[seqfan] Re: Davide Rotondo's conjecture

[Tomasz Ordowski]

Dear "Seqfans"!

Yes, Robert's proof is brilliantly simple. He was the first to cleverly
prove my hypothesis recently.
Then I found only one small error (oversight). Now I can't fault anything,
it's OK.

When I read the first message in this thread, I started to think whether it
can be proven.
When I came to the conclusion that it was probably true and I was about to
do something in this direction seriously,
I now saw that the matter was already over. My congratulations!

With spring greetings (at the full moon before our Lord's resurrection),

Tomasz Ordowski

______________________
Original message (in Polish)

Drodzy "Seqfani"!

Tak, dowód Roberta jest genialnie prosty.
To właśnie on pierwszy udowodnił sprytnie moją hipotezę niedawno temu.
Wtedy znalazłem tylko jeden mały błąd (niedopatrzenie). Teraz do niczego
nie mogę się przyczepić, jest OK.
Gdy przeczytałem pierwszą wiadomość tego wątku, zacząłem myśleć, czy to da
się udowodnić. Gdy doszedłem do wniosku, że chyba tak i już miałem coś
robić w tym  kierunku na poważnie, zobaczyłem teraz, że już jest po
sprawie. Moje gratulacje!

Z wiosennymi pozdowieniami (przy pełni Księzyca przed zmartwychwstaniem
naszego Pana),

Tomasz Ordowski

wt., 26 mar 2024 o 20:50 <israel at math.ubc.ca> napisał(a):

> This is simply a consequence of the fact that if x < y, 0 <= pi(y) - pi(x)
> <= y - x and the inequality on the right is strict if y-x > 1 except for
> the case of 1 and 3.
>
> Thus we start with b(0) - b(1) = pi(n). While |b(i) - b(i+1)| > 2
> we get |b(i+1) - b(i+2)| = |pi(b(i+1)) - pi(b(i+2))| < |b(i) - b(i+1)|.
> Eventually we must either reach |b(j+1) - b(j)| = 0 or
> |b(j+1) - b(j)| = 1.
> If it's 0, i.e. b(j+1) = b(j), then clearly b(k) = b(j)
> for all k > j.
> If b(j+1) = b(j) + 1 = n - pi(b(j)), then
>  b(j+2) = n - pi(b(j)+1) = b(j+1) or b(j+1)-1.
> If b(j+1) = b(j) - 1, then
>  b(j+2) = n - pi(b(j)-1) = b(j+1) or b(j+1)+1.
> Thus from this point on we either get a 2-cycle or a 1-cycle.
>
> Cheers,
> Robert
>
>
>
>
>
> On Mar 26 2024, Neil Sloane wrote:
>
> >Dear Sequence Fans,
> >
> >On March 23 2024, Davide Rotondo sent me an email with the following
> >interesting conjecture.
> >
> >(I've simplified it a bit.)
> >
> >For a positive integer n, define a sequence b by b(0) = n; b(i) = n -
> >pi(b(i-1)) for i >= 1,
> >
> >where pi(x) = number of primes <= x.
> >
> >The conjecture is that after some initial terms, b becomes periodic with
> >period length 1 or 2,
> >
> > and the n for which the period is 2 are 3 together with A095116, that is,
> >2, 3, 4, 7, 10, 15, 18, 23, 26, 31, 38, 41, 48, ...
> >
> >I've checked a few thousand terms and it seems to be true. Here is a
> random
> >example: for n = 2954, the b sequence is 2954, 2529, 2585, 2578, 2579,
> >2578, 2579, ..., with period 2.
> >
> >There is probably a simple proof, but I don't have time to work on it.
> >
> >I'm going to add it as a comment to A095116, and if anyone has a proof, I
> >will add it there.
> >
> >
> >Best regards
> >Neil
> >
> >Neil J. A. Sloane, Chairman, OEIS Foundation.
> >Also Visiting Scientist, Math. Dept., Rutgers University,
> >Email: njasloane at gmail.com
> >
> >--
> >Seqfan Mailing list - http://list.seqfan.eu/
> >
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>