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    The question is: Let the number of n-digit numbers such that no three consecutive numbers are the same be  a_n .

    Explain why  a_{n+2} = 9(a_{n+1} + a_n) .

    Struggling!
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    (Original post by Jooooshy)
    The question is: Let the number of n-digit numbers such that no three consecutive numbers are the same be  a_n .

    Explain why  a_{n+2} = 9(a_{n+1} + a_n) .

    Struggling!
    Consider the digits as a sequence.

    For a sequence of n+1, we can put 9 different digits on (different in that they are not the same as the last digit in the n+1 sequence), hence the 9a_{n+1}

    We can only put the same digit on, if the now last two digits are different to the n'th digit, and there are 9a_n of them.

    Hence,,,,
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    (Original post by ghostwalker)
    Consider the digits as a sequence.

    For a sequence of n+1, we can put 9 different digits on (different in that they are not the same as the last digit in the n+1 sequence), hence the 9a_{n+1}

    We can only put the same digit on, if the now last two digits are different to the n'th digit, and there are 9a_n of them.

    Hence,,,,
    Ah so we kind of partition on whether or not the last two digits are the same? That makes sense! I also forgot 0 could be appended to the end

    Thank you!
 
 
 
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