Proof that $\pi$ has infinitely many $1$s

Question

I understand that there is currently no proof in any base for whether $\pi$ is normal or disjunctive.

Are there any proofs of whether specific sequences of numbers are infinitely occurring in $\pi$? For instance, is there a formal proof that 1 is infinitely occurring in the decimal expansion of $\pi$?

Does this answer your question? What is the state of our ignorance about the normality of pi? or Does pi contain 1000 consecutive zeroes (in base 10)? — Timothy Chow, Apr 29 '22 at 16:28
It's not known whether the decimal digits of $\pi$ are eventually all $0$s and $2$s. — mathworker21, Apr 29 '22 at 17:52
We do know that there exists a $10^{100}$-length sequence of digits that appears infinitely often in the decimal expansion of $\pi$ (and $10^{100}$ can be replaced by any other positive integer), but this is true for any infinite sequence of decimal digits . . . — Dave L Renfro, Apr 29 '22 at 19:01
@SamHopkins In base 2, every real number has a binary expansion which contains infinitely many 1's, since $\sum_{k=1}^\infty2^{-k}=1$. — Z. M, Apr 29 '22 at 19:10
@Gerry Myerson: Means every finite digit sequence appears (equivalently, appears infinitely often) -- see here. Related is "lexicon" -- see The typical number is a lexicon by Calude/Zamfirescu (1998). While normal numbers are full measure (large in one way) and meager (small in another way), the disjunctive numbers are measure zero and meager and smaller still -- see this 19 February 2003 sci.math post. — Dave L Renfro, Apr 30 '22 at 06:28
@Dave, if a number is normal, then every finite digit sequence appears, so it's disjunctive. The normals are a subset of the disjunctives, no? — Gerry Myerson, Apr 30 '22 at 12:46
@Gerry Myerson: Yes. I wondered whether I should be writing something like this after 4 hours of sleep, but I did it anyway. What's small (measure zero & meager, even $\sigma$-porous) is the set of non-disjunctive numbers. There is a dichotomy of Lebesgue measure and Baire category for normal numbers: Most numbers ARE normal (Lebesgue measure) and most numbers are NOT normal (Baire category; indeed, "NOT" in a very strong way -- see Olsen and Stylianou). However, most (both ways) numbers are disjunctive. — Dave L Renfro, Apr 30 '22 at 17:27
Compare a related problem for $\sqrt{2}$, which is a bit more accessible: https://arxiv.org/abs/1711.01722. — Kurisuto Asutora, May 02 '22 at 08:22

Proof that $\pi$ has infinitely many $1$s

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