2} we study the reverse situation, where $i+j>2.$
Here the use of the primes
dividing $N(N{-}\<1)(N{-}\<2)$ will prove significantly stronger than
the use of primes dividing $N(N{-}\<1),$ but the cancellation
of factors by the denominator $i!\,j!$ will take its toll.
That problem is not serious for low values of $i+j:$
we will find that there can be no counterexamples with $2*2} below cannot make use of them.
Perhaps some reader will succeed in doing so.
\section{The case $i=j=1.$}\label{S.i=j=1}
The proposition below gives the first of the two results
previewed above, and a bit more.
In the proof, and the remainder of this note,
by ``the prime-powers factors of $N$'' we shall mean the factors
occurring in the decomposition of $N$ into positive powers of
{\em distinct} primes.
(So in this usage, $4$ is among the ``prime-power factors''
of $12,$ but $2$ is not.)
\begin{proposition}\label{P.(N-2)+1+1}
Suppose $N$ is a positive integer having decompositions with positive
integer summands,
%
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.aaabbb}
$N\ =\ (N-2)\ +\ 1\ +\ 1,\qquad
N\ =\ a_1\ +\ a_2\ +\ a_3,\qquad N\ =\ b_1\ +\ b_2\ +\ b_3,$
\end{minipage}\end{equation}
%
such that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.oAp}
for every prime $p,$ at least one of the decompositions
of~\textup{(\ref{d.aaabbb})} is $\!p\!$-acceptable.
\end{minipage}\end{equation}
Then $N\ \geq\ 1726\ =\ 2^6\cdot 3^3\,-\,2.$
If $N$ is even, then in fact $N\ \geq\ 6910\ =\ 2^8\cdot 3^3\,-\,2.$
In either case, $N{-}\<1$ is divisible by at least $3$ distinct primes.
\end{proposition}
\begin{proof}
From the discussion in the last section, we see that
for every prime $p$ dividing $N$ or $N{-}\<1,$ one of the last
two decompositions of~(\ref{d.aaabbb}) must be $\!p\!$-acceptable.
Thus, if $N$ is divisible by $p^d,$ then
looking at the last $d$ digits of the base-$\!p\!$ expression of $N$
in the light of Definition~\ref{D.accept}, we see that
$p^d$ must divide all three summands in one of those two decompositions;
i.e., either all three of $a_1,\,a_2,\,a_3$ or all
three of $b_1,\,b_2,\,b_3.$
Hence, $N^3\,|\,a_1\,a_2\,a_3\,b_1\,b_2\,b_3.$
In the same way, we see that each prime power factor of $N{-}\<1$ must
either divide two of $a_1,\,a_2,\,a_3$ or two of $b_1,\,b_2,\,b_3,$
hence $(N{-}\<1)^2\,|\,a_1\,a_2\,a_3\,b_1\,b_2\,b_3.$
Since $N$ and $N{-}\<1$ are relatively prime, this gives
%
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.N^3(N-1)^2}
$N^3(N{-}\<1)^2\ |\ a_1\,a_2\,a_3\,b_1\,b_2\,b_3.$
\end{minipage}\end{equation}
On the other hand, it is easy to verify that for any positive real
number, the decomposition into three nonnegative summands having the
largest product is the one in which each summand is one third of the
total; so $a_1\,a_2\,a_3\leq(N/3)(N/3)(N/3);$ and likewise for the
$b_i:$
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.N^3/27}
$a_1\,a_2\,a_3\ \leq\ N^3/\,3^3,\qquad b_1\,b_2\,b_3\ \leq\ N^3/\,3^3.$
\end{minipage}\end{equation}
At this point, we could combine~(\ref{d.N^3(N-1)^2})
and~(\ref{d.N^3/27}) to get a lower bound on $N.$
But let us first strengthen each of~(\ref{d.N^3(N-1)^2})
and~(\ref{d.N^3/27}) a bit, using
some special considerations involving the prime $2.$
That prime necessarily divides $N(N{-}\<1);$ assume
without loss of generality that $a_1+a_2+a_3$ is $\!2\!$-acceptable.
Then by~(\ref{d.2vals}), the powers of $2$
dividing $a_1,\ a_2$ and $a_3$ are distinct.
If $2\,|\,N,$ occurring, say, to the $\!d\!$-th power,
this means that $a_1\,a_2\,a_3$ must be divisible not merely by
the factor $2^d\,2^d\,2^d=2^{3d}$ implicit in our derivation
of~(\ref{d.N^3(N-1)^2}), but by $2^d\,2^{d+1}\,2^{d+2}=2^{3d+3}.$
If, rather $2\,|\,N{-}\<1,$
again, say, to the $\!d\!$-th power, we can merely
say that $a_1,\ a_2$ and $a_3$ include, along with one odd term,
terms divisible by $2^d$ and $2^{d+1},$ giving a divisor $2^{2d+1}$ in
place of the $2^{2d}$ implicit in~(\ref{d.N^3(N-1)^2}).
Thus, we can improve~(\ref{d.N^3(N-1)^2}) to
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.more_2}
$2\,N^3(N{-}\<1)^2\ |\ a_1\,a_2\,a_3\,b_1\,b_2\,b_3,$\quad and if
$N$ is even,\quad
$8\,N^3(N{-}\<1)^2\ |\ a_1\,a_2\,a_3\,b_1\,b_2\,b_3.$
\end{minipage}\end{equation}
To improve~(\ref{d.N^3/27}), on the other
hand, consider three real numbers
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.r1r2r3}
$r_1\ \geq\ r_2\ \geq\ r_3$
\end{minipage}\end{equation}
%
(which we shall assume given with specified
base-$\!2\!$ expressions, so that, e.g., $1.000\dots$ and $0.111\dots$
are, for the purposes of this discussion, distinct),
subject to the condition that
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.nocarry}
there is no need to carry when $r_1,\,r_2,\,r_3$ are added in base~$2;$
\end{minipage}\end{equation}
%
and suppose we want to know how large the number
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.ratio}
$r_1\,r_2\,r_3\,/\,(r_1+r_2+r_3)^3$
\end{minipage}\end{equation}
%
(regarded as a real number, without distinguishing between
alternative base-$\!2\!$ expansions if these exist) can be.
