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 + 1/2 = 3/2
1 + 1/(2 + 1/2) = 7/5
1 + 1/(2 + 1/(2 + 1/2)) = 17/12
1 + 1/(1 + 1/(2 + 1/(2 + 1/2))) = 41/29
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 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: 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.
In [1]:
def convergent(L):
a = L[0]
if len(L) == 1:
return (a,1)
x,y = convergent(L[1:])
return (a*x+y, x)
L = [2]
for i in range(1,34):
L.extend([1,2*i,1])
ans = convergent(L)[0]
print(sum(map(int,str(ans))))
In [ ]: