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Computing terms of a recurrence

Solve the first problem 16 problems

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  1. Fifth term of Alicia's sequence ●○○○○ 10 ptsAlicia starts a sequence with m=3m=3. Her algorithm is: 1. Write down the number mm as the first term. 2. If mm is even, set n=12mn=\frac{1}{2}m; if mm is odd, set n=m+1n=m+1. 3. Write down the number m+n+1m+n+1 as the next term. 4. Set mm equal to the value of the term just written in Step 3. 5. Repeat Steps 2, 3 and 4 until she has five terms, then stop. What is the fifth term of her sequence?
  2. Fifth term of a recursive sequence ●○○○○ 10 ptsIn the sequence {an}\{a_n\}, a1=2a_1=2 and an+1=2an−1a_{n+1}=2a_n-1. Find a5a_5.
  3. First four terms of a recurrence ●○○○○ 10 ptsFind the first four terms of the sequence defined by tn=2tn−1+4t_n=2t_{n-1}+4 with t1=8t_1=8.
  4. Evaluating the Ackermann function ●●○○○ 20 ptsDefine a function A(m,n)A(m, n) by A(m,n)={n+1if m=0,A(m−1,1)if m>0 and n=0,A(m−1,A(m,n−1))if m>0 and n>0.A(m,n) = \begin{cases} n+1 & \text{if } m = 0, \\ A(m-1, 1) & \text{if } m > 0 \text{ and } n = 0, \\ A(m-1, A(m, n-1)) & \text{if } m > 0 \text{ and } n > 0. \end{cases} Compute A(2,1)A(2, 1).
  5. First term of a recurrence with a constant subtraction ●●○○○ 20 ptsA sequence a1,a2,a3,…a_1, a_2, a_3, \ldots satisfies an=3an−1−xa_n = 3a_{n-1} - x for every n≥2n \ge 2, where xx is an integer that is the same for all terms. Given that a2=7a_2 = 7 and a4=43a_4 = 43, find a1a_1.
  6. Third term of a power-plus-constant sequence ●●○○○ 20 ptsLet {an}\{a_n\} be the sequence given by an=an+ma_n = a^n + m for positive integers nn, where a<0a < 0 and mm are constants. Given a1=2a_1 = 2 and a2=4a_2 = 4, find a3a_3.
  7. Sequence with a quadratic shift relation ●●●○○ 50 ptsLet the sequence a0,a1,a2,…a_0, a_1, a_2, \ldots satisfy a0=a1=11a_0=a_1=11 and am+n=12(a2m+a2n)−(m−n)2a_{m+n}=\frac{1}{2}\left(a_{2m}+a_{2n}\right)-(m-n)^2 for all integers m,n⩾0m, n \geqslant 0. Find a45a_{45}.
  8. Initial terms of a recursive sequence ●●●○○ 50 ptsLet the sequence of positive integers {an}\{a_n\} satisfy an+3=an+2(an+1+2an)a_{n+3}=a_{n+2}\left(a_{n+1}+2a_n\right) for n=1,2,…n=1,2,\ldots, and a6=2288a_6=2288. Find a1a_1, a2a_2, a3a_3.
  9. Possible second terms of a sequence ●●●●○ 100 ptsA sequence a1,a2,…a_1, a_2, \ldots of non-negative integers is defined by the rule an+2=∣an+1−an∣a_{n+2}=|a_{n+1}-a_n| for n≥1n\geq1. If a1=999a_1=999, a2<999a_2<999, and a2006=1a_{2006}=1, how many different values of a2a_2 are possible?

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