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        riddles >> easy >> Triangular Numbers.
        
(Message started by: rloginunix on Jan 9^th, 2015, 7:54am)

Title: Triangular Numbers.
Post by rloginunix on Jan 9^th, 2015, 7:54am

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 (T_n is a triangular number):

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_tn.png

(one can generalize further for square, pentagonal, hexagonal numbers, etc.)

Title: Re: Triangular Numbers.
Post by jollytall on Jan 11^th, 2015, 9:54am

I would assume that for the n-th number it is 1+6*(n-1)

Title: Re: Triangular Numbers.
Post by towr on Jan 11^th, 2015, 1:04pm

I'd add a bit of pi to that.

Title: Re: Triangular Numbers.
Post by rloginunix on Jan 11^th, 2015, 4:30pm

I think it's just a typo.

Title: Re: Triangular Numbers.
Post by jollytall on Jan 11^th, 2015, 9:20pm

I wanted to be too quick :-). 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

Title: Re: Triangular Numbers.
Post by rloginunix on Jan 11^th, 2015, 10:39pm

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))