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.

Written by gamwe6 who lives and works in San Francisco building useful things. You should follow him on Twitter