[seqfan] Can you help me to find flaws in this reasoning ?

[Thomas Scheuerle]

Hi,

A093545 Is a permutation of the nonnegativ integers.
I found m = a(n) + floor((1+a(n))/5) is the smallest solution to A014682(m) = n.
(There exists either only one m or to possible m as solutions).
Let f_a(n) = a(n) + floor((1+a(n))/5) and define f_a(n)^2 = f_a(f_a(n)).
Then if you apply f_a(n) on the set of nonnegativ integers, you remove 1/5th of numbers of it. Chose any number n > 4 and there exists a k where f_a(n)^k would remove this number from the set. right ?
Therefore f_a(n)^k can not have finite cycles for n > 4 right ?

Let f_b(n) = 2*n.
Let f_x(n) be an arbitrary combination of f_a(n) and f_b(n) like for example f_a(f_b(n)).

Then f_x(n)^k can not have finite cycles for n > 4 right ?

For any A014682(n)^k = m exists a f_x(m) such that f_x(m) = n.  right ?

As we know that f_x(n)^k has no finite cycles for n > 4, we know now that the same must hold for A014682(n)^k ?

This would mean A014682(n) has no finite cycles for n > 4 only question open would be
if it is all connected, if not fully connected it could have cycles of infinite length.

Writing this at 3AM and drunk so, are there any correct parts in it ?

Thank you very much.

-- 
Kind regards

Thomas Scheuerle