Sequence

Question

Conjecture a simple formula for a_n if the first few terms are 

1,2,2,2,2,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4

Comments

No of times $n^{th}$ no appears in sequence is  is $n^{{2^{n}}}$

---

also the range of $n^{th} $ term is  $\left [ \frac{\left ( n-1 \right )n\left ( 2n-1 \right )}{6}+1, \frac{\left ( n \right )\left ( n+1 \right )\left ( 2n+1 \right )}{6} \right ]$

---

but simple formula for a_n?

---

1*1=1, 1 appears 1 time, 2*2=4, 2 appears 4 times, 3*3=9, 3 appears 9 times, so a number repeats n*n times

Answer 1

1 square is 1 so 1 is written once

2 square is 4 so 2 is written 4 times

3 square is 9 so 3 is written 9 times and it goes so on