Conference rusure::math

    I think more definition is needed.
    
    Suppose (2,4,6) is a color.  Is this the same color as ADDING a
    constant to each, i.e. (7,9,11) ?  Or must we multiply, for
    instance (10,20,30) ?
    
    And why did you suggest omitting (0,0,0) ?  Perhaps we should
    omit (0,0,1) too ?  What's the rule here ?
    
    Also, I'm having trouble seeing why you calculate P(2) by subtracting
    P(1) from it.  What does that have to do with anything ?
    
    I'd count P(2) as

    000,001,002,010,011,012,020,021,022,100,101,102,110,111,112,120,121,122,
    200,201,202,210,211,212,220,221,222.
    
    Then, I would think 111 and 222 are "the same" because they both
    represent even RBG's.  So, 000 is probably the same too, although
    that gets a bit philisophical, since it's black.  (kind of like,
    when bear shits in woods with no one else there, does it stink?)
    
    I'm not sure I could eliminate anything else though.
    
    /Eric

    Re .1:
    
    Two colors, a and b, are the same iff a = kb or ka = b for some k.
    
    Here, the individual components of the triple tell how much red light,
    green light, or blue light to emit.  Changing the intensity of the
    light ("multiplying the triple by a number") changes only the total
    amount of light emitted and not the proportions of it.  So (2,2,2) is
    the same _color_ as (1,1,1), although they are different intensities of
    that color, and (0,0,0) is, in a degenerate sense, the same color.  And
    (2,2,0) is the same as (1,1,0). 
    
    
    				-- edp 

    The colors (a,b,c) and (k*a,k*b,k*c) are considered the same,
    for k > 1.  If k = 1, they *are* the same.

    I'd count P(2) as

        001,    010,011,012,    021,    
    100,101,102,110,111,112,120,121,122,
        201,    210,211,212,    221.

    so P(2) = 19.

    I'd count P(3) as

        001,        010,011,012,013,    021,    023,    031,032,   
    100,101,102,103,110,111,112,113,120,121,122,123,130,131,132,133,
        201,    203,210,211,212,213,    221,    223,230,231,232,233,
        301,302,    310,311,312,313,320,321,322,323,    331,332.   

    so P(3) = 49.

>   And why did you suggest omitting (0,0,0) ?

    Because it makes the above 'chart' for P(2) look symmetric.

If you forget the "0" for a moment, then you are looking for all (x,y,z)
such that gcd(x,y,z) = 1. There is a total of 10^3 (x,y,z)'s. From this
you have to subtract:
 * The amount of (x,y,z)'s such that gcd(x,y,z)=2
 * The amount of (x,y,z)'s such that gcd(x,y,z)=3
 * The amount of (x,y,z)'s such that gcd(x,y,z)=4
 * The amount of (x,y,z)'s such that gcd(x,y,z)=5
    .  .  .
Each line by itself is easier to figure out - only beware of not subtracting
the same thing twice (for example, the "=2" and the "=4" triplets).

As for the "0", for (x,y,0) it's the same story, only with gcd(x,y) and so on.

Note: this is not a complete solution - just an idea.
Nitsan

Let's look at this algorithmically.

    For n <- 1 to 100 do

	! Determine the number T of triples (x,y,z) with 0 <= x,y,z <= n
	! such that gcd(x,y,z) = 1, and (x,y,z) not equal to (0,0,0).

	T <- (n+1)**3 - 1		! This is our initial estimate

	For m <- 2 to n do

	    ! Subtract triples that have gcd(x,y,z) = m

	    ! Consider the prime factorization of m.

	    ! Have we already subtracted the triples with gcd(x,y,z) = m?
	    ! We have, only if the exponents in the factorization all equal 1.

	    If m is square-free
	    Then
		! We've already subtracted these triples

	    ! Perhaps we've subtracted these more than once already.
	    ! This can happen, for example, if m = 10 -- we subtracted
	    ! these once for gcd(x,y,z) = 2, and once for gcd(x,y,z) = 5.

	    Else
		T <- T + ( [n/m]**3 - 1 ) * d(m) 	! [x] is the floor of x
		! where d(m) = 1 - 2*(k mod 2), and m has k prime factors

To show that this is correct, we'd like to show that triples with
gcd(x,y,z) = m are subtracted from T exactly once.  Redefining d(m)
slightly, let

		-1 if m is square-free with an odd number of prime factors
	d(m) =   0 if m is not square-free
		+1 if m is square-free with an even number of prime factors
		      (thus d(1) = +1)

We need to show that

	Sum d(r) = 0

where the sum is over all (possibly non-prime) factors of m, including 1 and m.
(note that we chose to include d(1) = 1 in this summation -- that's why the
sum is not -1).

          q1    q2          qk
Letting p1  * p2  * ... * pk   be the prime factorization of m, this sum is:

	        i1    i2          ik
	Sum d(p1  * p2  * ... * pk  )
	0 <= i1 <= q1
	0 <= i2 <= q2
	    ...
	0 <= ik <= qk

But d(...) is zero if any of the exponents is 2 or greater.  So the sum is:

	        i1    i2          ik
	Sum d(p1  * p2  * ... * pk  )
	0 <= i1 <= 1
	0 <= i2 <= 1
	    ...
	0 <= ik <= 1

This is clearly true for k = 1, since d(1) + d(p1) = +1 + -1 = 0.  Note
that by our definition, d(...) only depends on the parity of i1+i2+...+ik.
So the sum equals

	    Sum      e(i1+i2+...+ik)
	0 <= i1 <= 1
	0 <= i2 <= 1
	    ...
	0 <= ik <= 1

where e(s) = 1 if s is even, and e(s) = -1 if s is odd.

                    a!{Letting C(a,b) = --------, we see that there are C(k,s) elements in the above
                 b!(a-b)!
sum having i1+i2+...+ik = s.  Thus, we can rewrite the sum as:

	Sum C(k,s) * e(s) = Sum C(k,s) * (-1)^s
	0 <= s <= k         0 <= s <= k

A simple application of the binomial theorem shows the sum does indeed equal 0.

In summary,
	        n
	T(n) = Sum ( [n/m]**3 - 1 ) * d(m)
	       m=1
where
		-1 if m is square-free with an odd number of prime factors
	d(m) =   0 if m is not square-free
		+1 if m is square-free with an even number of prime factors
		      (thus d(1) = +1)

Folks may find the discussion of colors, color systems and color
VAXstations in topic 854 of BULOVA::VWS (q.v.) of interest.

Note also, that while the stated problem is of mathematical interest,
it is not a good model of the real world.  I can assure you that the
RGB settings 10/20/40 and 20/40/80 are quite distinct colors, at least
on my VAXstation.
				/AHM

    Is there some better way to evaluate the T(n) given in 685.5?

I noticed a problem with a noisy line in 685.5.  By the bottom, it should read:

		    a!
Letting C(a,b) = --------, we see that there are C(k,s) elements in the above
                 b!(a-b)!

    The d(m) in .5 is often called mu(m).