#OK to post homework

#Zidong Zhang, 3/28/2021, Assignment 17



#CheckMordell(N):that verifies that 而(p)*而(q)=而(pq) for all primes p,q, between 2 and N. Please use RseqFF(24,N), with N as large as your computer would agree to.


CheckMordell:= proc(N)   local tau , flag ,  p , q , i , j, ip , k:

tau:= RseqF(24,N):
flag:=0:
p:=[]:

#create a list that contains all the primes less than N
for i from 1 to N do
	ip:=ithprime(i):
	if ip <=N then
		p:=[op(p), ip]:
	else
		break:
	fi:
od:		
q:=p:

#verifies that 而(p)*而(q)=而(pq) for all primes p,q, between 2 and N. if not equal it should return 0, otherwise return 1.
for k in p do
	for j in q do
		if k*j<=N then	
			if  tau[k]*tau[j]=tau[k*j] then
				#RETURN(FAIL):
				flag:=1:
			
			fi:
		else
			break:
		fi:
	od:
od:
flag:
end:





#CheckDeligne(N):that outputs the primes between 2 and N such that |&tau[(p)|/2 p11/2 is as small as possible, followed by its value (that hopefully is less than 2). Who is the lucky prime? What is the value of |而(p)|/2 p11/2 for that lucky prime?

###my output of CheckDeligne(N) is a list with 2 element. the first one is the lucky prime. and the next one is the |而(p)|/2*p^11/2.



CheckDeligne :=proc(N)    local tau , p, i, w, L, k, w1, ip:

tau := RseqF(24,N):
p:=[]:

#create a list that contains all the primes less than N
for i from 1 to N do
	ip:=ithprime(i):
	if ip <=N then
		p:=[op(p), ip]:
	else
		break:
	fi:
od:
#p:
#we compute every |tau(p)|/2*p^(11/2) and 

#tau1:=[]:
w:=evalf(abs(tau[p[1]])/(2*p[1]^(11/2))):
L:=[p[1], w]:
	for k from 2 to nops(p) do
		#w:=abs(tau[p[k]])/(2*p[k]^(11/2)):
		w1:=evalf(abs(tau[p[k]])/(2*p[k]^(11/2))):
		if w1<w then
			L[1]:=p[k]:
			L[2]:=w1:
			w:=w1:
		fi:
	od:
L:
end:


#	CheckDeligne(20)=[7, 0.1882738483]
# 	CheckDeligne(40)=[31, 0.1657561056]
#	CheckDeligne(100)=[43, 0.008882639437]
#	CheckDeligne(500)=[43, 0.008882639437]
#	CheckDeligne(1000)=[907, 0.002282222814]
#	CheckDeligne(4000)=[2927, 0.0004943967857]



####################

####[seq(FZ(i,10000),i=1..10)]=[4, 8, 3, 10, 1561, 6, infinity, 4, infinity, 7]

#####[seq(FZ(i,10000),i=1..20)]=[4, 8, 3, 10, 1561, 6, infinity, 4, infinity, 7, infinity, infinity, infinity, 5, 54, infinity, infinity, infinity, infinity, infinity]


#####[seq(FZ(i,10000),i=1..30)]=[4, 8, 3, 10, 1561, 6, infinity, 4, infinity, 7, infinity, infinity, infinity, 5, 54, infinity, infinity, infinity, infinity, infinity, infinity, infinity, infinity, infinity, infinity, 10, infinity, infinity, infinity, infinity]

#####it's not in OEIS, and I don't think it should be in OEIS since I'm not seen the pattern of how many time that infinity occures between 2 number.






############from C17.txt
RseqF:=proc(r,N) local f,i:
f:=1:
for i from 1 while (3*i+1)*i/2<=N do
 f:=f+(-1)^i*q^((3*i-1)*i/2) +(-1)^i*q^((3*i+1)*i/2):
od:

 
f:=taylor(q*f^r,q=0,N+1):
[seq(coeff(f,q,i),i=1..N)]:

end: