Superintelligenz ist auch F.U.D. (OAI:The-irrationality-exponent-of-pi-is-2-September-24-2026)

From	Mild Shock <janburse@fastmail.fm>
Newsgroups	sci.logic, comp.lang.prolog, sci.math
Subject	Superintelligenz ist auch F.U.D. (OAI:The-irrationality-exponent-of-pi-is-2-September-24-2026)
Date	2026-10-10 00:34 +0200
Message-ID	<11abq5p$17a99$2@solani.org> (permalink)
References	(6 earlier) <10ssije$13umd$3@solani.org> <11478b2$fv28$2@solani.org> <119omnu$pkou$2@solani.org> <11ab8sg$16ucq$1@solani.org> <11abbds$170ap$1@solani.org>

Cross-posted to 3 groups.

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Hi,

An Elegant Continued Fraction for π
L. J. Lange - Pages 456-458 | Volume 106, 1999
The American Mathematical Monthly
https://www.tandfonline.com/doi/pdf/10.1080/00029890.1999.12005070

Meine Vermutung der jeweils zu einer bestimmten
Tiefe berechnete Kettenbruch stimmt mit den
partiellen Summen der Nilakantha Reihe überein.

Aber ich weiss nicht ob man mit der Nilakantha
Reihe so direkt rho=2 erhält, wie es den Anschein
macht für den Kettenbruch.

Bye

See also this AI slop from a "frontier model":

https://github.com/openai/math/tree/main/preprints/The-irrationality-exponent-of-pi-is-2-September-24-2026

https://github.com/openai/math/blob/main/reasoning_traces/irrationality-exponent-of-pi.pdf

