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\bf
\centerline
{
The Number of Ways of Walking in $x_1 \geq \dots \geq x_k \geq0$
for $n$ Days, Starting and Ending at the Origin,
}
\centerline
{
Where at each Day you may either Stay in Place
or Move One Unit in any Direction, 
}
\centerline
{
Equals the Number of
$n$-Cell Standard Young Tableaux with $\leq 2k+1$ Rows.
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\centerline{ {\it
Doron 
ZEILBERGER}\footnote{$^1$}
{\eightrm  \raggedright
Department of Mathematics, Rutgers University (New Brunswick),
Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway,
NJ 08854-8019, USA.
%\break
{\eighttt zeilberg  at math dot rutgers dot edu} ,
\hfill \break
{\eighttt http://www.math.rutgers.edu/\~{}zeilberg} .
Dec. 6, 2007.
Exclusively published in the Personal Journal of Ekhad and 
Zeilberger
{\eighttt http://www.math.rutgers.edu/\~{}zeilberg/pj.html} .
Supported in part by the NSF.
}
}

{\bf Theorem}: See title .

{\bf Proof:} Consider the number of ways
of walking in $x_1 \geq \dots \geq  x_k \geq 0$, 
from the origin back to the origin, in $n$ days, without the lazy option, i.e. where
at each day one is obliged to move one unit in any (legal) direction.
The exponential generating function for this sequence
is given by a certain determinant
due, in more general form, to
D. Grabiner and P. Magyar, and spelled-out, for our case, in
Eq. (9) of {\tt http://arxiv.org/pdf/math/05012230}
(by Chen et. al). Obviously, the exponential generating function for the
same quantity for lazy walkers, who sometimes rest, 
is $e^t$ times that, since the latter is the binomial transform of
the former.

Surprise! This is exactly Gessel's  formula for the exponential generating 
function
for the number of Standard Young Tableaux with $n$ cells and $\leq 2k+1$ rows,
spelled out in Eq. (24) of
R. Stanley's paper
{\tt http://arxiv.org/pdf/math/050125035} . \halmos 

It would be interesting to find a {\it nice} bijective proof.

{\bf Acknowledgement}: I wish to thank Guoce Xin for pointing out the 
(a posteriori) trivial proof (modulo Gessel and Grabiner-Magyar).

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