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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?

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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 $281$.

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  • $\begingroup$ 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. $\endgroup$ Oct 17 at 17:01
  • $\begingroup$ I will say that the answer itself is correct, but the why is wrong. $\endgroup$ Oct 17 at 17:07

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