How to obtain an upperbound for knots up to k crossings? I think I've found something which involves the genus but I'm not sure.
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6$\begingroup$ Paolo- This is a really vague and unhelpful question. The reader is left completely unclear on what you are looking for. Are you looking for references? For theoretical upper bounds? For actual data? You'll get much better answers if you're more specific. $\endgroup$– Ben Webster ♦Commented Mar 29, 2010 at 18:19
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1$\begingroup$ I mean theoretical upper bounds depending on the number of crossing $\endgroup$– Paolo AcetoCommented Mar 29, 2010 at 21:01
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$\begingroup$ this spares OEIS entry in principle should have the info you are looking for. Perhaps you can update it with your findings: oeis.org/A086825 $\endgroup$– Sidharth GhoshalCommented Nov 17, 2023 at 0:28
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2 Answers
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There are some known exponential bounds on the number. For example, if kn is the number of prime knots with n crossings, then Welsh proved in "On the number of knots and links" (MR1218230) that
2.68 ≤ lim inf (kn)1/n ≤ lim sup (kn)1/n ≤ 13.5.
The upper bound holds if you replace kn by the much larger number ln of prime n-crossing links.
Sundberg and Thistlethwaite ("The rate of growth of the number of prime alternating links and tangles," MR1609591) also found asymptotic bounds on the number an of prime alternating n-crossing links: lim (an)1/n exists and is equal to (101+√21001)/40.
answered Mar 29, 2010 at 19:29
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Dowker codes can be used to get an (over)estimate for the number of knots with $k$ crossings. Hoste has written a few, extremely clear, papers on using Dowker codes for enumeration of knot tables. I don't see how genus could be used - crossing number is an invariant defined in terms of diagrams while genus is much more topological... Very curious!
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$\begingroup$ The over-estimate is way too large! Once you get beyond 17 crossings the probability that the code will not be planar is way too large. $\endgroup$ Commented Mar 29, 2010 at 22:47
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$\begingroup$ @Scott - I did say it was an (over)estimate. :) My bound is of the order $n^n$ while the reality of the situation is more like an exponential (which I wasn't aware of). I've voted up Steven's answer. Hmmm. I will point out that my answer is elementary... $\endgroup$– Sam NeadCommented Mar 30, 2010 at 1:18