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Topic: Triangular Numbers. (Read 2358 times) 

rloginunix
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Triangular Numbers.
« on: Jan 9^{th}, 2015, 7:54am » 
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Triangular Numbers (a small generalization I came up with). Express the length of a rubber band stretched over tangent unit circles forming an equilateral triangle as a function of the number of circles. A sample formation for T_{3} = 6 is shown below (T_{n} is a triangular number): (one can generalize further for square, pentagonal, hexagonal numbers, etc.)


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jollytall
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Re: Triangular Numbers.
« Reply #1 on: Jan 11^{th}, 2015, 9:54am » 
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I would assume that for the nth number it is 1+6*(n1)


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rloginunix
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Re: Triangular Numbers.
« Reply #3 on: Jan 11^{th}, 2015, 4:30pm » 
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I think it's just a typo.


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jollytall
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Re: Triangular Numbers.
« Reply #4 on: Jan 11^{th}, 2015, 9:20pm » 
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I wanted to be too quick . So it is 1 full circle + the 6*(n2) straight diameters. It is therefore not only a bit of Pi, but 2 of them: 2Pi+6*(n1). And reading the question, it has to be expressed in the number of circles, so the above formula is correctly 2Pi+6*(Tn2). And Tn=n*(n+1)/2, n=(sqrt(8Tn+1)1)/2 And thus the solution is 2Pi+3*sqrt(8Tn+1)9


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rloginunix
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Re: Triangular Numbers.
« Reply #5 on: Jan 11^{th}, 2015, 10:39pm » 
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That's right. Another, less proper but may be intuitively more clear way to put it is to recall that the sum of any two consecutive triangular numbers is a perfect square: 2(Pi + 3(sqrt(T_{n} + T_{n+1})  2))


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