Maths Olympiad Prep

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Problem 1034

AMC 12 late, AIME early
Algebra Difficulty 4.9 Find the answer HMMT November

Let a1,a2,a_{1}, a_{2}, \ldots be a sequence of positive integers such that for integers n>2,an=n>2, a_{n}= 3an12an23 a_{n-1}-2 a_{n-2}. How many such sequences {an}\left\{a_{n}\right\} are there such that a201022012a_{2010} \leq 2^{2012} ?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Consider the characteristic polynomial for the recurrence an+23an+1+a_{n+2}-3 a_{n+1}+ 2an=02 a_{n}=0, which is x23x+2x^{2}-3 x+2. The roots are at 2 and 1 , so we know that numbers aia_{i} must be of the form ai=a2i1+ba_{i}=a 2^{i-1}+b for integers aa and bb. Therefore a2010a_{2010} must equal to a22009+ba 2^{2009}+b, where aa and bb are both integers. If the expression is always positive, it is sufficient to say a1a_{1} is positive and aa is nonnegative, or a+b>0a+b>0, and a0a \geq 0. For a given value of a,1ab22012a22009a, 1-a \leq b \leq 2^{2012}-a 2^{2009}, so there are 22012a22009+a2^{2012}-a 2^{2009}+a possible values of bb for each aa (where the quantity is positive). aa can take any value between 0 and 232^{3}, we sum over all such a in this range, to attain 922012(1+2+3+4+5+6+7+8)22009+(1+2+3+4+5+6+7+8)9 \cdot 2^{2012}-(1+2+3+4+5+6+7+8) 2^{2009}+(1+2+3+4+5+6+7+8), or 36(22009)+3636\left(2^{2009}\right)+36, which is our answer.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.