[seqfan] Re: The mysterious Layman sequences

[Robert Gerbicz]

Let c=0.36704, I'll prove the linear recursion only for this c value. Maybe
you could generalize this.

We'll use [just to simplify the calc]:
T1: for even n: a(n)=3*a(n-1)-a(n-2)
T2: for odd n:  a(n)=4*a(n-1)-a(n-2)
It's a very easy proof with induction.

Furthermore: let R=sqrt(5+2*sqrt(6)) then for n>=6 it holds:
for even n: 0.30618*R^n<a(n)<0.30619*R^n
for odd n:  0.35355*R^n<a(n)<0.35356*R^n
What could be provable "easily" if you solve the linear recursion [notice
that R and -R is also a root of the characteristic polynomial].
WolframAlpha or other programs are easily giving the coefficients needed
for this proof.

We claim that for each n we can see in the generations the following
subsequence from the head:
a(1),a(2),a(3),...,a(n),x

where x=a(n)-a(n-1) and after 2 or 3 steps [that depends on the parity of
n] we will insert a(n+1) after a(n), and immediately leaving y=a(n+1)-a(n)
after a(n+1) in the sequence. So with induction the convergence of the
sequence will be a(1),a(2),a(3),...,a(n),a(n+1),...

When we insert something between a(n) and x one after another insert
p*a(n)+x for p=1,2,3,..

What is the maximal p value for that we can still insert (p+1)*a(n)+x ?

we can insert only if c*(p*a(n)+x)<=a(n)
so we can still insert (p+1)*a(n)+x iff p<=(1-c*x/a(n))/c is true.

S1: for even n:
x=a(n)-a(n-1)
0.19380*R^n<0.30618*R^n-0.35356*R^(n-1)<x<0.30619*R^n-0.35355*R^(n-1)<0.19382*R^n

from this: 2.09147<(1-c*x/a(n))/c<2.09156

hence the maximal p value is 2, finally we insert
3*a(n)+x=3*a(n)+a(n)-a(n-1)=4*a(n)-a(n-1)=a(n+1)

Before that we inserted
y=2*a(n)+x=2*a(n)+a(n)-a(n-1)=3*a(n)-a(n-1)=a(n+1)-a(n)
so the induction step is good for even n!!!

S2: for odd n:
x=a(n)-a(n-1)
0.25623*R^n<0.35355*R^n-0.30619*R^(n-1)<x<0.35356*R^n-0.30618*R^(n-1)<0.25625*R^n

from this: 1.9997<(1-c*x/a(n))/c<1.9998

hence the maximal p value is 1, finally we insert
2*a(n)+x=2*a(n)+a(n)-a(n-1)=3*a(n)-a(n-1)=a(n+1)

Before that we inserted y=a(n)+x=a(n)+a(n)-a(n-1)=2*a(n)-a(n-1)=a(n+1)-a(n)

What we needed, the proof is complete.