[seqfan] Re: A076725 independent sets recurrence constant
M. F. Hasler
seqfan at hasler.fr
Fri May 19 14:24:41 CEST 2017
On Thu, May 18, 2017 at 7:47 AM, Kevin Ryde wrote:
> In A076725, Vaclav Kotesovec notes limit c2 * c1^(2^n). Factor
> c2=A263719 is clear. Where would I learn more about the power
> part c1=1.289751..? Is it harder, or did I miss something easy?
>
Kevin,
If you accept a(n) ~ c2 * c1^(2^n)
then obviously log a(n) ~ 2^n log c1
so c1 = exp lim (log a(n))/2^n.
If you want to establish the asymptotic law,
and accept a(n+1) ~ c2 a(n)^2
then log a(n+1) ~ c3 + 2 log a(n)
or log a(n+1) + c4 ~ 2 ( log a(n) + c4 )
[being a bit sloppy here]
so log a(n) + c4 ~ 2^n (log a(0) + c4) =: 2^n c5
or a(n) ~ exp( c5 * 2^n - c4 ).
Thus c1 = exp(c5) with c5 = lim (log a(n))/2^n
as above.
Maybe there's a better way to get this constant,
but this quick hack gives you already about 20 decimals:
(PARI)
{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)])))}
%= [62, 1.2897512927198122074876157691907584655]
(Only about half the decimals are OK (up to "207"), as you can see if you
print the values progressively. It stops due to an overflow at n=62 [where
it tries to compute a(63)].)
- Maximilian
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