6] - 1 = [(N-4)/6]
>
>using a rather simple principle to count composites, as [N/3] gives
>the count of naturals that have 3 as a factor, while [N/6] gives the
>count that have 6 as a factor, so subtracting that count off subtracts
>off the evens, and 1 is subtracted to handle 3 itself, as it is prime.
>
>That relation was shown to be true by a poster in my previous thread,
>and I noted that I'd be interested if someone could derive the
>relation, but no one has.

First of all, the relation is only true for even N. Don't forget to
include that bit of information.

Secondly, I have no idea why and how you would derive that particular
form, since there is a far more logical form (IMHO) that works for odd
N as well. Here's how you 'derive' it (in a completely
nonproffessional way).

k is a positive integer.
Start with (This can easily be seen by drawing the functions):
floor(N/k) = floor(N/(2*k)) + floor((N+k)/(2*k))
floor(N/k) - floor(N/(2*k)) = floor((N+k)/(2*k))

subtract -1 from both sides:
floor(N/k) - floor(N/(2*k)) - 1 = floor((N+k)/(2*k)) - 1
floor(N/k) - floor(N/(2*k)) - 1 = floor((N+k)/(2*k) - (2*k)/(2*k))
floor(N/k) - floor(N/(2*k)) - 1 = floor((N-k)/(2*k))

Now substitude k for 3:
floor(N/3) - floor(N/6) - 1 = floor((N-3)/6)

>I further noted that you can find a formula for 5 that gives the count
>of primes that are odd that do not have 3 as a factor that have 5 as a
>factor,

Primes? With five as a factor? Is this some new definition of a prime
number or am I missing something obvious?

>and it is
>
>[(N-16)/10] - [(N-16)/30], with even N>6,
>
>and I then have that
>
>[N/5] - [N/10] - [N/15] + [N/30] - 1 = [(N-16)/10] - [(N-16)/30]
>
>for even N>6, and notice here that I have to add numbers back in on
>the left with [N/30] to handle a double subtraction from [N/10] and
>[N/15] for numbers that are multiple of 30.

Why not use [(N+5)/30] + [(N-5)/30]? It works for odd N as well...

>Clearly there's a good deal of compression on the right side, as 5
>terms on the left are required versus only two on the right.
>
>The relations work out to positive infinity, so the compression on the
>right is rather effective, but how does it work?
>
>I don't know if anyone else can even derive that expression,

No, just like before, I cannot derive THAT expression...

> and have
>yet to see anyone derive the one for 3 though I have seen it proven to
>be correct.
>
>That is, when I gave [(N-4)/6] and pointed out that it is the count of
>odd composites that have 3 as a factor, then someone else was able to
>prove that assertion to be correct, but, how did I get the formula?

I don't know... why don't you tell us...

>I'll see if anyone out there that bothers to read
>alt.math.recreational or sci.math can derive anything I've shown here,
>and if not, then I think I have yet another paper I can right up to
>add to my growing pile (all others are currently at journals STILL
>under review).
>
>Some additional information can be found at my blog:
>
>http://mathforprofit.blogspot.com/
>
>
>James Harris