replying to myself :: netiquette for dummies. I thought that c=1/2 in a(n+1)=a(n)+a(n)^c was somewhat magical, in the sense that it's as low as you can go without primeless intervals ** at big n ** Some simple observational math (ahum) has changed my mind. Even sticking to the a(n+1)=Floor[a(n)+a(n)^c] with c=1/E=~ 0.37 shows the cloud rising from the 0-line: at n=10000, I have between 4 and 20 primes per interval, and rising. The last zero (primeless interval) was at n=1132, and the last 1 (mono-prime interval) at n=5296. Substituting Li[n] for PrimePi[n] gives an indication: density in the 'Conway intervals' with c=1/E rises with about 7 per 10000 at n=10000. Question: what value of c (or what interval definition other than n+n^c) would make its prime-density non-increasing? An interval of size 1/Li(n), no? Ah, meddling in matters way above my head again. This only serves to illuminate how big the gap is between the pro's and the computationally-endowed-but-otherwise-illiterate mainstream down here. just smile now, Wouter. -----Original Message----- From: Meeussen Wouter (bkarnd) [mailto:wouter.meeussen@vandemoortele.com] Sent: dinsdag 10 juni 2003 14:07 To: 'math-fun' Subject: RE: [math-fun] a prime in each row? since nesting n+n^(1/2) on 1 produces A002984,(a(n) = a(n-1)+[sqrt a(n-1)]) and changing floor to ceiling, we get A002620 (a(n) = Floor[n^2 /4] ), we should count primes in intervals from Floor[n^2 /4] to Floor[(n+1)^2 /4] giving, I recon : {0, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 2, 2, 1, 4, 1, 2, 2, 2, 3, 3, 2, 2, 2, 4, 2, 4, 3, 1, 4, 2, 4, 3, 3, 3, 4, 4, 3, 4, 3, 2, 4, 4, 5, 4, 4, 4, 3, 4, 4, 4, 5, 4, 4, 4, 4, 5, 5, 5, 4, 6, 4, 4, 5, 5, 5, 7, 2, 3, 6, 6, 6, 6, 5, 8, 4, 5, 6, 5, 4, 7, 5, 4, 7, 6, 7, 7, 3, 6, 7, ... (not yet EIS) de amatoribus nihil sed bonum. W. -----Original Message----- From: John Conway [mailto:conway@Math.Princeton.EDU] Sent: donderdag 5 juni 2003 17:07 To: math-fun Subject: Re: [math-fun] a prime in each row? On Thu, 5 Jun 2003 asimovd@aol.com wrote:
David Wilson writes:
<< Propp's conjecture* would imply a prime between n^2 and (n+1)^2, which conjecture I believe is stil outstanding.
Question: Where does the conjectural territory first begin? [2n+1,3n] ? [3n+1,4n] ? Etc.
--Dan
All of these are fine - we know for any positive epsilon that there's a prime between n and n(1+epsilon) for all sufficiently large n (and explicit bounds can be given for how large n need be). The right question to ask is for which c there's necessarily a prime between n and n + n^c (for all sufficiently large n). The Riemann hypothesis would tell us that this is true for any c > 1/2. Ingham proved it for c = 3/5, remarking that his methods would give a strictly smaller number, and later people have explicitly produced values that are strictly smaller, but not by very much. John conway _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.