Proof of forumla for upper s-Wuthoff sequence

During my work on I bump into generated formulas that doesn’t exist in the Online Encyclopedia of Integer Sequences ( Lately, I’ve started to submit them whenever I got some time over.

I recently proposed a formula for

A184517 Upper s-Wythoff sequence, where s=4n-2.

The discovered formula is simple


Where Δ\Delta is the difference between the terms, to rewrite it without an explicit delta function we can rewrite it as

a(n)=n2ϕ2(n1)2ϕ2=ϕ2(n2(n1)2)a(n)=\lceil n^2\phi^2 - (n-1)^2\phi^2\rceil = \lceil \phi^2(n^2 - (n-1)^2)\rceil so we get

a(n)=ϕ2(2n1))a(n)=\lceil \phi^2(2n-1))\rceil

This is the formula I proposed after testing it against 10^6 first terms of A184517. An editor asked me if I could prove it which is a very good question since I hadn’t marked it as an empirical finding. Since the formula involves decimals, floor and ceiling it would certainly be comforting with a proof!

Looking in the program section of the entry, we can find the current formula used to generate the sequence, a bit simplified it can be written as

b(n)=12(20+6)(n22+20b(n)=\left\lfloor\frac{1}{2}(\sqrt{20}+6)(n-\frac{2}{2+\sqrt{20}} \right\rfloor

Now assuming this formula is correct, we need to prove that the new formula, a(n)=b(n)a(n)=b(n).

Let’s start by simplifying b(n)b(n) by setting 20=25\sqrt{20}=2\sqrt5 this is our connection to the Golden Ratio, phi=1+52phi=\frac{1+\sqrt5}{2}.

b(n)=12(25+6)(n22+25b(n)=\left\lfloor\frac{1}{2}(2\sqrt{5}+6)(n-\frac{2}{2+2\sqrt{5}} \right\rfloor

Then we simplify it further, by removing some obvious cancellable terms


Focusing on whatever is inside the floor


With some further reworking we get to

n(3+5)ϕn(3 + \sqrt5) - \phi

Now we make a leap of faith and hope that whatever is inside the floor and ceiling is exactly 11 apart

n(3+5)ϕ1=ϕ2(2n1)n(3 + \sqrt5) - \phi-1=\phi^2(2n-1)

Recall that ϕ+1=ϕ2\phi+1=\phi^2 so on the lhs we get


where ϕ2=3+52\phi^2=\frac{3+\sqrt5}{2}

Factoring out out 3+53+\sqrt5 gives us

Dividing it by two and multiplying the n part with 2 finally gives us


So we have proven that the expression inside the floor of b(n)b(n) minus 1 equals the ceiling of a(n)a(n), hence they are the same.

I am sure this took far too many steps and detours before arriving at the conclusion - but I don’t want to spend too much time on this.

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