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CTK Exchange
Kenneth Ramsey
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Dec-20-06, 08:20 AM (EST) |
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"Fibonacci Number Identity"
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Challange Question Prove that F(i)*F(i+1)*F(i+1)*F(i+2) is an oblong number, i.e. is equal to x(x+1) where F(i), F(i+1) and F(i+2) are three consecutive Fibonacci numbers and x is an integer. Hint F(i+2) always equals F(i)+F(i+1). Also F(0) = 0 and F(1) = 1. |
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Ramsey2879
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Dec-25-06, 07:32 AM (EST) |
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2. "RE: Fibonacci Number Identity"
In response to message #1
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Yes but that's just a pea in the pod. Let T(i,j) be the jth element in the ith row of the square Wythoff array Sloane's sequence A035513 in the online encyclopedia of integer sequences https://www.research.att.com/~njas/sequences/table?a=35513&fmt=312 and A(i) be the ith element of Allan Wechsler's "J determinant" sequence A0022344 https://www.research.att.com/~njas/sequences/A022344. Let = the floor of x.Then If j is odd then <3*T(i,j)^2/2>-^2 -A(i) equals a square (x*x) if T(i,j) is even or an oblong, x(x-1), if T(i,j) is odd. If j is even then <3*T(i,j)^2/2>-^2 (plus) A(i) equals a square (x*x) if T(i,j) is even or an oblong, x(x-1), if T(i,j) is odd. |
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Ramsey2879
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Dec-25-06, 07:32 AM (EST) |
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3. "RE: Fibonacci Number Identity"
In response to message #1
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>Even better, you can find x = function of ( F(i 1) ) fairly >easily. >Hint: two cases depending on whether i is odd or even! To correct and improve my earlier message and clarify my reference units for T(i,j) and A(i). Let T(1,1) = 1, T(1,2) = 2, T(2,1) = 4 in the Wythoff Table and in the Wechsler's sequence let A(1) = 1, A(2) = 5The Table is at https://www.research.att.com/~njas/sequences/A035513 The sequence is https://www.research.att.com/~njas/sequences/A022344 If T(i,j) is even then (5(T(i,j))^2)/4 - A(i)*(-1)^i = x(i,j)*x(i,j) where x(i,j) is an integer If T(i,j) is odd then (5*T(i,j)^2 - 1)/4 - A(i)*(-1)^i = x(i,j)*x(i,j) x(i,j) where x(i,j) is an integer x(i,j) = x(i,j-1) plus x(i,j-2) if T(i,j) is odd x(i,j) = x(i,j-1) plus x(i,j-2) plus 1 if T(i,j) is even I don't have enough time to tabulate the x(i,j) table |
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alexb
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Dec-26-06, 10:36 AM (EST) |
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4. "RE: Fibonacci Number Identity"
In response to message #0
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Well, the identity Fi-1Fi+1 - Fi2 = (-1)i has been proved somewhere at this site and in numerous places elsewhere. Is it not what you are looking for? |
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John Warner
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Dec-26-06, 03:10 PM (EST) |
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5. "RE: Fibonacci Number Identity"
In response to message #4
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Yes, Prove this identity and you have my solution. I had forgotten ( did not know? ) the above identity, but it dropped out pretty quickly as what was really in need of being proved. Above is proved by induction. |
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Kenneth Ramsey
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Dec-28-06, 11:17 AM (EST) |
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6. "RE: Fibonacci Number Identity"
In response to message #4
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>Well, the identity > >Fi-1Fi+1 - Fi2 = >(-1)i > > has been proved...Look my last post again. I think that those who try to come up with a relationship between primes should check it out. I mean the A(i) terms seem to be prime as long as T(i,j) and T(i,j+1) are coprime and are otherwise are a prime multiplied by the square of the greatest common denominator. >Is it not what you are looking for? It was but I found an even more impressive identity after reviewing your post. |
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Ramsey2879
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Dec-30-06, 10:42 AM (EST) |
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8. "RE: Fibonacci Number Identity"
In response to message #0
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>Challange Question >Prove that F(i)*F(i+1)*F(i+1)*F(i+2) is an oblong number, >i.e. is equal to x(x+1) where F(i), F(i+1) and F(i+2) are >three consecutive Fibonacci numbers and x is an integer. >Hint F(i+2) always equals F(i)+F(i+1). Also F(0) = 0 and >F(1) = 1. The proof can be found be the property of determiants: the value of a determinant is not changed by adding or substracting the values of one column to or from the corresponding values of another column |F(n) F(n+1)| |F(n)+ F(n+1) F(n+1) | |F(n-1) F(n) | = |F(n-1)+ F(n) F(n) |
But F(n)+F(n+1) = F(n+2) and F(n-1) + F(n) = F(n+1), Thus |F(n) F(n+1| |F(n+2) F(n+1)| |F(n-1) F(n) | = |F(n+1) F(n) | Thus F(n)^2 - F(n+1)*F(n-1) = F(n)*F(n+2) - F(n+1)^2 Since this value is 1 for n = 1, the the square a Fibonacci number differs from the product of its two adjacent Fibonacci numbers by plus or minus 1. Thus the product in question equals an oblong number. |
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mr_homm
Member since May-22-05
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Jan-01-07, 02:44 PM (EST) |
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9. "RE: Fibonacci Number Identity"
In response to message #8
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Or more directly, F(n+1)*F(n-1) + F(n)*F(n+2) = F(n+1)*{F(n+1) - F(n)} + F(n)*{F(n) + F(n+1)} = F(n+1)^2 + F(n)^2, which rearranges into the formula you want for your recursion. Cheers, Stuart Anderson |
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Ramsey2879
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Jan-02-07, 11:38 AM (EST) |
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10. "RE: Fibonacci Number Identity"
In response to message #9
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Yes I see that F(n+1)^2 + F(n)^2 = F(2n+1), but my recursion seems distinct from that. I have (5*F(n)^2 -(If F(n) is odd:1,0))/4 = x*(x + (If F(n) is odd:1,0))more directly since 4* a square is an even square and since 4* an oblong number + 1 is an odd square my recursion simplifies to F(n)^2 + 4*F(n-1)*F(n+1) = a square This too appears to not be well known if it is known at all Note that it works for all Fibonacci like sequences in Wythoff's table This result can be used to find Pythagorean Triangles even to derive the formula for Pythagorean triples. How is this so? Simple just make F(n-1) and F(n+1) coprime squares! |
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