Problem Description

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

√2 = 1 +	1
	2 +	1
		2 +	1
			2 +	1
				2 + …

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

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

1 +	1	= 3/2
	2

1 +	1		= 7/5
	2 +	1
		2

1 +	1			= 17/12
	2 +	1
		2 +	1
			2

1 +	1				= 41/29
	2 +	1
		2 +	1
			2 +	1
				2

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

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].

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

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

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

Analysis

Continue the series for the numerator to 100 terms and print the sum of the digits.

Solution

Runs < 1 second in Python.

n, d, i = 2, 1, 100
while i>0:
  c = 1
  if i%3 == 0: c = 2*i/3
  n, d, i = d, c*d+n, i-1
print "Answer to PE65 = ", sum(int(i) for i in str(d))

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