Problem:
The sequence (an) satisfies a0=0 and an+1=58an+564n−an2 for n≥0. Find the greatest integer less than or equal to a10.
Solution:
The recursion formula is equivalent to an+1=2(54an+534n−an2), which resembles a sum-of-angles trigonometric identity with an=2nsinα. Let θ=sin−1(53). Then cosθ=54 and
an+1=2(2ncosθsinα+sinθ4n(1−sin2α))={2n+1sin(α+θ), if cosα>02n+1sin(α−θ), if cosα<0.
Because cos45∘<54<cos30∘, it follows that 30∘<θ<45∘. Hence the angle increases by θ until it reaches 3θ, after which it oscillates between 2θ and 3θ. Thus
an=⎩⎪⎪⎨⎪⎪⎧2nsin(nθ),2nsin(2θ),2nsin(3θ), if n=0,1,2 if n>2 and even if n>1 and odd.
Thus a10=210sin2θ=1024⋅2524=2524576, and the requested answer is 983.