Mild Shock schrieb:
> Hi,
> 
> Now with integers, to really produce some results:
> 
> % pi2(+Integer, +Rational, -Rational)
> pi2(0, P/Q, R/Q) :- !, R is P-3*Q.
> pi2(N, P/Q, S) :-
>     H is 6*P+(2*N-1)^2*Q,
>     M is N-1,
>     pi2(M, H/P, S).
> 
> Now I see:
> 
> ?- between(10,20,N), pi2(N,6/1,X), Y is X, write(N-X-Y), nl, fail; true.
> 10-475113942765/151242416325-3.1414067184965018
> 11-11922290683065/3794809718700-3.1417360992606653
> 12-322869019821075/102776096548125-3.141479689004255
> 13-9388645795842075/2988412653476250-3.1416831892077552
> 14-291703390224616125/92854250304440625-3.1415189855952756
> 15-9646071455650881825/3070380543400170000-3.141653394197426
> 16-338203386739761387075/107655217802968460625-3.1415419859977827
> 17-12533792135642378629875/3989575718580595893750-3.1416353566793886
> 18-489501901570061970946125/155815096120119939628125-3.1415563302845726
> 19-20095772843114788169975625/6396619735457555416312500-3.1416238066678384
> 20-865107029346752986828909875/275374479611447760672253125-3.141565734658547 
> 
> true.
> 
> And further:
> 
> % roth2(+Integer, -Float)
> roth2(N, X) :-
>     pi2(N, 6/1, A/_),
>     M is N+1,
>     pi2(M, 6/1, B/_),
>     X is 1+msb(B)/msb(A).
> 
> Now I see, no more alternating:
> 
> ?- between(10,20,N), roth2(N,X), write(N-X), nl, fail; true.
> 10-2.1315789473684212
> 11-2.116279069767442
> 12-2.104166666666667
> 13-2.0943396226415096
> 14-2.086206896551724
> 15-2.079365079365079
> 16-2.0735294117647056
> 17-2.0684931506849313
> 18-2.0769230769230766
> 19-2.0595238095238093
> 20-2.056179775280899
> true.
> 
> And for big values pretty much looks like 2+C/N:
> 
> ?- roth2(1000,X).
> X = 2.0011523151058035.
> ?- roth2(10000,X).
> X = 2.000116750597374.
> ?- roth2(100000,X).
> X = 2.0000111335844575.
> 
> Bye
> 
> BTW, because of the way I compute
> the continued fraction, it takes some time:
> 
> ?- time(roth2(100000,X)).
> % Zeit 11258.941 ms, Benutzer 100 %, Lips 53 k
> X = 2.0000111335844575.
> 
> Mild Shock schrieb:
>> Hi,
>>
>> Take this continued fraction:
>>
>> % pi(+Integer, +Float, -Float)
>> pi(0, R, S) :- !, S is R-3.
>> pi(N, R, S) :-
>>     H is 6+(2*N-1)^2/R,
>>     M is N-1,
>>     pi(M, H, S).
>>
>> Nice little corridor for pi:
>>
>> ?- between(10,20,N), pi(N,6,X), write(N-X), nl, fail; true.
>> 10-3.1414067184965013
>> 11-3.1417360992606653
>> 12-3.141479689004255
>> 13-3.141683189207755
>> 14-3.141518985595275
>> 15-3.1416533941974265
>> 16-3.141541985997783
>> 17-3.141635356679389
>> 18-3.1415563302845726
>> 19-3.1416238066678384
>> 20-3.1415657346585473
>> true.
>>
>> We can als compute roth:
>>
>> % roth(+Integer, -Float)
>> roth(N,X) :-
>>     pi(N, 6, A),
>>     M is N+1,
>>     pi(M, 6, B),
>>     X is 1+log(B)/log(A).
>>
>> Now we have a corridor for 2:
>>
>> ?- between(10,20,N), roth(N,X), write(N-X), nl, fail; true.
>> 10-2.0000915947815723
>> 11-1.9999287043438398
>> 12-2.0000565883650765
>> 13-1.999954341912648
>> 14-2.000037375237465
>> 15-1.9999690217963195
>> 16-2.000025963563826
>> 17-1.9999780257693796
>> 18-2.000018763059092
>> 19-1.9999838523285578
>> 20-2.000013997167758
>> true.
>>
>> How proof limsup is 2?
>>
>> Proof Sketch:
>>
>> Lets check the formula 1+log(An+1)/log(An) .
>> If An+1 < pi and An > pi. We know that
>> 1+log(An+1)/log(An) < 1+log(pi)/log(An)
>> and 1+log(An+1)/log(An) < 1+log(An+1)/log(pi).
>>
>> So we get limsup 1+log(An+1)/log(An) =<
>> limsup 1+log(An+1)/log(pi) = 1+log(pi)/log(pi) = 2.
>> But since pi is irrational the lowerbound
>>
>> is anyways 2, so that Liouvielle-Roth
>> irrationality measure = 2 follows.
>>
>> Q.E.D.
>>
>> Bye
>>
>> Mild Shock schrieb:
>>> Hi,
>>>
>>> OpenAI has announced enhancing their Math Swarms
>>> by an interpretability accelerator prompt recently
>>> discovered in a Terrence Tao workshop.
>>>
>>> So they are now takling:
>>>
>>> Mathematician W. Hugh Woodin Explains Continuum Hypothesis
>>> https://www.youtube.com/watch?v=ZC7wglkBWMM
>>>
>>> Infinite fun with infinite worlds
>>> https://ethz.ch/en/news-and-events/eth-news/news/2020/08/infinite-fun-with-the-infinite-worlds.html 
>>>
>>>
>>> A partial Lean 4 formalization is already in
>>> the making. But the hurdles are high. Some
>>> speaker of the effort said:
>>>
>>> The computer might find a new tamagochi knottheory
>>> lemma, which will then allow SpaceX to fly to
>>> workholes, but the hurdle will be hight that
>>>
>>> a single human mathematician mind might
>>> ingest, understand and interpret it.
>>>
>>> Bye
>>>
>>>
>

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