>! 110 110 110 110 110 110 110 110 1 <br>
>! 101 101 101 101 101 101 101 101 1

Explanation

>! The sum of N whole numbers being odd or even only depends on each of the numbers being odd or even. So we use 0 and 1, but:<br>
>! <ul><li>any 0 can be replaced with any even number
>! <li>any 1 can be replace with any odd number.
>! </ul>
>! <p>Since we need any 3 successive numbers to be even, we need a repeating pattern of 3 numbers.<br>
>! The possible values of a 3 digit-pattern using only 0 and 1 are:<br>
>! **`000` even** <br>
>! `001` odd <br>
>! `010` odd <br>
>! **`011` even** <br>
>! `100` odd <br>
>! **`101` even** <br>
>! **`110` even** <br>
>! `111` odd <br>
>! <br>
>! That leaves us with `000`, `011`, `101` and `110`:<br>
>! `000 000` -> the sum of any 3 consecutive digits (always `000`) is even.<br>
>! `011 011` -> the 3 consecutive digits can be either `011`, `110` or `101`, all 3 sums are even.<br>
>! `101 101` -> it's the same 3 groups just offset by 1<br>
>! `110 110` -> ditto<br>
>! <br>
>! Another way of getting this list is considering that to get an even sum of 3 digits, we need either:<br>
>! <ul><li>Only even digits (0 -> `000`)
>! <li>Two odd digits (1, the sum of which will be even) and one even digit (0) -> `011`, `101` or `110`.
>! <p><br>
>! Now, by repeating 8 times either of these 4 patterns, we get a series of 24 numbers where the sum of any 3 consecutive numbers is even. Clearly the sum of all 24 numbers is even as well.<br>
>! <br>
>! We still need to add one digit, and for the sum of all numbers to be odd. This condition means the last digit must be 1.<br>
>! Since we need to continue repeating our pattern, it means the pattern must start with 1, which only leaves us `101` and `110`.