[seqfan] Re: Reproducing A064690 (sign changes of a rational fraction iteration)

Kevin Ryde user42_kevin at yahoo.com.au
Sun Jul 23 13:35:25 CEST 2023

Hugo Pfoertner <yae9911 at gmail.com> writes:
> if someone could independently verify this.

I agree with those, and make next block starting a(126) = 3297298.
A fixed-point interval arithmetic of 1000 bits gets there, and
more bits can go further.  I iterated x -> x - 1/(x+1) which
I think helps keep down "widening" of the interval at each step.

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