We have the source code and from that we can figure out that each of the factors p,q,r somehow depends on i.

i = getRandomNBitInteger(1024)
d = getRandomNBitInteger(8)
for _ in range(d):
i = find_next_prime(i)
p = find_next_prime(i)
If we run the script, we can notice that r-q and q-p is less than 200000 which implies that: q=p+x and r=p+x+y, where x,y are integers less than 200000 Therefore, n=pqr=pow(p,3)+... Using gmpy2.iroot(n,3) will give us the number closest to 'i'. Running their script on 'i' will give us p,q,r. Checkout the script given below:

from Crypto.Util.number import *
import gmpy2
def find_next_prime(n):
if n <= 1:
return 2
elif n == 2:
return 3
else:
if n % 2 == 0:
n += 1
else:
n += 2
while not isPrime(n):
n += 2
return n

n=2739699434633097765008468371124644741923408864896396205946954196101304653772173210372608955799251139999322976228678445908704975780068946332615022064030241384638601426716056067126300711933438732265846838735860353259574129074615298010047322960704972157930663061480726566962254887144927753449042590678730779046154516549667611603792754880414526688217305247008627664864637891883902537649625488225238118503996674292057904635593729208703096877231276911845233833770015093213639131244386867600956112884383105437861665666273910566732634878464610789895607273567372933766243229798663389032807187003756226177111720510187664096691560511459141773632683383938152396711991246874813205614169161561906148974478519987935950318569760474249427787310865749167740917232799538099494710964837536211535351200520324575676987080484141561336505103872809932354748531675934527453231255132361489570816639925234935907741385330442961877410196615649696508210921
i=gmpy2.iroot(n,3)[0]
#i=139926822890670655977195962770726941986198973494425759476822219188316377933161673759394901805855617939978281385708941597117531007973713846772205166659227214187622925135931456526921198848312215276630974951050306344412865900075089120689559331322162952820292429725303619113876104177529039691490258588465409367076

while(True):
i=find_next_prime(i)
if(n%i==0):
print(i)
break
p=i
#p=139926822890670655977195962770726941986198973494425759476822219188316377933161673759394901805855617939978281385708941597117531007973713846772205166659227214187622925135931456526921198848312215276630974951050306344412865900075089120689559331322162952820292429725303619113876104177529039691490258588465409397803

while(True):
i=find_next_prime(i)
if(n%i==0):
print(i)
break
q=i
#q=139926822890670655977195962770726941986198973494425759476822219188316377933161673759394901805855617939978281385708941597117531007973713846772205166659227214187622925135931456526921198848312215276630974951050306344412865900075089120689559331322162952820292429725303619113876104177529039691490258588465409494847

r=n//(p*q)

e=65537
c=2082926013138674164997791605512226759362824531322433048281306983526001801581956788909408046338065370689701410862433705395338736589120086871506362760060657440410056869674907314204346790554619655855805666327905912762300412323371126871463045993946331927129882715778396764969311565407104426500284824495461252591576672989633930916837016411523983491364869137945678029616541477271287052575817523864089061675401543733151180624855361245733039022140321494471318934716652758163593956711915212195328671373739342124211743835858897895276513396783328942978903764790088495033176253777832808572717335076829539988337505582696026111326821783912902713222712310343791755341823415393931813610365987465739339849380173805882522026704474308541271732478035913770922189429089852921985416202844838873352090355685075965831663443962706473737852392107876993485163981653038588544562512597409585410384189546449890975409183661424334789750460016306977673969147

phi=(p-1)*(q-1)*(r-1)
d=inverse(e,phi)
print(long_to_bytes(pow(c,d,n)))