Conference rusure::math

                      103!     1    104!     1   1   104!
        C[103,40] = ------- = --- ------- = --- -- -------
                    40! 63!   104 40! 63!   104 40 39! 63!
        
                   1   1 64 65   104!     64 * 65
        	= --- -- -- -- ------- = -------- C[104,39]
                  104 40  1  1 39! 65!   40 * 104
        
                  8 * 8 * 13 * 5
        	= -------------- C[104,39] = C[104,39]
                  8 * 5 * 13 * 8
        
        Dan

    I conjucture(sp?) that there are infinte m,n such that
    
    C[m,n] = C[m+1,n-1]
    
    (one does't have to proof what they conjucturate (sp?) , right?)
    
    /nasser

The conjecture is well founded. The conditions can be massaged into:

	5(2m-3n+3)^2 - (5n-1)^2 = 4

call it 	5x^2 - y^2 = 4

These Pell-type equations, when they have a solution, have an infinite number
of solutions. The chain of solutions can be produced using a unimodular 2*2
matrix whose coefficients are obtained by solving the associated equation

	 	5x^2 - y^2 = 1

In this particular case the slight hiccup is that not all x and y values are
soluble for integer m and n. Here's an interesting way to show the solutions:

	x	y	m	n
	1	1
	2	4	1	1	(you can argue about (-1)!)
	5	11
	13	29	14	6
	34	76
	89	199	103	40	(your one, Lynn)
	233	521
	610	1364	713	273	(the next one)
	...
	...

Each x{r+2} = 3x{r+1} - x{r}
and  y{r+2} = 3y{r+1} - y{r}

Dick

    Ref 290.0, particularly paragraph 3
    
    Andrew
    
    PS: Strange as it may seem, this topic is a spin-off of an important
    sales opportunity!

    >call it 	5x^2 - y^2 = 4  (I)
    >
    >These Pell-type equations, when they have a solution, have an infinite
    >number of solutions. The chain of solutions can be produced using a
    >unimodular 2*2 matrix whose coefficients are obtained by solving the
    >associated equation
    >
    >	 	5x^2 - y^2 = 1  (II)
                             ^

    First of all this should have been 5x^2 - y^2 = 5. I didn't explain how to
    derive the matrix that generates the solutions of (I). A suitable solution
    to (II) is (3/2,5/2). The matrix is completed as follows:

    		/ 3/2 1/2 \
    		\ 5/2 3/2 /

    and the solution generation goes:

    	/ x{r+1} \ = / 3/2 5/2 \ * / x{r} \
    	\ y{r+1} /   \ 5/2 3/2 /   \ y{r} /

    The generation of m's and n's from this goes:

    	/ m{r+1} \ = / 8 -3 \ * / m{r} \ + / 9 \
    	\ n{r+1} /   \ 3 -1 /   \ n{r} /   \ 4 /

    >David Singmaster has investigated this problem and has found that
    >C(n+1,k+1)=C(n,k+2) is an identity if n=F[2i+2]F[2i+3]-1 and
    >k=F[2i]F[2i+3]-1 where F[i] are the Fibonacci numbers beginning with
    >F[0]=0.

    Can anyone prove that my solution is equivalent to his?

    >PS: Strange as it may seem, this topic is a spin-off of an important
    >sales opportunity!

    OK, I'll fall for it. What was the sales opportunity?
    
    Dick

Last week I got a message from Martti Hekinen in Valbonne, who is trying 
to sell a couple of V9K's to a University in Spain. They are running an old 
copy of REDUCE on whatever they currently have, and submitted a REDUCE 
program as part of a benchmark for the sale. The program looked for matches
among the C(m,n). 

We couldn't get in touch with the REDUCE people quickly enough to respond, 
so I rewrote the algorithm in MAPLE and sent it back with the result in .0,
which pops out very quickly. Martti is going to try to sell the MAPLE
solution to them. 

In the meantime, I have found that REDUCE is available on VMS for $500, in 
case anyone is interested.

I got some invaluable help from the MAPLE folk who are demoing here at the 
ICIAM conference. Also, there are a couple of new Symbolic Math products
on display there that look interesting.

Re .3: Very good, Dick! Now we have

> binomial(713,273);
353783517152276505700698314852071849495718735701142713669137522738808260668458\
303266608833496206146190109047751319782133000090617056558704082023644438947070\
1575515092325417606033095416151914090271577807800

= binomial(714,272).

    If I read .5 correctly, the next solution should be
    
    C(4894,1870) = C(4895,1869)
    

Right. Anyone with MAPLE can verify this; I'm not going to show the 1412 digits.

        You don't need MAPLE to verify whether C(4894,1870)
        = C(4895,1869).  The first divided by the second is
        
           4894!      1869! 3026!   3025 * 3026   9153650
        ----------- * ----------- = ----------- = ------- = 1
        1870! 3024!      4895!      1870 * 4895   9153650
        
        where the last two multiplications can be done at VMS or
        ULTRIX command level
        
        	$ write sys$output 3025 * 3026
        	% echo '1870 * 4895' | bc
        
        (or just factor the terms and cancel instead of doing the
        multiplication)
        
        Dan

Right! I have this big hammer, so everything tends to look like a nail...

:-)

Lynn Yarbrough