The number of binary sequences of length n that avoid the appearance of 1111 is equal to the n+4th tetranacci number.
I can see the relation between the sequence and tetranacci number but where does the +4 factor come in?
Proving number of 0-1 sequences of length n that avoid 1111 equal to tetranacci (n+4)
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sequences-and-series
recurrence-relations
fibonacci-numbers
2 Answers
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Assuming you have derived the relation: $$a_{n} = a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4},$$ for $n\geq 4.$ Also, $a_0=T_3=1, a_1 = 2=T_4, a_2 = 4=T_5, a_3 = 8 = T_6.$ Hence, by induction, you can see that $a_n = T_{n+3}$.
I guess your problem says $T_{n+4}$ because it's using a slightly different definition of the Tetranacci numbers.
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0I can now see that the problem does give a different definition of the Tetranacci numbers to that which I am used to. I haven't derived the first equation, which is what I'm trying to do. Could you give any hints without giving away the answer please. Thank you – 2017-02-27
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Hint: You can find your answer in the one of exercises of the valuable book A first course in discrete mathematics by Ian Anderson. The exercise and hint for solution of it is as follows
