[seqfan] Re: a(n) divides the concatenation

[David Wilson]

Let two adjacent be a(n) = x and a(n+1) = y.

By your definition, y | concat(x, y) = 10^d * x + y where d is the number of digits in y.

But this implies that y | 10^d * x, that is, a(n+1) | 10^d * a(n).

Now we can show that every a(n) is of the form  2^j * 5^k.

Clearly, the base case a(0) = 1 = 2^0 * 5^0 is true.

Now suppose a(n) is of the form 2^j * 5*k. Then a(n+1) | 10^d * a(n) = 2^(j+d) * 5^(k+d), so a(n+1) is also of the form 2^j * 5^k.

I would conjecture that every 2^j * 5^k eventually shows up in your sequence.

But 3 clearly does not.

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Eric
> Angelini
> Sent: Monday, April 01, 2013 6:11 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] a(n) divides the concatenation
> 
> 
> Hello SeqFans (again, sorry to invade, last post of April First);
> 
> We look for the smallest integer a(n)
> no yet present in S such that a(n) divides the concatenation of a(n-1) and
> a(n).
> S starts with a(1)=1
> 
> S is infinite, of course, and seems to
> look like this (hope no bugs were introduced by hand):
> 
> S=1,2,4,5,10,20,8,16,25,50,40,32,64,80,100,125,...
> 
> (we see for example here that 2 divides 12, that 4 divs 24, that 5 divs 45, that
> 10 divs 510, etc.)
> 
> Is there a pattern somewhere? Will the integer "3" appear at some point?
> Is S interesting?
> 
> Best,
> É.
> 
> 
> 
> 
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