#ATTENDANCE QUIZ FOR LECTURE 19 of Dr. Z.'s Math454(02) Rutgers University

# Please Edit this .txt page with Answers

#Email  ShaloshBEkhad@gmail.com 
#Subject: p19
#with an attachment called
#p18FirstLast.txt
#(e.g. p19DoronZeilberger.txt)

#Right after finishing watching the lecture but no later than Nov. 13, 2020, 8:00pm


THE NUMBER OF ATTENDANCE QUESTIONS WERE: 8

PLEASE LIST ALL THE QUESTIONS FOLLOWED, AFTER EACH, BY THE ANSWER


QUESTION #1:

(i) What is the explicit expression for the sequence, a(0) = 0, a(1) = 0, a(2) = 0, a(n) = 1 for n=3,4,...
(ii) Use Maple to find an explicit expression for the generating function of 0,1,8,27,64,125,..., a(n)=n^3

ANSWER:

(i) f(x) = 0 + 0 + 0 + x^3 + x^4 + x^5 + ..., is x^3*(1+x+x^2+x^3+...) = x^3*(1/1-x)
(ii) Running sum(n^3*x^n, n = 0 .. infinity); we get:
                          / 2          \
                        x \x  + 4 x + 1/
                        ----------------
                                   4    
                            (x - 1)     
QUESTION #2:

Find the EGF of a(n)=n for 0<=n<=5 and a(n)=0 for n>=6.

ANSWER:

x+x^2+x^3/2+x^4/6+x^5/24

QUESTION #3:

What is the EGF of a(n)=0 for n=0,1,2,3,4,5 and a(n)=1 for n>=6

ANSWER:

f(x) = x^6/6! + x^7/7! + ... = e^x - (1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120)

QUESTION #4:

Find the EGF of a(0)=0, a(1)=0, a(n)=(n-2)!

ANSWER:

f(x) = x^2 + x^3/6 + x^4/12 + x^5/20 + ...
According to Maple, sum(x^n/(n*(n - 1)), n = 2 .. infinity) gives:
-x*ln(-x + 1) + ln(-x + 1) + x

QUESTION #5:
(i) What is the A-number of 1,1,4,38,728,26704
(ii) How many digits does the number of labeled connected graphs with 150 vertices has?

ANSWER:
(i) A001187
(ii) f := taylor(log(add(2^(n*(n - 1)/2)*x^n/n!, n = 0 .. 151)), x = 0, 151)
Then, length(150!*coeff(f, x, 150)) gives the number of digits which is,
3365

QUESTION #6:

Use the technique of Weight Enumerator to find the exact number of sequences where each a[i] is in {3,4,7} that add up to 1001

ANSWER:

By the Weight Enumerator, we have (x^3 + x^4 + x^7)^1001

QUESTION #7:

Is [{{1,3,4}, {6,7}}, 52] a member of X(7)?

ANSWER:

Yes, because |{1,3,4} U {6,7}| = 5 and the size of the permutation is 2, so 5+2=n=7.