[seqfan] Re: The mysterious Layman sequences

[Robert Gerbicz]

Found and basically proved the interval that contains c=0.36704 and is
giving the same convergent sequence of a[]=1,3,11,30,109,297,1079,...

If -2*sqrt(6)/3+2<=c<=3/8 then the convergence of the sequence will be a[].
( so 0.367006838144547...<=c<=0.375, we were quite close to the lower bound
with the Layman's c=0.36704 value, but it was quite obvious from the tight
1.9997<(1-c*x/a(n))/c<1.9998 bound from the previous proof ).

If c>3/8 then we even won't start with 1,3,11.
If c<-2*sqrt(6)/3+2 then again it will give another sequence, but the first
difference could be at a much higher index
Say for c=367/1000 the convergent seq starts with 1, 3, 11, 30, 109,
406,... the first difference is at index=6.

For all n>1 let u=a[n-1]/a[n] then the proof's induction works iff
for even n:    1/(4-u)<c<=1/(3-u)  [to get max p=2]
for odd n:   1/(3-u)<c<=1/(2-u)     [to get max p=1]
Then the induction will be true, the convergent sequence is a[] for the c
value.

Let r[n]=a[n-1]/a[n], then

for n=2 we see that r[2]=1/3 is giving c<=1/(3-1/3)=3/8 the sharpest upper
bound.
for even n we get 1/(3-r[n])<0.3 that is a very weak restriction, odd n
values are giving sharper bounds:

(n is odd)
r[n]=a[n-1]/a[n]=a[n-1]/(4*a[n-1]-a[n-2])=(3*a[n-2]-a[n-3])/(11*a[n-2]-4*a[n-3])

we claim that r[n] is an increasing sequence, so for odd n values:
r[n-2]<r[n] (from this 1/(3-r[n]) is also increasing)

a[n-3]/a[n-2]<(3*a[n-2]-a[n-3])/(11*a[n-2]-4*a[n-3]) (used the recursion
for the a[] sequence)

11*a[n-2]*a[n-3]-4*a[n-3]^2<3*a[n-2]^2-a[n-2]*a[n-3]

So we need E=3*a[n-2]^2-12*a[n-2]*a[n-3]+4*a[n-3]^2>0

But notice that E=3 (if n is odd), because
3*a[n-2]^2-12*a[n-2]*a[n-3]+4*a[n-3]^2=3*(11*a[n-4]-4*a[n-5])^2-12*(11*a[n-4]-4*a[n-5])*(3*a[n-4]-a[n-5])+4*(3*a[n-4]-a[n-5])^2=
=3*a[n-4]^2-12*a[n-4]*a[n-5]+4*a[n-5]^2=3 (with induction)

>From the explicit formula of a[] we could "easily" get that for odd n:
1/(3-r[n]) goes to (-2*sqrt(6)/3+2) for n->inf.
Proving the statement.