# Sequence of numbers 6

Fill in the correct number in this sequence:

$$\begin{array} & 3 & 4 & 2^2 & 5 & 10 & 60 & ? \end{array}$$

The options are

$$\begin{array} 60/10 & 6 & 12/2 & 7 & 49/7 \end{array}$$

So, apparently, the answer is either $$6$$ or $$7$$, but why? Also, I do not understand why the 4 is written in terms of a square and is present 2 times in the sequence. I have no clue how to solve it.

Source: a publicly available practice test in a book for an IQ-test I got via a friend in the Netherlands.

• Well I think I might have noticed one thing: ROT13 Gur cnggrea fgnegf jvgu guerr, gura fxvccvat gjb fgrcf tbrf gb svir, fb gura fxvccvat gjb fgrcf ntnva, vg fubhyq tb gb frira. Abj jr whfg unir gb svaq n eryngvbafuvc orgjrra gur erfcrpgvir fxvcf fgrccrq naq gur ahzore cerprqvat rnpu bs gurz, V guvax... jung qb lbh erpxba? – Mr Pie Dec 7 '18 at 15:23
• I have edited my comment (hope you understand)?I have to get to bed anyways. I hope this is right! Ooh, fingers crossed... g'night :D – Mr Pie Dec 7 '18 at 15:39
• In the original puzzle, does it show 2^2 or $2^2$? – Dr Xorile Jan 30 '19 at 3:38
• @Xorile The second one. – Dennispuz Jan 30 '19 at 5:04
• Has a useful answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful. – Rubio Mar 7 '19 at 4:45

the sequence is $$3,4,X,5,X,X,6,X,X,X,7,\dots$$. That is, start with $$3,4,5,6,7\dots$$ and insert $$a_n-3$$ $$X$$'s after each entry. The $$X$$'s can be anything - they only distract.
We notice that $$\begin{array}{c|c}\text{number}&0&1&2&3&4&5&6\\\hline\#\text{occurences}&2&1&2&1&1&1&1\end{array}$$ Going through each choice, the pattern of occurrences becomes $$\begin{array}{c|c}\text{choice}&60/10&6&12/2&7&49/7\\\hline\small\text{pattern}&\small4,2,2,1,1,1,2&\small2,1,2,1,1,1,2&\small2,2,4,1,1,1,1&\small2,1,2,1,1,1,1,1&\small2,1,2,1,2,1,1,1,0,1\end{array}$$ and only the first two seem to have a pattern in the pattern of occurrences, since the first one has one $$4$$, two $$2$$s and three $$1$$s, and rises back up; the second one has odd numbers of consecutive $$1$$s broken by the $$2$$s. If we look back at the sequence, $$3$$ and $$4$$ are together, separation of $$0$$, and $$4$$ and $$5$$ are separated by $$1$$ - the $$2^2$$ as it is not written as a single-digit integer. Therefore, the obvious prediction is that two spaces along, it is a single-digit integer and is the next one along from $$5$$. So it is $$6$$, and the answer is the second choice.