Given the sequence 1,3,3,5,7,11,...,43,k,111,...; find k.

Of course there are an infinite number of infinite sequences that start off with the given finite sequence. Here is one way to get k=69:

Extend the Fibonacci sequence:

...,-1,1,0,1,1,2,3,5,8,13,21,... so that `F_(-2)=-1,F_(-1)=1,F_(0)=0,F_1=1,F_2=1` , etc...

Then the given sequence can be written as a combination of terms of the Fibonnaci sequence:

`T_(n+3)=F_n+F_(n+3)+1,n>=-2`

So `T_1=F_(-2)+F_1+1=-1+1+1=1`

`T_2=F_(-1)+F_2+1=1+1+1=3`

`T_3=F_0+F_3+1=0+2+1=3`

`T_4=F_1+F_4+1=1+3+1=5` and so on...

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`T_9=F_6+F_9+1=8+34+1=43`

`T_(10)=F_7+F_10+1=13+55+1=69`

`T_(11)=F_8+F_11+1=21+89+1=111`

**So k is the tenth term of the sequence and k=69.**

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Perhaps a more direct route is to note that this sequence is very much like a Lucas sequence (the Fibonacci sequence is a special case of the Lucas sequences). Each term is 1 away from the sum of the previos two terms:

First term set to 1

Second term set to 3

1+3-1=3

3+3-1=5

3+5-1=7

5+7-1=11

7+11-1=17

11+17-1=27

17+27-1=43

27+43-1=69

43+69-1=111