The following is from Clark Kimberling’s problems page.

**Note: Clark Kimberling labels the rows 1, 2, 3, ….
For this solution it is more convenient to label the rows 0, 1, 2, …,
and I have made this change when copying the following statement of the problem.
Also, I use Φ for (1 + sqrt(5))/2. – MB.
**

Begin an array [row 0] by writing the Fibonacci numbers: 1, 2, 3, 5, 8, ... Start row 1 with the least unused positive integer, which is 4; follow 4 by 6, and finish the row using the Fibonacci recurrence (i.e., add the two most recent numbers to produce the next, so that row 1 starts with 4, 6, 10, 16, 26, 42). Start row 2 with the least unused, which is 7, follow 7 by 12, and then use the recurrence to produce 19, 31,.... Continue in this manner (the choice of second term in each new row is revealed below), getting rows 0 to 3 starting like this:

1 2 3 5 8 13 . . .

4 6 10 16 26 42 . . .

7 12 19 31 50 81 . . .

9 14 23 37 60 97 . . .
.

.

.

Here is the recipe for the second number in each row: let Φ be the golden ratio, (1+sqrt(5))/2, let i be the number of the row [i = 0 for top row], and let x be the first number in the row; then the second number is [Φx] if i is odd, and it is [Φx] + 1 if i is even.

For example, row 4 starts with the least unused, which is x=11, and this is followed by [11Φ] + 1, which is 18.

**Is every number is column 2 even?**

In anticipation that the answer is yes, the array is introduced as the Even Second Column Array in

**C. Kimberling,** "The First Column of an Interspersion,"
The Fibonacci Quarterly 32 (1994) 301-315.

The answer is yes.

The solution, plus some alternative proofs of known results, is published in The Fibonnaci Quarterly,
**50(2)** (May 2012),
106–118. The full paper is available on the
Fibonacci Quarterly website.
NB: Access to recent issues of the FQ is by subscription only.

A version containing the solution only in PDF format is available on this website.

An extended version of Table 2 is also available.