Is $\{ a_{m}(n) | a_{m}(n)$ with all even digits$\}$ finite set for all $m$?

Asked May 01 '16 at 10:09

Active May 01 '16 at 15:04

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For $m \ge 1$, define $a_{m}(n)$ as the following,

$a_{m}(1) = m$,

$a_{m}(n + 1) = 2 a_{m}(n).$

Consider the set $X_m = \{ a_{m}(n) | a_{m}(n)$ with all even digits$\}$.

For example,

$X_1 = X_2 = \{ 2, 4, 8, 64, \cdots \}$,

$X_3 = X_6 = \{ 6, 24, \cdots \}$,

$X_4 = \{ 4, 8, 64, \cdots \}$,

$X_5 = \{ 20, 40, 80, 640, \cdots \}$,

$X_7 = \{ 28, 224, \cdots \}$.

Is $X_m$ finite set for all $m$?

P.S.

edited Apr 13 '17 at 12:19

Community

asked May 01 '16 at 10:09

TOM

Do you even know any $m$ for sure where $X_m$ is finite? – Hagen von Eitzen May 01 '16 at 10:30
Even for m = 2, it is difficult problem. See https://oeis.org/A068994. – TOM May 01 '16 at 10:32
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So a simpler(?) preliminary question might be: Is there any $m$ for which it is known whether $X_m$ is finite or infinite? – Hagen von Eitzen May 01 '16 at 13:32
Sorry, I don't know. I think it isn't known whether $X_{m}$ is finite or infinite for any $m$. – TOM May 01 '16 at 13:56

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