# [seqfan] Boomerangs

Ed Jeffery ed.jeffery at yahoo.com
Mon Oct 10 19:14:11 CEST 2011

```Seqfans,

I was having fun with this useless idea until it gave me a headache:

For n=1,2,3,..., define the array T(n,k), k=0,1,2,..., in which T(n,0)=n and, for k>0,

T(n,k)=Append the digits of floor(R(n,k-1)/2) to R(n,k-1).

Thus, for all k>0, T(n,k) begins with the digits of T(n,k-1).

A term T(n,j) is said to be a "boomerang" if its digit sequence ends with the digits of a k-th term T(n,k), j>k. For a k-th term of the n-th sequence T(n,k), the digits following those of the embedded (k-1)-th term are called the "trajectory" of T(n,k-1) or the k-th trajectory of T(n,0). This is supposed to be the trajectory of the boomerang (if it exists) and which is all in fun of course. The digit sequences of trajectories are also obviously highly repetitive. I was expecting it to be easy, but sadly I have yet to detect a boomerang.

This sequence begins as A={1,2,10,3,21,105,4,31,2110,...} when read from the anti-diagonals of the table
T(n,k)=
{1, 10,  105,    10552, ...}
{2, 21, 2110, 21101055, ...}
{3, 31, 3115, 31151557, ...}
{4, 42, 4221, 42212110, ...}
etc.

Now I have some questions about T, since I don't have the ammunition to calculate large terms:

Do boomerangs exist for this sequence?
Does a generating function exist?

Suppose, for each n, that we precede the digits of each term with a decimal point; e.g. for row 2, we would have the sequence {.2, .21, .211, .21101055, ...} with D(2)->.21101055..., as k->infinity. (However, I doubt that any such limit (in the usual sense) can be calculated.) Are there D(n) which are:

(*)   Rational?

(**)  Algebraic?
(***) Transcendental?

Regards,

Ed Jeffery

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