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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?

 See The Solution Submitted by broll No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Unproven formula | Comment 1 of 3
Playing around with the numbers and using some intuition I got the answer:

Rewrite the '5x^2-4 is a square' etc as fractions and the pattern becomes clear:
k=3: (5/1)x^2 - (4/1)
k=4: (6/2)x^2 - (4/2)
k=5: (7/3)x^2 - (4/3)
k=6: (8/4)x^2 - (4/4)
k=7: (9/5)x^2 - (4/5)
[I haven't shown any of these to be true but I've checked the first several terms of each.]
the pattern emerges for any k

((k-2)x^2 - 4)/(k-2) is a square for the generalized sequence.

Again, I haven't done the work to prove it but I've checked the first few terms:
a(0)=a(1)=1.  Formula gives 1^2
a(2)=k-1.  Formula gives (k+1)^2
a(3)=k^2-k-1.  Formula gives (k^2+k-1)^2
a(4)=k^3-k^2-2k+1
a(5)=k^4-k^3-3k^2+2k+1

Next step: find a formula for a(n) then check that my formula actually yields a square.  It looks messy and I'm not actually sure how to do it.

Edited on December 19, 2016, 8:22 pm
 Posted by Jer on 2016-12-18 12:39:53

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