Problem 318
2011 nines

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 p q, 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.

These problems are part of Project Euler and are licensed under CC BY-NC-SA 2.0 UK