[robertk]

By the pigeonhole principle, there is a sentence that writes out its entire SHA256 representation this way. Alternatively, the map from these kinds of sentences with 256 terms to 2^256 given by SHA256 admits a fixed point.

[anderskaseorg]

The pigeonhole principle does not say that. It can be used to show that there are two different sentences with the same hash as each other (among any collection of 2^256 + 1 sentences), but it tells you nothing about hashes that agree with the content of the sentence. The probability that a random hash function on a collection of 2^256 sentences has a fixed point is about 1 - 1/e, and it approaches 1 as you add more variations to grow the collection infinitely. But SHA-256 isn’t actually random, so the only way to know this for sure would be to find an example.

[TimWolla]

I don't believe this is necessarily true. Unless I'm misunderstand you, each of the possible variants of spelling out 32 hexadecimal characters could theoretically SHA-256 into the spelled-out hash + 1 (looping around at ff…ff).

[delecti]

I don't see how pigeonhole principle applies to that situation. It could well be that "zero" hashes to 1, "one" hashes to 2... and "f" hashes to 0, extended out to the hash's length.