Moo Of Doom said:
I think he means that it extends infinitely to the right and infinitely downward, but has an edge on the top and on the left - basically like N^2.
An example of such a lattice is the following, where we just fill diagonals with successive numbers:
1 2 4 7 .
3 5 8 .
6 9 .
10 .
.
Another might be based on prime decomposition (1 is placed arbitrarily):
1 2 4 6 8 . . .
3 9 15 21 . . .
5 25 35 55 . . .
7 49 77 91 . . .
11 121 143 . . .
. . .
Yes something like that is what I had in mind. I don't like the second since I don't think the numbers below are defined by what I consider to be "a basic operation" on the upper numbers. For the first example we have a_{i,j} = a_{i-1,j+1} + 1 = a_{i-1,j}*2 - a_{i-2,j} + 1.
I have formed an infinite number of lattices by taking as the first row sequence A000201 from Sloane's Online Encyclopedia of sequences starting with tne nth term but subtracting
a_n -1 from each term. Then each underlying term is uniquely defined be the term immediately above by the operations of addition, multiplication and subtraction where the amounts of subtraction and addition are based upon the values of n and a_n.
To wit sequence A000201 is 1,3,4,6,8,9,11,12,14,16,... and starting at term a_4 5 is subtracted from each term to give 1,3,4,6,7,9,11,...
Regardless of the starting term the terms on the rows below are all below are then defined based upon n as follows:
A_{i,j} = \lfloor (A_{i-1,j}+a_{n} -n -1)*(phi + 1) - a_{n} +1 \rfloor phi = sqrt(5) + 1)/2
I checked for n = 1 to 107 that all integers less than 54002 appear in each such lattice once and only once. And also that each term in the lower rows equals 3 times the next upper term minus the 2nd next upper term. I believe that the method works for all n such that each integer from 1 to infinity populates the lattice once and only once, but am still working on the proof.
My lattice based upon the 4th term as the starting term is
1,03,04,06,07,09,11...
2,08,10,15,18,23,29...
5,21,26,39,49,60,76...
...