Hi, I have determined, heuristically, that the sum of primes < n ~ n^2/(2*log(n)-1). I did this by doing the "Noble" thing and supporting evidence of sums of...
Hey Cino, Sorry for the layman comments, as probably no one will react onto your posting. Thanks for doing the great job of trying to find a connection/link ...
... I like how the cranks will just high-five each other, rather than doing a simple Google search, which would have revealed that someone has already studied...
Here it is again. Sorry for the jumble in the last post. Drat hotmail to yahoo conversion. I deleted the bad one and did this one in the yahoo primeform group....
to me anyway... let x = ((3+sqrt(5))/2); let n be some odd number; iff (x^n (rounded up)) mod n == 3 and (x^((n+1)/2) (also rounded up)) mod 2 == 1, (odd...
... Indeed. Here's a much better formula: est6(n)=n^2/2/(log(n)-c-d/log(n)-e/log(n)^2) with c=1/2,d=1/4,e=2/5 gives: est6(10^10)/2220815456739024823-1. =...
Hi Max, Glad you enjoyed tweeking my beautiful formula est5 =n^2/(2*log(n)-1). Take it easy. Just kidding. Seriously, though, I will use your formula when new...
... er... not sure the 5 should be squared... ... I think I tried to do a copy-paste from your post... ... ah, nostalgy... but.... wasn't it rather 48 K of...
... sorry,... it should be... this statement [(x^((n+1)/2) (also rounded up)) mod n] mod 2 == 1, (odd result), now it's right... proofreading... I hate it....
look at my last corrected post... now it should get them all if you calculate it correctly ... From: mad_math_max <oeis-2008@...> Subject:...
... ********************************** I think the proof may be as follow Let P prime = 2*n+1 , n=(P-1)/2 The sum of odd numbers from 1 to P is (n+1)^2 or...
Hi, ... prime ... (log ... Yep, this is pretty much the way I did it in http://tech.groups.yahoo.com/group/primeform/message/8982 In a later post, I...
You wrote: Re: interesting primality test: "it's from an article on T3 sequences (primality test)". Could you point me to the article, or email me a copy? I'm...
Re: interesting primality test, Based on the listing below, the 2nd part of the test appears to have a consistency: Excluding the primes n=2 which is even, and...
Here's a computational look at a T3 sequence. say you choose 17 then calc. terms Tn T0 = 2, T1 = 3, T2 = 7 or T2 = T1^2-2 = 3^2-2 = 7 mod 17 = 7, ...
... which is a weak Lucas test. The recursion T(n)= if(n==0,2,if(n==1,3,if(n%2,T((n-1)/2)*T((n+1)/2)-3,T(n/2)^2-2))) is solved by the Lucas sequence ...
it fails with 9, even, and 15, etc. but this can be ruled out if the last step before. let's say T9 for the case of T17 were EVEN 15 does fail the...
sorry for getting uptight w/you Mr. Broadhurst. here's my final attempt to explain away the skepticism. take a number, say 23; compute all the following terms...
... Try it with n=377, where T((n+1)/2)%n is odd. ... Here is a relevant benchmark: On an 2.2 GHz Opteron machine, the BPSW pseudoprimality test of Pari-GP...
Mr. Chickhurst... j/k wow,... you use a lot of jargon that I'm familiar with, but is it really necessary ??? I noticed that 9 and 15 won't pass the T3 sequence...
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