Has anyone conjectured if there exists a covering set for every base so giving rise to Sierpinski and Riesel numbers for all bases? Intuitively it seems...
... David: This is wonderfully accurate agreement. However, I can't help continually being bothered by that <S(T)> = 1/2 "fudge" I helped to introduce. I just...
... If you change 1/2 - 7/8 = -3/8 to my original 1 - 7/8 = 1/8 in ... you /do/ lose 6 decimal places, with an error: 3.004107511 E-11 [Try it!] I take this to...
... Whoops, that should be sav(n1,n2)=sum(n=n1,n2,s(n))/(n2-n1+1) but the picture is unchanged. Averaging S /at the primes/ we see that <S> ~ 1/2 just as Mike...
\\ Here I include the s.d's as well as the means: \p50 \rzerosdml.inp s(n)=n-7/8-ls[n]/2/Pi*log(ls[n]/2/Pi/exp(1)) s1(n1,n2)=sum(n=n1,n2,s(n))/(n2-n1+1) ...
... I don't know why I missed them, but I have just re-visited that marvellous site http://pmmac03.math.uwaterloo.ca/~mrubinst/L_function_public/ZEROS/DEG ...
... Yuk! GP refuses to index vectors of length more than 2^24 - 2 =~ 16.8 M flaking like this: tstit=vector(2^24-2); flake=vector(2^24-1); *** vector: length...
... Dear Andrey, Indeed! But I have a much more intuitive way of saying it: Ignoring noise, S(T) must be close to +0.5, just after a zero, and close to -0.5,...
... I used the low precision data from Waterloo on the first 35M zeros, to compute, million by million, the mean and s.d. (at the zero + epsilon, as Andrey...
Hello David, ... it's about log(log(T)), as shown in chapter XIV, p. 24 of Titchmarsh. If we assume RH, of course :) ... I'm going to calculate them with...
... Thanks, Andrey! It surely seemed to be a slower growth than constant*sqrt(log(T)/log(log(T))) which I took (perhaps wrongly!) as an upper bound from Hugh...
... I got this ... but it comes with bastard accuracy: using 19 digits for the first 1M DeMichel zeros but only good 11 Waterloo digits thereafter :-( If we...
... I guessed as much :-) I was looking at http://web.mala.bc.ca/pughg/thesis.d/masters.thesis.pdf ... t ~ 6 M corresponds to n =~ 12 M, only a third of the...
... Robert: I don't have time to look at this now, but your heuristic argument that there is a covering set in every base smells right. [Jack might like to...
... About the existence of covering sets for any base... Note that b^2-1 will necessarily have at least one prime factor > 2, unless b = 3. Then note that...
... David: I like that "know"... but aren'y you jumping the gun a bit there... Obviously it would be nice if the Waterloo zeros had more than 11- digit...
... Jack, many thanks for this. So it does seem likely but will be hard to prove. It was interesting as I mapped out what the covering set would look like 1...
... of ... downhill ... I have at last written a quite decent Eratosthenes-sieve program (in Pascal) for this problem, and after a run time of 5 GHz-hrs have ...
... If one manages to get a PDF copy of the Bays-Hudson paper and turns up the magnification to 1600% on the top left seismograph of Fig. 1a, it is possible to...
... Making a more delicate detector, using the first 10^6 zeros, at high precision, I suggest that a Ralph-beater is near 10^(18.122218 +/- 0.000012). David...
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