OFFSET
0,2
COMMENTS
There is no maximum number of steps and for any value of n, there MUST be a term a(n) that reduces in n steps. This is demonstrable as follows: take any term in the above sequence and convert it to base 2. The resulting value, if interpreted as a base 10 value will require one additional step to reduce. The resulting value may not be the FIRST value to resolve in that many steps, however, so it may not belong in this sequence.
LINKS
Bert Dobbelaere, Table of n, a(n) for n = 0..100
Bert Dobbelaere, Backtracking program (Python)
Chuck Seggelin, Interesting Base Conversions.
EXAMPLE
a(0) = 1 because 1 is the first term that reduces to an unchanging value in zero steps (i.e. 1 is already fully reduced.) a(1) = 10 because 10 reduces in one step (10 in base 2 is 2, 2 does not reduce further.) a(8) = 88 because 88 reduces in 8 steps: 88 --> 80 --> 72 --> 58 --> 53 --> 33 --> 15 --> 11 --> 3.
PROG
(Python)
def A091049(n):
k = 1
while True:
m1 = k
for i in range(n+1):
m2 = int(str(m1), 1+max(int(d) for d in str(m1)))
if m1 == m2:
if i == n:
return k
else:
break
m1 = m2
k += 1 # Chai Wah Wu, Jan 07 2015
CROSSREFS
Cf. A054055 (largest digit of n) A068505 (n as base b+1 number where b=largest digit of n) A091047 (a(n) = the final value of n reached through repeated interpretation of n as a base b+1 number where b is the largest digit of n) A091048 (number of times n must be interpreted as a base b+1 number where b is the largest digit of n until an unchanging value is reached).
KEYWORD
base,nonn
AUTHOR
Chuck Seggelin (barkeep(AT)plastereddragon.com), Dec 15 2003, Jul 09 2008
EXTENSIONS
a(30)-a(31) from Chai Wah Wu, Jan 14 2015
STATUS
approved