To begin with, since we are only interested in the parity, we can calculate each subsequent value modulo 2, returning a 0 for an even number and a 1 for an odd number:

1,0,0,1,0,1,1,1,0, etc (let’s call this sequence b(n)).

Since each value only depends on the previous 6 values, we can return a 7 digit binary number for each value (using the fact that b of negative n is 0, to calculate the first few):

0000001, 0000010, 0000100, 0001001, 0010010, 0100101, etc

And then convert these into their decimal equivalents:

1,2,4,9,18,37,75,23,etc (let’s call this sequence c(n)).

Since whenever a particular number appears in this sequence it will be followed by the same number, and since there are only 128 possible values, from 0 to 127, using the pigeonhole principle, we know that the sequence must develop into a repeating loop, and in at most 128 steps.

We know that c(0)=1, and it doesn’t take much work on a spreadsheet or even by hand, to show that c(63)=1 also. So the sequence c(n), and therefore also the sequence b(n), repeats every 63 steps.

All that remains is to look at 63 consecutive values in b(n) and count how many are 0 and how many are 1.

31 are 1 and 32 are 0, and so the proportion of numbers in the a(n) Rugby sequence that are odd is exactly 31/63, or about 49.2%.