[seqfan] Re: Fibonacci concatenated

[israel at math.ubc.ca]

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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