it fixes the issue We need to show that, for every constant c, there exists n such that 2^(2^ceil(log2(n))) > c * 2^n. Let's consider only n = 2^k + 1 for some integer k > 1; this is our right, since we are not trying to prove the statement for all n. The desired inequality becomes

code :

```
2^(2^ceil(log2(2^k + 1))) >? c * 2^(2^k + 1).
```

```
ceil(log2(2^k + 1)) = k + 1
2^(2^ceil(log2(2^k + 1))) = 2^(2^(k + 1)).
```

```
2^(2^(k + 1)) >? c * 2^(2^k + 1).
```

```
2^(2^(k + 1) - 2^k - 1) = 2^(2^k - 1) >? c.
2^k - 1 >? log2(c)
2^k >? log2(c) + 1
k >? log2(log2(c) + 1).
```