What is the $68^{th}$ term of the sequence- $1, 4, 5, 16, 17, 20,....................?$

Question

Given sequence- $1, 4, 5, 16, 17, 20, .........$

I have two guesses here.

Firstly, we see that the differences between two consecutive terms are $3, 1, 11, 1,$ and $3$. So accordingly if we maintain a "sequence of differences" it will turn out to be $3, 1, 11, 1, 3, 1, 11, 1, 3,.......$

thus, by adding the differences we can find the $68^{th}$ term of the main sequence to be $273$.

I thought a second way to solve it.

in the given sequence $1, 4, 5, 16, 17, 20, ........$

if we separate the $odd^{th}$ $(1st , 3rd, 5th,...etc)$ and even-th $(2nd, 4th, 6th,...etc)$ terms, we see that

the difference between two consecutive odd-th term keeps increasing by $4 [ 1, 5, 17,....],$

Similarly, a difference between two consecutive even terms keeps decreasing by $4.$ $[ 4, 16, 20......]. $

In this way, as $68^{th}$ term is an even-th term, so we follow the above rule and get, $68^{th}$ term $= -8036$

I am not completely sure whether any of my strategies is correct. So I need a correct answer

Answer 1

Representing the terms in base $4$, we have

$$1_{10} = 1_{4}$$ $$4_{10} = 10_{4}$$ $$5_{10} = 11_{4}$$ $$16_{10} = 100_{4}$$ $$17_{10} = 101_{4}$$ $$20_{10} = 110_{4}$$

so to get the $68th$ term in the sequence we represent $68$ in binary as $1000100$, pretend it's a base $4$ representation, then convert back to decimal to get $4112.$

EDIT: Ah, I should have read the comments first. @Amin$235$ beat me to this by $5$ hours.

Answer 2

Using the "sequence of difference" hypothesis, the pattern has a period of $4$ and the difference between every fourth term is $16$. Then the $68^{th}$ term is $$1+17\times16-1=\color{green}{272},$$ as you start from $1$, add $68/4$ periods and backtrack by one term (correction $-1$).

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Without this hypothesis and without further information, the $68^{th}$ term can be proven to be $42$.