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{\bf Proof of a Conjecture of Neil Sloane Concerning Claude Lenormand's }
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{\bf ``Raboter" Operation (OEIS sequence A318921)} 
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{\it Doron ZEILBERGER}
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\qquad\qquad\qquad
{\it Dedicated to Neil James Alexander Sloane (born   Oct. X, MCMXXXIX) on  his turning $1001111$ years young
($100101$ days late)}

Yesterday, Neil Sloan gave a great talk([S1]), where among many other fascinating results and open problems, he mentioned
a conjecture that he made less than two months ago, concerning the ``planing" operation ({\it raboter}) introduced in
2004 by sequence enthusiast Claude Lenormand, and described in  [S2]. He also mentioned that
he has no idea how hard it is to prove, and made a `meta-conjecture' that it may be a `low-hanging fruit'. 
I will now show that both conjecture and meta-conjecture are true.

By examining the definition of  {\tt http://oeis.org/A318921} in terms of the binary representation of $n$, it is readily seen that
$r(n)$ may be defined recursively as follows
$$
r(n) \, := \,
\cases{ 
2\, r(n/2) \, , \quad if \quad n \equiv 0 \, (\, mod \, 4 \, ) \, ; \cr
\cr
\, r(\frac{n-1}{2}) \, , \quad if \quad n \equiv \, 1 \, (\, mod \, 4\, )  \, ; \cr
\cr
\, r(n/2) \, , \quad if \quad n \equiv \, 2 \, (\, mod \, 4\, )  \, ; \cr
\cr
\, 2\, r(\frac{n-1}{2})\, + \, 1  \, , \quad if \quad n \equiv \, 3 \, (\, mod \, 4\, )  \, ,
}
$$
subject to the initial conditions $r(0)=0$ and $r(1)=0$.

Sloane defined, for, integers $k$, $k \geq 1$,
$$
L(k):=\sum_{n=2^k}^{2^{k+1}-1} \, r(n) \quad ,
$$
and conjectured the following

{\bf Fact:} $L(k)\,= \, 2 \, \cdot \, 3^{k-1} \, - \, 2^{k-1}$ .

Indeed, breaking up the sum into the four congruence classes modulo four, we have
$$
L(k) \, = \, \sum_{n=2^k}^{2^{k+1}-1} \, r(n) 
$$
$$
\, = \, 
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(4m) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(4m+1) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(4m+2) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(4m+3) 
$$
$$
\, = \, 
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, 2 \, r(2m) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(2m) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(2m+1) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, ( \, 2 \, r(2m+1) \, + \, 1    \, )
$$
$$
\, = \, 
3\, \sum_{m=2^{k-2}}^{2^{k-1}-1} \,  r(2m) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(2m+1) \, + \,
2\, \sum_{m=2^{k-2}}^{2^{k-1}-1} \,  r(2m+1)   
\, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \, 1   
$$
$$
\, = \, 
3\, \left ( \sum_{m=2^{k-2}}^{2^{k-1}-1} \, r(2m) \, + \,
\sum_{m=2^{k-2}}^{2^{k-1}-1} \,  r(2m+1)   \right ) \, + \,2^{k-2} 
$$
$$
\, = \, 
3\, \left ( \sum_{m=2^{k-1}}^{2^{k}-1} \,  r(m) \, \right ) \, + \, 2^{k-2} 
\, = \,
3 \,L(k-1)\, +\,  2^{k-2} \quad .
$$
Hence the sequence $L(k)$ satisfies the first-order inhomogeneous recurrence with constant coefficients
$$
L(k) \, - \, 3\, L(k-1) \, = \, 2^{k-2} \quad .
$$
But $R(k):= \,  2 \, \cdot \, 3^{k-1} \, - \, 2^{k-1}$ also satisfies the same recurrence, i.e.
$$
R(k) \, - \, 3\, R(k-1) \, = \, 2^{k-2} \quad ,
$$
(check!), and the fact follows  by induction on $k$, since it is true for $k=1$. \halmos


{\bf References}


[S1] N.J.A. Sloane, {\it Coordination Sequences, Planing Numbers, and Other Recent Sequences}, talk given in
Rutgers University Experimental Mathematics Seminar, Nov. 15, 2018.  \hfill \break
part 1: {\tt https://vimeo.com/301216222} \quad ; \quad part 2: {\tt https://vimeo.com/301219515} \quad . \hfill \break
slides: {\tt http://sites.math.rutgers.edu/\~{}my237/expmath/EMNov2018.pdf} \quad .

[S2] N.J.A. Sloane, {\it The On-Line Encyclopedia of Integer Sequences}, {\bf Sequence A318921}, \hfill\break
{\tt http://oeis.org/A318921} \quad .

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Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt DoronZeil at gmail  dot com}   \quad .
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Exclusively published in the Personal Journal of Shalosh B. Ekhad and Doron Zeilberger.
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Nov. 16, 2018.
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