David, have you taken a look at my findings on Viswanath's constant, that I sent you? Or are you kindly ignoring me. http://myspacetv.com/index.cfm? ...
... I had looked and thought that silence might indeed be kinder. However, since I am now asked in public, here is my response. Viswanath's constant is defined...
David you've got it quite wrong. You've failed to explain why you've imagined that characteristic words have anything to do with what I'm saying. Don't imply...
I didn't input the Fibonacci word to begin with. That was just the output that was observed. Are you assuming a fixed recurrence, with the disagreeing bits? ...
... But that was what told me that your conjecture concerned characteristic words, which are notoriously poor at delivering decimal digits. ... I did so only...
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...
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