Euler program (python) problem 65

Convergents of Eproblem 65

The square root of 2 can be written as an infinite continued fraction.

√2 = 1 +	1
	2 +	1
		2 +	1
			2 +< /td>	1
				2 + ...

The infinite continued fraction can be written,√2 = [1; (2)] and (2) indicates that 2 repeats ad infinitum. In a similar way,√23 = [4; (1,3,1,8)].

It turns out this sequence of the partial values of continued fractions for square roots provide the best rational approxi Mations. Let us consider the convergents for√2.

1 +	1	= 3/2
	2

1 +	1		= 7/5
	2 +	1
		2

/table>

1 +	1			= 17/12
	2 +	1
		2 +< /td>	1	& nbsp;
			2

 

1 +	1				= 41/29
	2 +	1
		2 +	1
		&nbs P;	2 +	1
				2

Hence the sequence of the first ten convergents for√2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...

What's most surprising are that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1, ...,K, 1, ...].

The first ten terms in the sequence of convergents for e is:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

Python code:

def func (result,i):
T=I*RESULT[1]+RESULT[0]
RESULT[0]=RESULT[1]
Result[1]=t
return result


def func1 (x):
if (x%3) ==0:
k=2* (X/3)
Else
K=1
Return K


n=100
RESULT=[1,FUNC1 (N)]
For I in Range (n-1,1,-1):
Result=func (RESULT,FUNC1 (i))


RESULT[0]+=2*RESULT[1]
TEMP=STR (int (result[0]))
Result=0
K=len (temp)
For I in Range (0,k):
Result+=int (Temp[i])
Print (Result)

Time: <1s

Euler program (python) problem