If you would like to generate this sequence yourself, you can either Learn to program, or you can use the following bit of Python code. I wrote this to investigate a new feature of Python called generators, (a.k.a. co-routines), so it will look a little weird.
The code generates as many terms as you might want of the sequence detailed above - have fun playing with it! All computer science dorks (like myself) should note that the running time of this program is O(2n), but that's unavoidable because the sequence is doubly exponential, so the number of digits in the answer we have to write down grows exponentially every time. It does allow you to generate the sequence in any base you want, though - which is kind of cool.
#!/usr/bin/env python2.2
# Compute the audioactive sequence using coroutines. We can calculate it in
# any base, but since no number higher than 3 ever appears, it is all the same
# for bases 4 and higher
from __future__ import generators
def convertToBase(num, base):
rv = ""
while num > 0:
rv = ("%d" % (num % base)) + rv
num = num / base
return rv
def audioactive(base):
s = '1'
while 1:
yield s
snew = ""
count = 0
curr = ''
for i in s:
if i != curr:
if curr:
snew = snew + "%s%s" % (convertToBase(count,base),curr)
count = 1
curr = i
else:
count = count+1
s = snew + "%s%s" % (convertToBase(count,base),curr)
if __name__ == '__main__':
from sys import argv, exit
if len(argv) < 3:
print "Usage:", argv[0], " number base"
exit(1)
sequence = audioactive(int(argv[2]))
for i in xrange(int(argv[1])):
print sequence.next()
The first 10 terms of the sequence in base 2 look like:
1
11
101
111011
11110101
100110111011
111001011011110101
111100111010110100110111011
100110011110111010110111001011011110101
1110010110010011011110111010110111100111010110100110111011