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   Author  Topic: {n,k}:{n,k+1}:{n,k+2} = a:b:c  (Read 814 times)
ThudnBlunder
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{n,k}:{n,k+1}:{n,k+2} = a:b:c  
« on: Dec 17th, 2006, 4:19pm »		 Quote Quote Modify Modify
One I made up myself:
 
Let {n,k} = binomial coefficient
Let {n,k} : {n,k+1} : {n,k+2}  =  a  : b :  c (for some positive integers a,b,c)
eg. {14,4} = 1,001; {14,5} = 2,002; {14,6} = 3,003 Shocked  
 
Find necessary and sufficient conditions for there to be a solution.
 
If further c = a2, how many solutions are there? (I can find only three by brute force for a < 3000)
In this case, prove that 4a(k + 1)(k + 2) + 1 must be a square number.
 
« Last Edit: Dec 18th, 2006, 9:20am by ThudnBlunder » IP Logged
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #1 on: Dec 18th, 2006, 1:24am »		 Quote Quote Modify Modify
Any constraint on a:b:c ? like, e.g arithmetic progression?
Because otherwise you can always choose a:b:c to be {n,k}:{n,k+1}:{n,k+2}
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ThudnBlunder
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #2 on: Dec 18th, 2006, 7:13am »		 Quote Quote Modify Modify
on Dec 18th, 2006, 1:24am, towr wrote:

Because otherwise you can always choose a:b:c to be {n,k}:{n,k+1}:{n,k+2}

Not sufficient.
I was looking for n = f(a,b,c) and k = g(a,b,c)
 
Also, gcd(a,b,c) = 1
and a < b =< c  
 
So only the left-hand side of the triangle is to be considered, including the central column.
 
 
 
 
 
 
 
« Last Edit: Dec 18th, 2006, 11:51am by ThudnBlunder » IP Logged
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #3 on: Dec 18th, 2006, 11:39am »		 Quote Quote Modify Modify
on Dec 18th, 2006, 7:13am, THUDandBLUNDER wrote:
Also, gcd(a,b,c) = 1

Really?  Because in your example gcd(a,b,c) = 1,001...
 
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #4 on: Dec 18th, 2006, 11:53am »		 Quote Quote Modify Modify
on Dec 18th, 2006, 11:39am, SMQ wrote:

Really?  Because in your example gcd(a,b,c) = 1,001...
 
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Here a,b,c do not represent binomial coefficients.
a:b:c is merely a ratio, and if gcd(a,b,c) = 1 it will be in its simplest terms.
« Last Edit: Jan 25th, 2007, 1:56pm by ThudnBlunder » IP Logged
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #5 on: Dec 18th, 2006, 1:25pm »		 Quote Quote Modify Modify
Well, it's not really an answer yet, but:
 
Let A = xa, B = xb and C = xc for some integer x, then:
A = {n,k} = n!/[k!(n-k)!]
B = {n,k+1} = n!/[(k+1)!(n-k-1)!] = A[(n-k)/(k+1)]
C = {n,k+2} = n!/[(k+2)!(n-k-2)!] = B[(n-k-1)/(k+2)]
 
Solving for n and k gives:
n = (AB + 2AC + BC)/(B2 - AC)
   = (x2ab + 2x2ac + x2bc)/(x2b2 - x2ac)
   = x2(ab + 2ac + bc)/[x2(b2 - ac)]
   = (ab + 2ac + bc)/(b2 - ac)
And similarly, k = (AB + 2AC - B2)/(B2 - AC)
   = (ab + 2ac - b2)/(b2 - ac)
 
So I would say the necessary and sufficient conditions are that 0 < k < n-2 and the above equations for n and k have integer results...  but that's really only part way there.
 
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #6 on: Dec 18th, 2006, 1:51pm »		 Quote Quote Modify Modify
I agree with your n,k although I prefer
   
n = [a(b + c) + c(a + b)]/(b2 - ac)
 
k = [a(b + c)/(b2 - ac)] - 1
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #7 on: Dec 18th, 2006, 2:23pm »		 Quote Quote Modify Modify
Maybe I should have started with c = a + b, which works out nicely.
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #8 on: Dec 18th, 2006, 10:11pm »		 Quote Quote Modify Modify
on Dec 18th, 2006, 2:23pm, THUDandBLUNDER wrote:
Maybe I should have started with c = a + b, which works out nicely.

Quite nicely: b2-ac divides a(b+c) and c(a+b)=c2, hence also their difference bc-ab=b2.  Since a,b are relatively prime, b2-ac = 1 (not -1 since n>0).  We can rewrite this as
 
1  =  [b-a/2]2 - 5[a/2]2  =  5[b/2]2 - [a+b/2]2  =  [(a+3b)/2]2 - 5[(a+b)/2]2,
 
depending on whether a is even, b is even, or neither, respectively, and in either case solve Pell's equation.  The result is a=F2n, b=F2n+1, c=F2n+2 for some n, where Fn is Fibonacci.
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Re:  {n,k}:{n,k+1}:{n,k+2} = a:b:c  
« Reply #9 on: Dec 19th, 2006, 8:26am »		 Quote Quote Modify Modify
Yes, and  
n = [F(2m+2)F(2m+3)] - 1
k = [F(2m)F(2m+3)] - 1
 
First few solutions are
(a,b,c) = (1,2,3) gives {n,k} = {14,4}
(a,b,c) = (3,5,8) gives {n,k} = {103,38}
(a,b,c) = (8,13,21) gives {n,k} = {713,271}
(a,b,c) = (21,34,55) gives {n,k} = {4894,1868}
(a,b,c) = (55,89,144) gives {n,k} = {33551,12814}
 
The mth solution satisfies
F(2m)n = F(2m+2)k + F(2m+1) for m = 1,2,3…………..  
 
And nm/kmtends to Phi2 for large m.
 
« Last Edit: Dec 19th, 2006, 9:30am by ThudnBlunder » IP Logged
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