Puzzle of the Week #437 - Rugby Scores part 2

We saw last week how the number of ways of reaching a particular score by adding 3s, 5s and 7s can be found using the recurrence relation:

a(0)=1, a(n)=a(n-7)+a(n-5)+a(n-3) for n not equal to 0.

Now we go slightly abstract and just look at, for all the values of a(n) for n greater than or equal to 0, the parity (odd or even) of each value.

I’ll give a simpler example just to be clear. If a(n)=a(n-2)+a(n-1) and a(0)=1, you get this familiar sequence: 1,1,2,3,5,8,13,21,34,55, etc. Just looking at the parity we get: OOEOOEOOEO etc. It seems to be the case that two-thirds of the numbers in that sequence are odd, and indeed that is the case.

The rugby sequence begins 1,0,0,1,0,1,1,1,2,1,3,3,3,6,5,8,10,11,17…

Which is: OEEOEOOOEOOOOEOEEOO etc

If you continue that sequence forever, what proportion of the values are odd?