# An oddly formed sequence?

I've been playing around with sequences lately and came across one that was rather, odd.

$$101$$, $$123$$, $$147$$, $$189$$, $$191$$, $$213$$, $$217$$, $$279$$, $$...$$

##### Hints

Let $$N_i = 101$$...

$$N_{i - 4}$$ through $$N_{i - 1}$$ is $$11$$, $$33$$, $$77$$, $$99$$.
$$N_{i - 8}$$ through $$N_{i - 5}$$ is $$1$$, $$3$$, $$7$$, $$9$$.

Can you determine the next number in my sequence?

Not an extremely elegant solution, but

the complete sequence $$11, 33, 77, 99, 101, 123, 147, 189, 191, 213, 217, 279, ...$$ is made up of four interleaved arithmetic progressions, where three of them have an increment equal to $$90$$, and the other one (the third one, starting with $$77$$) has an increment equal to $$70$$. The subsequences are:
$$11, 101, 191,...$$ (increment $$90$$)
$$33, 123, 213,...$$ (increment $$90$$)
$$77, 147, 217,...$$ (increment $$70$$)
$$99, 189, 279,...$$ (increment $$90$$).

The next term is

the continuation of the subsequence $$11, 101, 191,...$$, which is $$281$$.

• Not quite, though I love this answer 🙂 +1 from me! I’ll add another hint in case you’d like to see why this answer doesn’t fit. Oct 17 '21 at 17:01
• I will say that the answer itself is correct, but the why is wrong. Oct 17 '21 at 17:07