duplicate hunting, pt. 19

[Andrew Plewe]

On 5/23/07, Max Alekseyev <maxale at gmail.com> wrote:

> > The last two integers contain all digits except "2". The sum of the next
> > integer n_imposiible and 415748410 + 1631058958 must end with a "2". So the
> > last digit of n_impossible has to be a "4", which is allready in 415748410.
>
> Agree.

I did not carefully read details of your proof (as I got the same
statement about the finiteness of this sequence by other means), and
did not notice a glitch in your proof at first glance.
Anyway, here is my proof:

Suppose that the next term is x, then the sum
s = 415748410 + 1631058958 + x = 2046807368 + x
may contain only decimal digit 2. Therefore, all solutions are given
by the formula:
x(k) = 2*(10^k-1)/9 - 2046807368
where k>=10.
It is easy to see that while x(10)=175414854 is smaller than
1631058958 (hence, it cannot be an element of our sequence),
all other x(k) contain decimal a digit 2 which is not allowed:
x(11) = 20175414854
x(12) = 220175414854
x(13) = 2220175414854
...
Therefore, there is no next term in this sequence. QED.

Max