Re: More number theory tidbits, paper?
From: Arthur J. O'Dwyer (ajo_at_nospam.andrew.cmu.edu)
Date: 08/02/04
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Date: Mon, 2 Aug 2004 12:03:46 -0400 (EDT)
On Sun, 1 Aug 2004, James Harris wrote:
>
> "Arthur J. O'Dwyer" <ajo@nospam.andrew.cmu.edu> wrote...
>>
>> How many odd composites less than N are divisible by 3?
[followed by the proof that it's [(N-4)/6], where [x] = floor(x)]
> But, think about it, how would you even know that these form[ula?]s
> *might* work unless you just had some really good wild guess or
> someone told you, like I did?
Why would I care, unless someone told me to derive it, as you did?
It's not like there's some Deep Truth hiding in the fact that six
equals two times three (which is really all you're saying, there).
> Now consider the more complex:
>
> [N/5] - [N/10] - [N/15] + [N/30] - 1 = [(N-16)/10] - [(N-16)/30], for
> even N>6
>
> Care to brute force that one?
Sure. It would of course be easier if we could remove all those
"floor" notations, though. And let's break it down into two parts:
R1 = [N/5] - [N/10] - [N/15] + [N/30] - 1
R2 = [(N-16)/10] - [(N-16)/30]
The first thing we notice (empirically) is that R1 \not= R2 only
for N=5 (mod 30) or N=25 (mod 30). So let's take that as a starting
point.
[(N-16)/10] = [N/10]-2 for N < 6 (mod 10)
[N/10]-1 for N >= 6 (mod 10)
[(N-16)/30] = [N/30]-2 for N < 16 (mod 30)
[N/30]-1 for N >= 16 (mod 30)
Now we can cancel corresponding terms:
R1 =?= R2 (N \not= 5, 25 (mod 30))
[N/5]-[N/10]-[N/15]+[N/30]-1 =?= [(N-16)/10]-[(N-16)/30]
/ [N/5]-[N/10]-[N/15]+[N/30]-1 =?= [N/10]-[N/30]-4 (N=0..4,10..15 m 30)
| [N/5]-[N/10]-[N/15]+[N/30]-1 =?= [N/10]-[N/30]-3 (N=6..9,20..24 m 30)
\ [N/5]-[N/10]-[N/15]+[N/30]-1 =?= [N/10]-[N/30]-2 (N=16..19,26..29 m 30)
/ [N/5] - 2[N/10] - [N/15] + 2[N/30] =?= -3 (N=0..4, 10..15 m 30)
| [N/5] - 2[N/10] - [N/15] + 2[N/30] =?= -2 (N=6..9, 20..24 m 30)
\ [N/5] - 2[N/10] - [N/15] + 2[N/30] =?= -1 (N=16..19, 26..29 m 30)
And of course it's easy to check this. At a glance, the reader can
see that [N/5] ~ 2[N/10] and [N/15] ~ 2[N/30] (where ~ indicates "is
approximately equal to"), so it makes sense that the sum should be
close to zero in all cases. (The slight discrepancies from zero are
due to the mismatched "steps" in the various graphs of [N/k].)
I don't know why JSH thinks this is an interesting family of
equations. It's pure adhockery on the level of Algebra I. Look,
here are five more:
[N/4] + [N/6] + [N/8] = 13*[N/24] for all N=0,1,2,3 (mod 24)
[N/100] = [(N-17)/100] for all N with the second-from-right
decimal digit >1
[(N+1)/2] = N/2 for all even N>42
[1] = N for all N=1
[[N]] = N for all integer N
-Arthur
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