def CF_of_sqrt(n):
""" Compute the continued fraction representation of the
square root of N.
The first element in the returned array is the whole
part of the fraction. The others are the denominators
up to (and not including) the point where it starts
repeating.
Uses the algorithm explained here:
http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/cfINTRO.html
In the section named: "Methods of finding continued
fractions for square roots"
"""
if isSquare(n):
return [int(math.sqrt(n))]
ans = []
step1_num = 0
step1_denom = 1
while True:
nextn = int((math.floor(math.sqrt(n)) + step1_num) / step1_denom)
ans.append(int(nextn))
step2_num = step1_denom
step2_denom = step1_num - step1_denom * nextn
step3_denom = (n - step2_denom ** 2) / step2_num
step3_num = -step2_denom
if step3_denom == 1:
ans.append(ans[0] * 2)
break
step1_num, step1_denom = step3_num, step3_denom
return ans
def problem66(): return max([(pell(D),D) for D in range(1,1000) if not isSquare(D)])[1]
