> #THE NUMBER OF ATTENDANCE QUESTIONS WERE: 9
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> 
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> #Q1
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> #Convert the following recurrence 6*f(n) + 12*f(n+1) + 18*f(n+3) = 0 to explicit form: 

> #6*f(n) + 12*f(n+1) + 18*f(n+3) = 0 
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> #                     18*f(n+3) = - 6*f(n) - 12*f(n+1) 
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> #                         f(n+3)= - 2/3*f(n+1) - 1/3*f(n)
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> 
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> #Q2
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> #f(n)=3*f(n-1)-4*f(n-2), f(0)=1, f(1)=-2, find f(5) and f(6)
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> 
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> #1,-2,-10,-22,-26,10,134
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> #f(5) := 10
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> #f(6) := 134
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> 
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> #Q3
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> #What is F(10^6)? How long it takes
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> #What is f(10^6)? How long it takes
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> 
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> #can't find due to the long computing time, maybe it takes infinite time
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> 
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> #Q4
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> #Consider the sequence that is defined by the recurrence f(n+1000) = f(n+999) + 5*f(n) 
> #What is the operator ope(N) such that ope(n) f(n) = 0
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> 
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> #ope := N^1000 - N^999 - 5N
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> 
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> #Q5
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> #Characterize the sequence(s) that satisfy a homogeneous recurrence of order zero. 
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> 
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> #10*f(n)=0
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> 
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> #Q6
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> #Prove that d(n)/n! -> 1/e3
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> 
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> #d(n)/n! = n * d(n-1) / n!
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> #        = n! * Sum((-1)^i/i!, i = 0..n) / n!
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> #        = Sum((-1)^i/i!, i = 0..¡Þ)
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> #lim = e^-1 = 1/e
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> 
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> #Q7
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> #What is the OEIS A number of this sequence?
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> 
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> #A000085
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> #Number of self-inverse permutations on n letters, also known as involutions; number of standard Young tableaux with n cells.
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> 
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> #Q8
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> #Find the operators in N and N^(-1) annihilated by w(n)
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> #ope = N^2 - N - (n + 1)
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> 
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> #Q9
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> # Do SumTools[Hypergeometric][ZeilbergerRecurrence]
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> 
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> #ZeilbergerRecurrence(T, n, k, f, l..u)
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> #T-hypergeometric term of n and k
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> #n-name
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> #k-name
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> #f-name of the recurrence function
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> #l..u-range for k
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