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\begin{document}

\noindent{\bf Lemma}.
Let $u \geq 3$ be odd. 
Then $A_{\sigma+\gamma} (3u)$ is never a power of\/ $3$.

\medskip
\noindent{\bf Sketch of Proof:}\quad 
Suppose that $A_{\sigma+\gamma}(d)$ is a power of 3 for odd
$d=3u$, $u\ge 3$, i.e., 
\begin{equation} \label{eq:power3}
A_{\sigma+\gamma}(d)=3^e,
\end{equation}
for some integer $e$. Then, 
since $A_{\sigma+\gamma}(d)$ is different from $3,9,27$ 
(see Table~1),
$A_{\sigma+\gamma}(d)$ must be divisible by 81. 
There are nine $d$ modulo 4212 with that property.

Next we consider \eqref{eq:power3} modulo $m=5, 8, 11, 13, 37, 41,
61$. For each fixed $m$, the numbers $A_{\sigma+\gamma}(d)$ modulo $m$ are
periodic (because $A_{\sigma+\gamma}(d)$ satisfies the recurrence
(4.32)), 
of length $d(m)$ say, and the powers $3^e$ modulo $m$ are 
periodic as well, of length $e(m)$ say. Thus, for each
fixed $m$, we obtain a number of solutions $(d,e)$ to
\eqref{eq:power3} mod $m$, the solutions $(d,e)$ being determined
modulo $(d(m),e(m))$ (which is short for: $d$ is determined modulo
$d(m)$ and $e$ is determined mod $e(m)$). Furthermore, if $3\mid
d(m)$, we may eliminate all solutions $(d,e)$ with 
$3\hbox{${}\not\kern2.5pt\mid{}$}d$. (Recall that $d=3u$.)

A computer search gives 
84 solutions $(d,e)$ to \eqref{eq:power3} mod 5, with each $(d,e)$ being
determined mod $(624,4)$, it gives
4 solutions $(d,e)$ to \eqref{eq:power3} mod 8, with each $(d,e)$ being
determined mod $(20,2)$, it gives
13 solutions $(d,e)$ to \eqref{eq:power3} mod 11, with each $(d,e)$ being
determined mod $(240,5)$, it gives
4 solutions $(d,e)$ to \eqref{eq:power3} mod 13, with each $(d,e)$ being
determined mod $(168,3)$, it gives
233 solutions $(d,e)$ to \eqref{eq:power3} mod 37, with each $(d,e)$ being
determined mod $(2736,18)$, it gives
21 solutions $(d,e)$ to \eqref{eq:power3} mod 41, with each $(d,e)$ being
determined mod $(480,8)$, and it gives
12279 solutions $(d,e)$ to \eqref{eq:power3} mod 61, with each $(d,e)$ being
determined mod $(453960,10)$. For example, the 4 solutions to
\eqref{eq:power3} mod 13 are $(d,e)$ =
$(3, 1)$, $(21, 2)$, $(63, 0)$, $(105, 1)$ mod $(168,3)$.

We may now determine a combined solution $(d,e)$, 
$d$ determined modulo 2173560480 (the least common multiple of
4212, 624, 20, 240, 168, 2736, 480, 453960), $e$ determined modulo 360
(the least common multiple of 4, 2, 5, 3, 18, 8, 10).
But, as a computer calculation shows, there is 
no combined solution, a contradiction.
\QED

\end{document}

