[seqfan] Re: A076725 independent sets recurrence constant

Kevin Ryde user42_kevin at yahoo.com.au
Sun May 21 01:26:55 CEST 2017

seqfan at hasler.fr (M. F. Hasler) writes:
> Thus c1 = exp(c5)  with  c5 = lim (log a(n))/2^n as above.

Hmm.  Yes, I'm happy enough there's a limit.  Logs would be fine if it
helped get there in a nice way, or whichever logs or m-th roots had a
name etc.

> {a=[0,0];for(n=1,99,iferr(a=[a[2],log(exp(a*[4,0;0,2])*[1,1]~)],E,return([n,exp(a[2]/2^n)])))}
> (Only about half the decimals are OK (up to "207"), ...

I suppose might have to worry how far earlier round-off extends, since
going by successive terms.

Is this constant, however reached, also some combinatorial measure of
freedom of choice of sets (in this case independent sets)?
Eg. unrestricted would be 2^m sets among m vertices, or totally
restricted to one set 1^m, and here somewhere in between at c^m.

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