Note that if, in one of the $r_i,$ we change a base-$\!2\!$ digit $1,$
other than the highest such digit, to $0,$ i.e., subtract $2^d$ for
appropriate $d,$ and simultaneously, for some $j>i$
(cf.~(\ref{d.r1r2r3}))
add $2^d$ to $r_j$ (change the corresponding
digit, which was $0$ by~(\ref{d.nocarry}), to $1),$ then,
proportionately, the decrease in $r_i$ will be less than the increase
in $r_j.$
From this it is easy to deduce that for fixed $r_1+r_2+r_3,$ we
will get the largest possible value for~(\ref{d.ratio}) by
letting $r_1$ contain only the largest digit $1$ of
that sum, $r_2$ only the second largest, and $r_3$ everything else.
(To make this argument rigorous, we have to know that~(\ref{d.ratio})
assumes a largest value.
We can show this by regarding the set of $\!3\!$-tuples of strings
of $\!1\!$'s and $\!0\!$'s, with a
specified number of these to the left of
the decimal point and the rest to the right, and with
$r_1\geq 1$ and no two
members of our $\!3\!$-tuple having $\!1\!$'s in the same position,
as a compact topological space under the product topology.
We can then interpret~(\ref{d.ratio}) as a continuous real-valued
function on that space, and conclude that it attains a maximum.)
More subtly, I claim that if some digit of $r_1+r_2+r_3$ after the
leading $1$ is $0,$ then the value~(\ref{d.ratio}) will be increased on
replacing that digit by $1$ in $r_3,$ and hence in $r_1+r_2+r_3.$
This can be deduced using the fact that the partial derivative
of~(\ref{d.ratio}) with respect to $r_3$ is positive, together
with some ad hoc considerations in the case where the digit
in question has value $>r_2.$
% the "positive derivative" condition actually holds as long
% as r_3 is < (r_1+r_2)/2. So increase r_3, without worrying
% about no-carrying, to the value of the first zero digit exceding
% r_2, at which point the no-carrying condition holds again; then
% reverse labeling of "r_2" and "r_3".
Since~(\ref{d.ratio}) is invariant under multiplication of
all $r_i$ by a common power of $2$ (``shifting the decimal
point''), it is not hard to deduce that it is
maximized when $r_1,\ r_2$ and $r_3$
have base-$\!2\!$ expansions $1_2,\ 0.1_2$ and $0.0111\dots\<_2\,.$
In that case, its value is $(1\cdot 1/2\cdot 1/2)/(1+1/2+1/2)^3=2^{-5}.$
Thus we can improve the first inequality of~(\ref{d.N^3/27}) to
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.N^3/32}
$a_1\,a_2\,a_3\ \leq\ N^3/\,2^5.$
\end{minipage}\end{equation}
%
(We cannot similarly improve the second inequality, since
the decomposition $b_1+b_2+b_3$ need not be $\!2\!$-acceptable,
i.e., need not satisfy the analog of~(\ref{d.nocarry}).)
If we now combine~(\ref{d.N^3/27}), so improved, with
the first assertion of~(\ref{d.more_2}), we get
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.N...leq}
$2\,N^3(N{-}\<1)^2\ \leq\ (N^3/2^5)(N^3/3^3).$
\end{minipage}\end{equation}
%
So
%
\begin{displaymath}\begin{minipage}[c]{33pc}
$(N{-}\<1)^2\ \leq\ N^3/\,(2^6\cdot 3^3),$\quad so\\[.5em]
$N^3/(N{-}\<1)^2\ \geq\ 2^6\cdot 3^3.$
\end{minipage}\end{displaymath}
Expanding the left-hand side in powers of $N{-}\<1,$ we get
$(N{-}\<1)\ +\ 3,$ plus terms whose
sum becomes $<1$ as soon as $N\geq 5.$
Since the above inequality certainly cannot be satisfied by
any integer $N$ with $1 2rt;$ and similarly we have $N>2su.$
Multiplying, we get $N^2>4\,r\,s\,t\,u=4N(N{-}\<1),$ which is
impossible.
\end{proof}
Remark: If one tries to extend the argument of the above paragraph
to the case where $N{-}\<1$ is a product of three prime powers,
one discovers one case which there is no obvious way to exclude:
One of our given decompositions, say $N=a_1+a_2+a_3,$ might
be $\!p\!$-acceptable for all three of those primes,
with one prime power factor dividing $a_1$ and $a_2,$
another dividing $a_1$ and $a_3,$
and the third dividing $a_2$ and $a_3.$
The product $t$ of the prime power factors of $N$ that divide all of
$a_1,\ a_2,\ a_3$ could then be nontrivial, though it would have to
be smaller than each of the three prime power factors of $N{-}\<1.$
\section{The case $i+j>2.$}\label{S.i+j>2}
Let us now consider a possible counterexample to Conjecture~\ref{Cj} for
$k=3$ of the contrary sort, where the member of our
triad of decompositions involving a summand $\geq p_{\max}^d,$
%
\begin{equation}\begin{minipage}[c]{35pc}\label{d.N-i-j_rep}
$N\ =\ (N{-}\*