req proof regarding A196

[Joseph Biberstine]

Max A. wrote:
<snip>
> Something is wrong with the statement.
<snip>
> Max

	Pardon me.  Begin with t=n and an empty list L and, while, t>0 append
floor(sqrt(t)) to L and set t=t-floor(sqrt(t)).  This gives the list
intended.  The request for a proof stands.  Here are the values
beginning with n=1:

{1}
{1,1}
{1,1,1}
{2,1,1}
{2,1,1,1}
{2,2,1,1}
{2,2,1,1,1}
{2,2,2,1,1}
{3,2,2,1,1}
{3,2,2,1,1,1}
{3,2,2,2,1,1}
{3,3,2,2,1,1}
{3,3,2,2,1,1,1}
{3,3,2,2,2,1,1}
{3,3,3,2,2,1,1}
{4,3,3,2,2,1,1}
{4,3,3,2,2,1,1,1}
{4,3,3,2,2,2,1,1}
{4,3,3,3,2,2,1,1}
{4,4,3,3,2,2,1,1}

	Relevant related sequences seem to be A196 (number of runs in L(n)),
A267 (length of L(n)), A33638 (after a point, n such that L(n) ends in
exactly three ones), and A80037 (after a point, n such that L(n) ends in
exactly two ones).  Formulas given at these entries suggest
straightforward relationships.

	Also of interest here is the obvious generalization of replacing
instances of floor(sqrt(n)) with another reduction function f() such as
floor(n/2) or floor(n/GoldenRatio).

	I would appreciate any help regarding proofs throughout.

-JRB

addenda:

- Relevant Mathematica code:
f[n_] := Floor[Sqrt[n]];
g[n_] := (t = n; s = {}; While[t > 0, s = Join[s, {f[t]}]; t = t -
f[t];]; s);
Table[g[k], {k, 1, 40}]

- Note that for certain f(), g needs to be amended as follows:
g[n_] := (t = n; s = {}; While[t > 1, s = Join[s, {f[t]}]; t = t -
f[t];]; Join[s, {1}]);

- Here's modified code for f(n)=floor(n/2), for which analogues of the
above mentioned properties (A196 et al) are mysteriously absent from
OEIS.. surely I've done something wrong?:
f[n_] := Floor[n/2];
g[n_] := (t = n; s = {}; While[t > 1, s = Join[s, {f[t]}]; t = t -
f[t];]; Join[s, {1}]);
Table[g[k], {k, 1, 30}]
Table[Length[Split[g[k]]], {k, 1, 80}]
Table[Length[g[k]], {k, 1, 80}]
Select[Range[3, 80], Take[g[#], -3] \[Equal] {1, 1, 1} &]
Complement[Range[First[%], Last[%]], %]
Table[Times @@ g[k], {k, 1, 30}]