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   Triangular Numbers.
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   Author  Topic: Triangular Numbers.  (Read 2358 times)
rloginunix
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Triangular Numbers.  
« on: Jan 9th, 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 T3 = 6 is shown below (Tn 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 11th, 2015, 9:54am »
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I would assume that for the n-th number it is 1+6*(n-1)
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towr
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Re: Triangular Numbers.  
« Reply #2 on: Jan 11th, 2015, 1:04pm »
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I'd add a bit of pi to that.
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rloginunix
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Re: Triangular Numbers.  
« Reply #3 on: Jan 11th, 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 11th, 2015, 9:20pm »
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I wanted to be too quick Smiley. So it is 1 full circle + the 6*(n-2) straight diameters.
It is therefore not only a bit of Pi, but 2 of them:
2Pi+6*(n-1).
 
And reading the question, it has to be expressed in the number of circles, so the above formula is correctly 2Pi+6*(Tn-2).
 
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 11th, 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(Tn + Tn+1) - 2))
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