I just had a Thai national test yesterday, and I have a question about proving this.

Given \(n\) seats,\(n \in I^{+} \cup \) {\(0 \)}. And infinite Englishmen.

Prove that the number of ways of Englishmen sitting on the seats is \(F_{n+2}\), such that no two Englishmen sit in an adjacent seats (no one sitting is also counted as 1 way).

Example. (0 is vacant. 1 is occupied)

n=0; \(\varnothing \rightarrow \) 1 way (no seats, no Englishmen)

n=1; 0 1\(\rightarrow \) 2 ways

n=2; 00 01 10 \(\rightarrow \) 3 ways

n=3; 000 100 010 001 101 \(\rightarrow \) 5 ways

n=4; 0000 1000 0100 0010 0001 1010 1001 0101 \(\rightarrow \) 8 ways

n=5; 00000 10000 01000 00100 00010 00001 10100 10010 10001 01010 01001 00101 10101 \(\rightarrow \) 13 ways

n=6; 000000 100000 010000 001000 000100 000010 000001 101000 100100 100010 100001 010100 010010 010001 001010 001001 000101 101010 101001 100101 010101\(\rightarrow \) 21 ways

etc.....

Note: \(F_{n}\) is the nth Fibonacci number, \(F_{1} = 1, F_{2} = 1, F_{n} = F_{n-1} + F_{n-2} \) for \(n \geq 3\).

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## Comments

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TopNewestI tried proving by strong induction. I can do the induction step, but I can't start the basis step. T__T

I use double counting for the induction step.

Find the number of ways given \(n\) seats

1st: nth position is vacant, (n-1) seats remaining, so there're \(F_{n+1}\) ways of sitting.

2nd: nth position is occupied, so (n-1)th seat can't have anyone sitting, (n-2) seats remaining, so there're \(F_{n}\) ways of sitting.

Therefore, there're \(F_{n} + F_{n+1} = F_{n+2}\) ways.

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Let number of ways of seating \( x \) people be \( T_x \)

To Prove:- \( T_n = F_{n+2} \)

Then,

Base Case: \( T_0 = F_2 = 1 , T_1 = F_3 = 2 \)

Inductive case : If true for \( n, n+1\),

Then, \( T_{n} + T_{n+1} = T_{n+2} \) ( Proved in Original Post)

Or, \( T_{n} + T_{n+1} = F_{n+2} + F_{n+3} = F_{n+4} \)

\( \Rightarrow T_{n+2} = F_{n+4} \)

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Thank you ^

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