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

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

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