Let a1,a2,… be a sequence of positive integers such that for integers n>2,an=3an−1−2an−2. How many such sequences {an} are there such that a2010≤22012 ?
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Consider the characteristic polynomial for the recurrence an+2−3an+1+2an=0, which is x2−3x+2. The roots are at 2 and 1 , so we know that numbers ai must be of the form ai=a2i−1+b for integers a and b. Therefore a2010 must equal to a22009+b, where a and b are both integers. If the expression is always positive, it is sufficient to say a1 is positive and a is nonnegative, or a+b>0, and a≥0. For a given value of a,1−a≤b≤22012−a22009, so there are 22012−a22009+a possible values of b for each a (where the quantity is positive). a can take any value between 0 and 23, we sum over all such a in this range, to attain 9⋅22012−(1+2+3+4+5+6+7+8)22009+(1+2+3+4+5+6+7+8), or 36(22009)+36, which is our answer.
Source: Omni-MATH,
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