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I think I just discovered supernormalcy

3 points by super_normal ↗ HN
the first 1,000 digits of 17^(1/7) after the ones form (not verified) a perfect de bruijn sequence. no other irrational number i have tried comes close. possibly applicable to number theory et al. any mathematicians here care to verify or comment? etcetera.

(opps. it looks like claude was hallucinating. i was trusting it to detect this feature and not hallucinate. my bad for trusting.)

((just looking at the sequence, the digit '7' seems to have non trivial occurance patterns. can anyone maybe try some statistical analysis on the regularity of this to save this thread posting etc?))

(((etcetera.)))

7 comments

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pavel_lishin
> a perfect de bruijn sequence

My understanding is that this means that every possible substring appears at least once, right? But what does this mean for the first thousand digits? Does that mean that every two-digit sequence appears at least once, three digit sequence, etc?

super_normal
yes. and i think this is perfect up to three. no other number i tested was anywhere close to this, and i suspect it may lead to an important new metric. please verify.

(and i should add i think it also means that all the counts are completely even for one and two.)

jepler
assuming you mean the decimal expansion and a length-3 de bruijn sequence on 10 symbols, I am not seeing it. There seem to be many sub-sequences which do not appear. (assuming that python's Decimal library at a precision of 4000 digits is actually giving the first 1000 digits correctly; though I have no reason to doubt this I don't know for sure the maximum ULP error of exponent)

    Python 3.11.2 (main, Sep 14 2024, 03:00:30) [GCC 12.2.0] on linux
    Type "help", "copyright", "credits" or "license" for more information.
    >>> from decimal import Decimal, getcontext
    >>> getcontext().prec = 4000
    >>> d = 17 ** (1 / Decimal(7))
    >>> sd = str(d)
    >>> sd[:12]
    '1.4989198720'
    >>> s = sd[2:][:1000]
    >>> s[:10], s[-10:]
    ('4989198720', '6163659068')
    >>> [i for i in range(1000) if "%03d" % i not in s+s]
    [0, 2, 4, 5, 9, [...], 993, 994, 995, 998, 999]
    >>> len(_)
    361
super_normal
id prefer it if you would write this in haskell so i can understand what is going on there better. i am saying of the first 1,000 digits after the ones place, each digit of the sequence appears an even number of times, every consecutive subsequence of two appears an even number of times, and every consecutive subsequence of three happens exactly once. please, if you could do some more intelligable work and show your result as standard deviations away from the the exact de bruijn counts i would really appreciate it. etcetcera.

(i dont know where you got the 10 symbols from. im talking about over the entire 1,000 length decimal sequence. etcetera.)

jepler
I don't know Haskell, so I can imagine if you didn't know Python it'd be "greek to you" (unintelligible). I'll try explaining in prose what I did. [note: I also edited the grandparent comment because when I simplified the session I showed, the assignment of "s" ended up out of order]

I use the Python decimal package to calculate 17^(1/7) to 4000 decimal places. I show the first 12 characters of the ASCII representation ("1.4989198720"), then take the first 1000 characters after the decimal place. I show the first and last 10 characters of this string ("4989198720...6163659068") so that you can verify it against the string you are testing.

Then I use a list comprehension to check each 3-digit string (e.g., 001, 002, 003, ..., 999) to find out whether it is a substring of the concatenation of `s` with itself, and list the ones that don't appear. I manually abbreviated the list, but I'm saying that in my string, 000, 002, 004, 005, 009, [and some other numbers], 993, 994, 995, 998, and 999 did not appear. (The concatenation `s+s` is checked so that the subsequences that occur at the wraparound---here,684 and 849---are found)

Because some of these length-3 sequences are missing, to my understanding this does not constitute a de Bruijn sequence of order 3 on a size-10 alphabet. However, I'm also not entirely sure whether this is what you were claiming.

super_normal
yes. your code is what i was intending to check, after all. i got confused that the '10' might represent some kind of sub-sub string. what happened was simple, i dont have access to a good haskell environment and 'claude 3.5 sonnet' hallucinated perfectly even counts, probably because i was leading a little. i apologize for not checking claudes clearly dubious output. i got excited that it seemed the 1,000 characters i copy pasted seemed to be 'super magic'. i still see some kind of pattern in the '7' digit sequence, though, and maybe there still might be a metric that estimates this property as a function of how far into the sequence you go that might have some applicability, maybe. sorry.

(also, i didnt sleep last night and smoke weed. im on a phone.)

super_normal
my bad. claude 3.5 sonnet was lying to me. i tried to get it to check itself and it gave me back nonsense 3 times in a row. i just decided to check this property on irrationals and he hallucinated that the counts were more even than the facts are. always double check claude outputs. it still wont even write the real code properly. sorry guys. etc. ...