#ATTENDANCE QUIZ FOR LECTURE 16 of Dr. Z.'s Math454(02) Rutgers University

# Please Edit this .txt page with Answers

#Email  ShaloshBEkhad@gmail.com 
#Subject: p16
#with an attachment called
#p16FirstLast.txt
#(e.g. p16DoronZeilberger.txt)

#Right after finishing watching the lecture but no later than Oct. 30, 2020, 8:00pm


THE NUMBER OF ATTENDANCE QUESTIONS WERE: 9

PLEASE LIST ALL THE QUESTIONS FOLLOWED, AFTER EACH, BY THE ANSWER
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1. Convert the following recurrences to explicit form:
   6*f(n)+12*f(n+1)+18*f(n+3)=0
A1. f(n+3) = (12*f(n+1)+6*f(n))/18

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2. First 7 terms satisfying the recurrence f(n+2)-3*f(n+1)+4*f(n)=0 and initial conditions 
   f(0)=1, f(1)=-2, f(2)=-10, f(3)=-22, f(4)=-26. Find f(5) and f(6).
A2.  Explicit format => f(n+2) = 3*f(n-1)-4*f(n) => f(n) := 3*f(n-1) - 4*f(n-2)
	f(5) := 3*f(5-1)-4*f(5-2) => -78-(-88) => 10
	f(6) := 3*f(6-1)-4*f(6-2) => 30-(-104) => 134

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3. What is F(10^6)? How long did it take?
   What is f(10^6)? How long did it take?
A3.
F(10^6);
2458745050599682757015378035481882035383438999154580592871756279\

  661111973933372432322733427701904766[...300831 digits...]48275\

  29846518999275111180114315872236327482873665581735845785170957\

  134521733463038891202865307177894


time(%);
                               0.


f := proc(n) option remember; if n = 0 then 1; elif n = 1 then -2; else 3*f(n - 1) - 4*f(n - 2); end if; end proc;

f(10^5);
1872528156411021969329905092373255197478513728585010253649543461\

  348426195124011733104305783418832036[...29904 digits...]600340\

  17338356412516035730487738166716195355236647791914666510875444\

  49117291696197756502799699177894


time(%);
                               0.

f(10^6) is too big to run
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4. 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
A4. f(n+1000)-f(n+999)-5*f(n) = 0
	ope(N):=N^1000 - N^999 - 5

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5. Characterize the sequence (S) that satisfies a (HOMOG.) recurrence of order zero. 
A5. 3*f(n) = 0

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6. Can you prove that d(n)/n! -> 1/e
A6. d(n) := (n-1)*d(n-1) + (n-1)*d(n-2) = 0 =>(n-1)(d(n-1)*d(n-2)) = 0  where d(0):= 1; d(1):=0
	n! := n*(n-1)*(n-2)*(n-3)*...*1
	d(n)/n! := (d(n-1)*d(n-2))/n
	d(n)/n! := 1/e

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7. What is the OEIS A number of the sequence: seq(w(n), n=0..40)
A7. OEIS A number : A000085

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8. Find the operators in N and N*(-1) annihilated by w(n). 
A8. 

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9. Look up the syntax of ZeilbergerRecurrence.
A9.
_*with(SumTools[Hypergeometric]);
T := (-1)^k*binomial(n - 5, k + 5)/binomial((a*k + 5)/a, k);
                        k                       
                    (-1)  binomial(n - 5, k + 5)
               T := ----------------------------
                                /a k + 5   \    
                        binomial|-------, k|    
                                \   a      /    

ZeilbergerRecurrence(T, n, k, f, 0 .. n);
                                                   5            
(-a n + 5 a - 5) f(n) + (a n - 9 a + 5) f(n + 1) = -- (n - 8) (n
                                                   24           

   - 7) (n - 6) (n - 5)


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