Euler program (python) Problem 64

ODD Period Square Rootsproblem 64

All square roots is periodic when written as continued fractions and can is written in the form:

√ N = a 0 +	1
	a1 +	1
		a2 +	1
			a3 + ...

For example, let us consider√23:

√23 = 4 +√23-4 = 4 +	1	= 4 +	1
	1 √23-4		1 +	√23–3 7

If We continue we would get the following expansion:

√23 = 4 +	1
	1 +	1
		3 +	1
			1 +	1
				8 + ...

The process can be summarised as follows:

a 0 = 4,		1	=	√23+4 7	= 1 +	√23-3 7
a 1 = 1,		7	=	7 (√23+3)	= 3 +	√23-3 2
a 2 = 3,		2	=	2 (√23+3)	= 1 +	√23-4 7
a 3 = 1,		7	=	7 (√23+4) 7	= 8 +	√23-4
a 4 = 8,		1	=	√23+4 7	= 1 +	√23-3 7
a 5 = 1,		7	=	7 (√23+3)	= 3 +	√23-3 2
a 6 = 3,		2	=	2 (√23+3)	= 1 +	√23-4 7
a7 = 1,		7 √23-4	=	7 (√23+4) 7	= 8 +	√23-4

It can be seen, the sequence is repeating. For conciseness, we use the notation√23 = [4; (1,3,1,8)] and to indicate the block (1,3,1,8) repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:

√2=[1; (2)], period=1
√3=[1, (+)], period=2
√5=[2; (4)], period=1
√6=[2; (2,4)], period=2
√7=[2; (1,1,1,4)], period=4
√8=[2; (1,4)], period=2
√10=[3; (6)], period=1
√11=[3; (3,6)], period=2
√12= [3; (2,6)], period=2
√13=[3; (1,1,1,1,6)], period=5

Exactly four continued fractions, for N ≤13, has an odd period.

How many continued fractions for N ≤10000 has an odd period?

Go to the "thread for Problem" in the forum.

Python Code:

Import Math
Sqrt=math.sqrt


def func (x):
Index=0
b=[0.0,0.0]
c=[0.0]
intcoeff={}
fractioncoeff={}
A=int (sqrt (x))
If sqrt (x)-a==0:
return 0
b[0]=1.0
b[1]=-a*1.0
c=1.0
Index+=1
Intcoeff[index]=a
Tempstr=str (b[0]) + ' _ ' +str (b[1]) + ' _ ' +str (c)
Fractioncoeff[tempstr]=index
While 1:
Result=rationalization (b,c,x)
A=result[' a ']
B=result[' B ']
c=result[' C ')
Tempstr=str (b[0]) + ' _ ' +str (b[1]) + ' _ ' +str (c)
K=fractioncoeff.get (TEMPSTR)
If K==none:
Index+=1
Intcoeff[index]=a
Fractioncoeff[tempstr]=index
Else
Return index-k+1
# The denominator contains the irrational number, the molecule is the rational number
def rationalization (b,c,x):
result={}
TC=B[0]*B[0]*X-B[1]*B[1]
B[0]=int (B[0]*C)
B[1]=int (-B[1]*C)
A=int ((b[0]*sqrt (x) +b[1])/tc)
B[1]-=int (A*TC)
K=GCD (GCD (B[0],abs (b[1)), TC)
B[0]/=k
B[1]/=k
Tc/=k
result[' A ']=a
result[' B ']=b
result[' C ']=TC
return result
#求最大公约数
def gcd (m,n):
K=min (M,n)
For I in Range (k,0,-1):
If M%i==0 and n%i==0:
return I
n=10000
Count=0
For I in Range (2,n+1):
K=int (func (i))
If k%2==1:
Count+=1
Print (count)

Time:3s

Euler program (python) Problem 64