# [seqfan] Help with proof strategy needed.

L. Edson Jeffery lejeffery2 at gmail.com
Tue Jul 14 21:27:31 CEST 2015

```I am trying to prove a conjecture related to the 3x+1 problem which arises
from the definition of a new sequence. I want to include the proof in a PDF
in the links section on the sequence page. I know of only one way to
structure the proof but have encountered a technical problem which I do not
know how to solve. I would of course like to complete the proof myself, so
in the following I generalize the problem and ask for your help with how to
proceed (questions are at the end of this note).

Let A and B be rectangular arrays (of natural numbers) with entries in row
n and column k denoted by A(n,k) and B(n,k), respectively, n,k >= 1. Let L
be the array with entries

L(n,k) = log(B(n,k))/log(A(n,k)).

We are given the following properties of the arrays.

1. A(n,k) and B(n,k) are defined as functions of n and k.

2. The rows {A(n,k)} and {B(n,k)}, k=1,2,..., are strictly monotonically
increasing sequences, and each is an arithmetic progression.

3. The column bisections {A(2*n-1,k)}, {A(2*n,k)}, {B(2*n-1,k)} and
{B(2*n,k)}, n=1,2,..., are strictly monotonically increasing sequences, and
each sequence of first differences is a geometric progression.

4. There exists an r such that B(n,k) < A(n,k), for all n <= r, and such
that B(n,k) > A(n,k), for all n > r.

5. For the rows of L, we have the repeated limits

lim_{n->infinity} lim_{k->infinity} L(n,k) = 1.

6. For c a known constant, for the column bisections of L we have the
repeated limits

(i) lim_{k->infinity} lim_{n->infinity} L(2*n-1,k)) = c;

(ii) lim_{k->infinity} lim_{n->infinity} L(2*n,k) = c.

I also proved for the main diagonal of L that

7. lim_{n->infinity} L(n,n)) = c.

Conjecture. L(n,k) < c, for each pair n,k.

I tried structuring the proof as follows.

For the r of property 4, for each n <= r, L(n,k) < 1 and the conjecture is
true. Suppose that n > r, so L(n,k) > 1. We must first prove:

(I) For each n > r, row n of L is a strictly monotonically decreasing
sequence.

If (I) is true, then it follows that

(II) For each n > r, row n of L contains a maximal element, namely L(n,1).

We must then prove:

(III) For column 1 of L, each of the bisections {L(2*n-1,1)} and {L(2*n,1)}
is a strictly monotonically increasing sequence.

If (III) is true, then it follows that

(IV) L(1,1) is the least element of sequence {L(2*n-1,1)}, and L(2,1) is
the least element of sequence {L(2*n,1)}.

To complete the proof, it remains to show that

(V) L(1,1) < c, and L(2,1) < c.

But (V) is easily verified numerically, so the result should follow.

Now, to prove (I) it must be shown that L(n,k) - L(n,k+1) > 0, for each k.
However, the expression

(1)  log(B(n,k))/log(A(n,k)) - log(B(n,k+1))/log(A(n,k+1))

seems to defy algebraic evaluation, as do the expressions L(n,1) -
L(n+2,1). (It seems that properties 2 and 3 should be useful here.) Could
someone please tell me how to evaluate the expression (1)? Or is there some
other approach to the proof that avoids this problem altogether?

Thank you for any help,

Ed Jeffery
```