IPV6 -Octillion-iteration v1.3

import ipaddress
import math
#for addr in ipaddress.IPv6Network('2001:db8:1000:0000:0000:0000::0/38'):
  #('2001:db00:0000:0000:0000:0000:0000::0/56'):
#  print(addr)

# ipv6no = ((2**112)-36)
ipv6nok = ((2**90)-90)

print("total IPV6 addresses 2 **90 - 90 embedded 0's:")
print(ipv6nok)
print("total IPV6 network addresses:")
#list(ipaddress.ip_network('2001:db8:1234::/128').hosts())

"""Please feel free to comment on this post and give your opinion, I would have liked to make the network 
addresses larger, however python module breaks at /38, leaving an Octillion unique addresses to use, which is
 less than 0.0000001% of total IPv6 addresses available!

 Based on my calculations from an Apple Macbook Pro running this script on 107128 ip addresses per second,  
 print(addr) will take three hundred sixty-six trillion, four hundred twenty-nine billion, one hundred 
 forty-two million, ninety-two thousand, seven hundred fifty-six years to complete.
 
 If your interested here is the link with the calculations 
 https://docs.google.com/spreadsheets/d/1iqnL-orQhXhzl7Gr9dSHxhjr_WuwnHteLQcQA6r5p4Q/edit?usp=sharing"""

ipv6noT = 3.4 * 10**38
print("total IPV6 addresses 3.4 * 10**38:")
print("340 undecillion", ipv6noT)