Lucas–Lehmer sequences and Heronian triangles with consecutive side lengths

a nicol · 2016-11-05, 14:18

I noticed this demonstration via Wolfram http://demonstrations.wolfram.com/In...veSideLengths/

Titled: Integral Grid Triangles with Consecutive Side Lengths

If we look at the solutions for 2,4,8,16,32..

We get Heronian triangles where the n +1 sides are of length: 14,194,37634,1416317954,2005956546822746114..

http://i.imgur.com/yHByck5.png

Obviously these are the same values as the classic Lucas-Lehmer sequence with a starting value of 4.

Is there any connection between the primality test for Mersenne numbers and these triangles? Do they help us visualise the proof in any way?

OlderOwl · 2016-11-08, 05:32

I notice the n+1 series follows the recursive formula S(n) = 4*S(n-1) - S(n-2) with S(0) = 2 and S(1) = 4 the Lucas-Lemer series LL(I) = S(2^(I))

2016-11-05, 14:18	#1
a nicol Nov 2016 29 Posts	Lucas–Lehmer sequences and Heronian triangles with consecutive side lengths I noticed this demonstration via Wolfram http://demonstrations.wolfram.com/In...veSideLengths/ Titled: Integral Grid Triangles with Consecutive Side Lengths If we look at the solutions for 2,4,8,16,32.. We get Heronian triangles where the n +1 sides are of length: 14,194,37634,1416317954,2005956546822746114.. http://i.imgur.com/yHByck5.png Obviously these are the same values as the classic Lucas-Lehmer sequence with a starting value of 4. Is there any connection between the primality test for Mersenne numbers and these triangles? Do they help us visualise the proof in any way? Attached Thumbnails

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2016-11-08, 05:32	#2
OlderOwl Nov 2016 1₈ Posts	I notice the n+1 series follows the recursive formula S(n) = 4*S(n-1) - S(n-2) with S(0) = 2 and S(1) = 4 the Lucas-Lemer series LL(I) = S(2^(I))