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

sqrt(N) = a0 + 1 / (a1 + 1 / (a2 + 1 / (a3 + ... )))

For example, let us consider sqrt(23):

sqrt(23) 
 = 4 + sqrt(23) - 4 
 = 4 + 1 / ( 1 / ( sqrt(23) - 4 ) )	
 = 4 + 1 / ( 1 + ( sqrt(23) - 3 ) / 7)

If we continue we would get the following expansion:

sqrt(23) = 4 + 1/(1 + 1/(3 + 1/(1 + 1/(8 + ... ))))

The process can be summarised as follows:

a	Step 1	Step 1	Step 1
a0	4, 1/(sqrt(23) - 4	1*(sqrt(23) + 4) / 7	1 + (sqrt(23) - 3)/7
a1	1, 7/(sqrt(23) - 3	7*(sqrt(23) + 3) / 14	3 + (sqrt(23) - 3)/2
a2	3, 2/(sqrt(23) - 3	2*(sqrt(23) + 3) / 14	1 + (sqrt(23) - 4)/7
a3	1, 7/(sqrt(23) - 4	7*(sqrt(23) + 4) / 7	8 + (sqrt(23) - 4)/1
a4	8, 1/(sqrt(23) - 4	1*(sqrt(23) + 4) / 7	1 + (sqrt(23) - 3)/7
a5	1, 7/(sqrt(23) - 3	7*(sqrt(23) + 3) / 14	3 + (sqrt(23) - 3)/2
a6	3, 2/(sqrt(23) - 3	2*(sqrt(23) + 3) / 14	1 + (sqrt(23) - 4)/7
a7	1, 7/(sqrt(23) - 4	7*(sqrt(23) + 4) / 7	8 + (sqrt(23) - 4)/1

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

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

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

Exactly four continued fractions, for N <= 13, have an odd period.

How many continued fractions for N <= 10000 have an odd period?