# [seqfan] Re: A puzzle from Emissary

franktaw at netscape.net franktaw at netscape.net
Wed Nov 30 22:37:54 CET 2011

```That isn't the way I would generalize it. In general, the iteration
stops when you reach a 1 digit number. Which 1-digit number you reach
depends only on the starting number, not on any intermediate choices.
(In fact, it depends only on n modulo b-1.) So I would ask, for each
base b, how many iterations it takes to reach a 1 digit number starting
with n.

The ones now are numbers

d000...0,

where d is any digit.

-----Original Message-----
From: Graeme McRae <g_m at mcraefamily.com>

One way to generalize this "number of steps needed to get to 1"
sequence  to
other bases, b, is as follows...

The base b sequence a_b(n) is the least number of steps it takes to get
from
(b-1)*n to (b-1), interpreting all numbers in base b for the purpose of
evaluating the expressions containing plus signs.

The 1's in this sequence would be for numbers of the form

90000...0

Where "9" represents b-1, and the 2's in this sequence include (but are
not
limited to) numbers of the form

899999...1,

where "8" represents b-2.

--Graeme McRae,
Palmdale, CA

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