[seqfan] Re: Fibonacci concatenated

[Eric Angelini]

Bravo Robert!
I had only 1 and 4:
1,4,5,9,14
Best,
É.


> Le 26 janv. 2016 à 05:43, "israel at math.ubc.ca" <israel at math.ubc.ca> a écrit :
> 
> If term n (which is F(n-1)*A + F(n)*B where term 0 is A and term 1 is B) is 
> the concatenation of A and B, namely 10^m*A + B where 10^m > B >= 10^(m-1), 
> that says A = (F(n)-1)*B/(10^m - F(n-1)). As long as F(n-1) <= 10^m - 
> 10^(m-1) you could take B = 10^m - F(n-1) and A = F(n)-1. For example, with 
> n = 5, F(4) = 3 <= 10^1 - 10^0 so A = F(5) - 1 = 4 and B = 10 - F(4) = 7. 
> The sequence goes 4, 7, 11, 18, 29, 47.
> 
> Cheers,
> Robert
> 
> 
>> On Jan 25 2016, Eric Angelini wrote:
>> 
>> 
>> Hello SeqFans,
>> Are there couples of integers A and B
>> such that the Fibonacci-like seq starting 
>> with A,B,(A+B),(A+2B),(2A+3B),(3A+5B),...
>> includes the concatenated integer [AB]?
>> Don't answer if this is old hat, please!
>> Best,
>> É.
>> 
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