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Title: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by THUDandBLUNDER on Dec 17th, 2006, 4:19pm 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 :o 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. |
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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by towr on Dec 18th, 2006, 1:24am 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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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by THUDandBLUNDER on Dec 18th, 2006, 7:13am on 12/18/06 at 01:24:05, towr wrote:
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. |
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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by SMQ on Dec 18th, 2006, 11:39am on 12/18/06 at 07:13:07, THUDandBLUNDER wrote:
Really? Because in your example gcd(a,b,c) = 1,001... --SMQ |
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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by THUDandBLUNDER on Dec 18th, 2006, 11:53am on 12/18/06 at 11:39:31, SMQ wrote:
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. |
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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by SMQ on Dec 18th, 2006, 1:25pm 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. --SMQ |
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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by THUDandBLUNDER on Dec 18th, 2006, 1:51pm 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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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by THUDandBLUNDER on Dec 18th, 2006, 2:23pm Maybe I should have started with c = a + b, which works out nicely. |
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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by Eigenray on Dec 18th, 2006, 10:11pm on 12/18/06 at 14:23:27, THUDandBLUNDER wrote:
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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Title: Re: {n,k}:{n,k+1}:{n,k+2} = a:b:c Post by THUDandBLUNDER on Dec 19th, 2006, 8:26am 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. |
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