Consider the real number 2+3.
When we calculate the even powers of 2+3 we get:
(2+3)² = 9.898979485566356...
(2+3)⁴ = 97.98979485566356...
(2+3)⁶ = 969.998969071069263...
(2+3)⁸ = 9601.99989585502907...
(2+3)¹⁰ = 95049.999989479221...
(2+3)¹² = 940897.9999989371855...
(2+3)¹⁴ = 9313929.99999989263...
(2+3)¹⁶ = 92198401.99999998915...

It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing.
In fact it can be proven that the fractional part of (2+3)²ⁿ approaches 1 for large n.

Consider all real numbers of the form p+q with p and q positive integers and pq, such that the fractional part of (p+q)²ⁿ approaches 1 for large n.

Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of
(p+q)²ⁿ.

Let N(p,q) be the minimal value of n such that C(p,q,n) 2011.

Find N(p,q) for p+q 2011.