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#1 |
Mar 2016
24·52 Posts |
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A peaceful morning for you,
if you regard the quotient between the primes p with p=f(n)=n²+1 and the primes with p | f(n) appearing the second times with increasing n the limes seems to go to 1 I calculate this sequence up to n=2^39 (Results under http://devalco.de/quadr_Sieb_x%5E2+1.php#4g and graphic) Is this a conjecture or are there any similiar proofs ? Greeting from the gaussian primes ![]() ![]() ![]() Bernhard |
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#2 |
Aug 2006
10111010110112 Posts |
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I can't tell what you're saying. At first I thought you were comparing
A002496, the primes of the form n^2+1 to A002313, the primes dividing n^2+1 for some n but I could find only evidence of the former in your numbers (the orange column). You clearly don't mean anything close to the usual definition of "all primes", and "appearing the second times" is quite opaque. |
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#3 |
Feb 2017
Nowhere
583610 Posts |
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I'm not sure what "limes" means here; lim sup perhaps?
If p == 1 (mod 4), there are exactly two integers 0 < n1 < n2 < p, with n2 = p - n1, for which n1^2 == n2^2 == -1 (mod p). If n1/p = k, then n2/p = 1 - k. I am guessing it is known that lim inf k = 0, so lim sup 1 - k = 1. |
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#4 |
Mar 2016
24·52 Posts |
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Thanks for your reply,
i try to improve my mathematical precision expressions: if you regard the quotient between A) the amount of primes p with p=f(n)=n²+1 (A002496) and B) the amount of primes with p | f(n) appearing the second times with increasing n [5|f(3), 13|f(8), 17|f(13), 29|f(17), 37|f(31)...] ( i did not find this sequence in OEIS ) the limes for n-> oo seems to go to 1 I calculate this sequence up to n=2^39 (Results under http://devalco.de/quadr_Sieb_x%5E2+1.php#4g and graphic) Is it possible to proof the limes mathematical correctly ? The sequence is monoton declining up from n=2^17 and i can proof that the amount for B) is infinitiv Greeting from the gaussian primes ![]() ![]() ![]() Bernhard |
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#5 |
"Curtis"
Feb 2005
Riverside, CA
14D716 Posts |
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What does "limes" mean? I thought it was a typo the first time you used it, but you used it again so now I think it's intentional.
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#6 |
Mar 2016
24·52 Posts |
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#7 |
"Rashid Naimi"
Oct 2015
Remote to Here/There
43328 Posts |
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It is the Latin word for limit according to google translate.
But I can't find any reference in English to use it as limit other than the fact that it is the root of the word limit. Below is everything you never wanted to know about the Latin limes. https://en.m.wikipedia.org/wiki/Limes ETA It can also mean limit in German. Last fiddled with by a1call on 2019-08-08 at 02:07 |
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#8 | |
Aug 2006
3×1,993 Posts |
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Because again, the primes p dividing f(n) = n^2+1 for some n are just A002313. So you're excluding 2 and 29 and lots of other values for some reason I don't understand. |
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#9 | ||
Mar 2016
1100100002 Posts |
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p | f(n) with n > (p-n), n < p Quote:
The 2 may be a special case which can be negligable if you consider the limes. 29 | f(12) (1.) and 29 | f(17) (2.) I consider the (2.) case. The number of the sequence A00231 are identical but the order / arrangement is different. Actually i do a sieve construction. You could sieve the primes p | f(n) either for the first time or for the second time. I have added two very simple sourcecode in C which are very easy in order to understand the construction: http://devalco.de/quadr_Sieb_x%5E2+1.php see point 3. j) k) Thanks a lot for your patience ![]() ![]() ![]() Bernhard |
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#10 |
Mar 2016
1100100002 Posts |
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If the romulus interpreter from Thailand is not present,
my expressions are poor and low. ![]() |
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#11 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×4,931 Posts |
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