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The sequence begins: $2,3,6,1,8,6,8,4,8,4,8....$

(See OEIS A093095.)

$2*3=6; 3*6=1,8; 6*1=6; 1*8=8; 8*6=4,8;$ and so on.

Will there ever be a $5$? Will the sequence ever repeat?

I tried doing this by hand, and so far the only numbers I have are $1,2,3,4,6,8$: none of which can be multiplied together to get a $5$.

Which digits never occur? How does one prove this in general?

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1 Answer 1

up vote 15 down vote accepted

Since you are always taking the product of single digit numbers, you might want to pull out a multiplication table: Multiplication Table Almost the only time $m*n$ for $m,n$ single digits produces a $0, 5, 7$ or $9$ is when one (or both) of $m$ and $n$ includes a $0, 5, 7$ or $9$.

The one exception is $3*3 = 9$, but no single digit numbers multiply to give $33$.

Using this reasoning, suppose one of those four digits appeared. In fact, consider the first time one of those digits appeared. Then it must have been preceded by one of those digits, which contradicts its firstness.

Thus, $0, 5, 7$ and $9$ never appear in this sequence.

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1  
Thank you! I had never heard of OEIS until now. –  EngieOP Apr 7 '13 at 6:21
1  
+1 for the old-school times-table with the '80s girl. –  machine yearning Apr 10 '13 at 14:16
    
I did awhile ago. –  EngieOP Jun 20 '13 at 22:01

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