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 One to rule them all. (Posted on 2016-12-18)

Consider these sequences in Sloane:

A001519, a(n) = 3*a(n-1) - a(n-2), '5x^2-4 is a square'
A001835, a(n) = 4*a(n-1) - a(n-2), '3x^2-2 is a square'
A004253, a(n) = 5*a(n-1) - a(n-2), 'x^2 - 5xy + y^2 + 3 = 0'
A001653, a(n) = 6*a(n-1) - a(n-2), 'Numbers n such that 2*n^2 - 1 is a square' etc.

Generally, a(n) = k*a(n-1) - a(n-2), with a(0) = 1, a(1) =1

As the quotes show, there is an exuberance of different algebraic forms given for the various sequences but there is actually a relatively straightforward formula that applies equally to all of them, with only the variable k at large.

What is it?

 Submitted by broll No Rating Solution: (Hide) Start by considering each sequence in terms of b^2, using improper fractions where necessary to bring out the underlying pattern: 5a^2-4 = b^2 3a^2-2 = b^2 (6/2a^2-4/2 = b^2) 7/3a^2-4/3 = b^2 2a^2-1=b^2 (8/4a^2-4 = b^2) 9/5a^2-4/5 = b^2 5/3a^2-2/3 = b^2 (10/6a^2-4/6 = b^2) etc. Now we have (1+4/k)a^2-(4/k) = b^2, or k(b^2-a^2)=4(a^2-1) When k = 1, the formula returns as a those Fibonacci Numbers of the form 5a^2-4 = b^2, which may be considered the most elementary form. The corresponding values of b are the bisection of the Lucas numbers A002878 in Sloane.

Comments: ( You must be logged in to post comments.)
 Subject Author Date re: Unproven formula broll 2016-12-18 21:36:10 re: Unproven formula Ady TZIDON 2016-12-18 15:19:21 Unproven formula Jer 2016-12-18 12:39:53
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