Xjyuk

Will this sequence ever have a 6 in it?

9, 1, 1, 1, 10, 3, 1, 1, 10, 5, 1, 1, 10, 1, 5, 2, 1, 1, 10, 1, 1, 1, 5, 4, 1, 1, 10, 3, 1, 1, 5, 1, 1, 1, 5, 2, 1, 1, 10, 5, 1, 1, 5, 3, 1, 1, 5, 4, 1, ...

First, we should determine what the sequence is.

The title provides a big clue - IX suggests Roman numerals (indeed, the sequence starts 9 which is IX in Roman numerals). Moreover, the second part - "Say" suggests the look and say sequence.

Indeed, if we reparse everything into Roman numerals we get the following:

IX, I, I, I, X, III, I, I, X, V, I, ...

Ignoring the IX which is the "seed" of the sequence, taking every subsequent pair of numbers gives a run-length enumeration of every symbol: the sequence starts with 1 I, 1 X, 3 I, 1 X, 5 I, etc.

Therefore, this seems like a reasonable definition of the sequence.

Now, does a 6 ever appear in this sequence?

Every other character must be a single-lettered Roman numeral, i.e. 1, 5, 10, (50, 100, 500, and 1000 are probably not going to show up). So 6 cannot be part of those elements. The only way that it can appear is in the quantity section.

Since the largest number of consecutive number of symbols in a single Roman numeral is 3 (III, VIII, etc.)... wait hold on can't IIII show up in clock faces? Ok ok fine let's recheck.

Quick detour

Fortunately 4 shows up in the sequence. Kind of late, but it'll have to do.

The surroundings are X, I, I, I, V, IV(?), I, I, X. Fortunately X can't be spontaneously formed, so this will described by the next set of X: 1 X, 3 I, 1 V, 1 I, 1 V, 2 I, 1 X. Ok good, 4 = IV, confirmed.

Back to the show

The largest number of consecutive symbols within a single Roman numeral is 3. This rules out ever forming another 10 outside of the original: the absolute maximum quantity we can come up with is 7: something like III, I, III, (V/X)...

This shows that we can't have two X's in a "period", let alone in a row, so in order to get to 6 we must have [II, I], [III, V] or [III, I], [II, V].

Can we ever have two V's next to each other? There are two cases: either there are [V, V] or [?, V], [V, ?]. Consider the earliest instance of such. The former implies that there was an earlier run of 5 V's, which implies an earlier [V, V] (and also, two consecutive [?, V], not possible). (This shows that there can't ever be 3 V's.) The latter implies [III, I], [II, V] <= [?, (V/X)], [I, I], [I, V], [V, I]. (If it was [II, V] then there was an earlier V, V.)

This is preceded by (V/X)IVIIIII. This must be [I, X], [I, V], [III, I], [I, V] (avoiding [II, V/X]). This is preceded by XVIIIV which must be [I, X], [V, I], [II, V] or [I, X], [VI, I], [I, V] both which violate the earliest instance provision.

Therefore we may conclude that a 6 does not appear in this sequence.