# [seqfan] Re: 6 Complementary Sequences

Paul D Hanna pauldhanna at juno.com
Tue Feb 3 11:05:15 CET 2009

```Seqfans,
These sequences are not as interesting as I thought at first.
After more calculation, it appears that
limit u(n)/n = limit v(n)/n = 5/2.
Given this is true, then the other limits fall out easily.

Let limit x(n)/n = x, then

1 = 1/x + 1/(x + 5/2) + 1/(x + 5)
or:
0  =  2*x^3 + 9*x^2 - 5*x - 25

So,
x = 1.644215131087862878324271116...
with
limit y(n)/n = x + 5/2 and limit z(n)/n = x + 5.

So, please disregard the original inquiry.
Nothing mysterious going on here.
Paul

-- "Paul D Hanna" <pauldhanna at juno.com> wrote:
Seqfans,
Consider the 6 permutation sequences x,y,z, and u,v,w,
where the union of x,y,z, forms the distinct positive integers
and the union of u,v,w, forms the distinct positive integers
and the elements of x,y,z, are chosen such that:
(1) x(n),y(n),z(n), are the least positive integers not appearing earlier in {x},{y},{z};
(2) u(n)=y(n)-x(n), v(n)=z(n)-y(n), w(n)=z(n)-x(n) = u(n)+v(n),
are the least positive integers not appearing earlier in {u},{v},{w}.

Below I give the initial terms of these sequences.

u,  v,  w;
n:  x,  y,  z;  y-x,z-y,z-x;
1:  1,  2,  4;   1,  2,  3;
2:  3,  7, 12;   4,  5,  9;
3:  5, 11, 18;   6,  7, 13;
4;  6, 14, 24;   8, 10, 18;
5;  8, 19, 31;  11, 12, 23;
6;  9, 23, 38;  14, 15, 29;
7; 10, 26, 43;  16, 17, 33;
8: 13, 26, 43;  19, 20, 39;
9: 15, 36, 58;  21, 22, 43;
10: 16, 40, 65;  24, 25, 49;
11: 17, 44, 70;  27, 26, 53;
12: 20, 48, 78;  28, 30, 58;
13: 21, 53, 84;  32, 31, 63;
14: 22, 56, 90;  34, 35, 69;
15: 25, 61, 98;  36, 37, 73;
16: 27, 67,105;  40, 38, 78;
...

This is done by hand, so there may be errors.

I am curious about what these limits tend to be as n grows:
* limit x(n)/n = ?
* limit y(n)/n = ?
* limit u(n)/n = ?
* limit v(n)/n = ?

and would like to get some idea of these limits at n=100 or so.

Would anyone be adept enough at programming to obtain more terms?

Thanks,
Paul

```