This reply is prompted by consumer f” who gave a touch:

Try changing to binary.

So I did…

1

1

2

10

5

101

10

1010

21

10101

42

101010

85

1010101

The sample turns into apparent at this level:

The sequence is a binary quantity n digits lengthy beginning with a 1 and alternating 1s and 0s (then transformed again to decimal, in fact).

This means the reply is:

10101010 which is 170 in decimal.

And to hyperlink it to the earlier solutions:

Double whether it is odd and double+1 whether it is even works as a result of when it’s odd meaning the final digit in binary is a 1. So to proceed the sequence we should add a zero to the top. This is equal to multiplying by 2. When it’s even we have to add a 1 on the top. We know that including a 0 on the top is multiplying by 2 so including a 1 on the top is multiplying by 2 and including 1.

and

My different reply of $f(n)=2^n+f(n−2)$ works by contemplating the ability growth of a binary quantity. For any given member of the sequence you can also make the factor two additional on by including a brand new energy of two which is 2 orders of magnitude larger than the earlier highest. So in case you take a look at $f(5) = 42 = 101010 = 2^5+2^3+2^1$ then you’ll be able to add a brand new energy of two to the start to provide $2^7+2^5+2^3+2^1 = 1010101 = 85$.