If $r<s\leqslant l$ and $\alpha_{r}>\alpha_{s}$, then the strictly-addable
nodes of $\lambda^{-0}$ in row $r$ lie in a longer ladder than the strictly-
addable nodes in row $s$. So using Proposition 4.7, if $a_{s}>0$, then
$a_{r}=2$. Similarly, if $a_{l+1}=2$ and $\alpha_{r}>0$, then $a_{r}=2$. In
particular, if $\alpha_{l}>0$ then $a_{l+1}=1$ (since otherwise
$a_{1}=\dots=a_{l}=2$, contrary to the last paragraph).

Combining these statements, we obtain $a_{r}=2$ for every $r$ for which
$\alpha_{r}>\alpha_{l}$.

#### Case 6.1

Suppose $a_{r}=1$ for a single value $r$, and $a_{s}=2$ for all $s\neq r$. So

$\lambda=(3l+1\stackrel{{\scriptstyle 3}}{{\dots.}}1)+3\alpha-\delta_{r},$

where $\delta_{r}$ is the composition which has a $1$ in position $r$ and $0$
elsewhere. The assumptions above mean that $\alpha_{r}=0$ and $a_{l+1}=2$. Now
we use a dimension argument. Let $\mu=(3l+1\stackrel{{\scriptstyle
3}}{{\dots.}}4)+3\alpha$. Then $\mu^{\mathrm{reg}}=\lambda^{\mathrm{reg}}$ and
$\operatorname{ddim}(\lambda)>\operatorname{ddim}(\mu)$ by Lemmas 5.2 and 5.13
as $\alpha_{r}=0$. So $\lambda$ is not homogeneous.

#### Case 6.2

Suppose that there exists $r$ such that $a_{r}=2$ and either $r=1$ or
$a_{r-1}>0$. Consider the partition $\lambda^{+1}$. We claim that this
partition has a strictly-addable $0$-node $(r,3(l-r+2+\alpha_{r}))$: this
follows since $\lambda_{r}=3(l-r+2+\alpha_{r})-2$, and $\lambda_{r-1}\geqslant
3(l-r+2+\alpha_{r})$ if $r\geqslant 2$, and $\lambda_{r-2}\geqslant
3(l-r+2+\alpha_{r})+2$ if $r\geqslant 3$.

If there is any $s\leqslant l$ for which $(a_{s},a_{s+1})$ equals $(0,2)$,
$(1,0)$ or $(1,1)$, then $\lambda^{+1}$ has a strictly-removable $0$-node in
either row $s$ or row $s+1$. Now the fact that $a_{r}=2$ and $a_{s}\leqslant
1$ and the argument at the start of Case 6 mean that
$\alpha_{s}\leqslant\alpha_{r}$, and therefore the strictly-removable $0$-node
in row $s$ or $s+1$ lies in a shorter ladder than the strictly-addable
$0$-node in row $r$. So by Proposition 4.7 $\lambda^{+1}$ is not homogeneous,
and hence neither is $\lambda$.

So we can assume that $a_{s}\geqslant 1$ for all $s$, and that there are no
two consecutive values of $s$ with $a_{s}=1$. As observed above, there is at
least one $1\leqslant s\leqslant l$ with $a_{s}=1$. If there is exactly one
$1\leqslant s\leqslant l+1$ for which $a_{s}=1$, then we are in case Case 6.1,
so assume instead that there are at least two values of $s$ for which
$a_{s}=1$.

Now we define a sequence of partitions $\nu^{(1)},\nu^{(2)},\dots$ by

$\nu^{(1)}=\lambda^{+1},\qquad\nu^{(k+1)}=((\nu^{(k)})^{+0})^{+1}\text{ for
}k\geqslant 1.$

Observe that $\lambda^{+1}$ has no parts congruent to $1$ modulo $3$, and so
by Lemma 7.2 $\nu^{(k)}$ has no parts congruent to $1$ modulo $3$, for any
$k$. Now we make the following observations, for $k\geqslant 1$.

Claim:

If $a_{1}=2$ or $a_{l+1}=2$ or the tuple $(a_{1},\dots,a_{l+1})$ contains $k$
consecutive $2$s, then $\nu^{(k)}$ has a strictly-addable $0$-node $(s,c)$ for
which $3s+c\geqslant 3(\alpha_{l}+l+k+1)$.

Proof:

If $a_{1}=2$, then we can take $(s,c)=(1,3(l+k+\alpha_{1}))$. If $a_{l+1}=2$,
then we can take $(s,c)=(l+k+1,1)$ by Lemma 7.3 and induction on $k$.

Assume now that $k<r\leqslant l$ such that $a_{r-k}=1$ and
$a_{r-k+1}=\dots=a_{r}=2$. Then $\alpha_{r-k}=\dots=\alpha_{r}=\alpha_{l}$, so
we can take $(s,c)=(r,3(\alpha_{l}+l+k-r+1))$ by Lemma 7.4.

Claim:

If there is $k\leqslant r\leqslant l$ such that
$(a_{r-k+1},\dots,a_{r+1})=(1,2,2,\dots,2,1)$, then $\nu^{(k)}$ has a
strictly-removable $0$-node $(s,c)$ for which $3s+c=3(\alpha_{l}+l+k)$.

Proof:

In this case $\alpha_{r-1}=\dots=\alpha_{r}=\alpha_{l}$. We can take
$(s,c)=(r,3(\alpha_{l}+l-r+k))$ by Lemma 7.4, since

$\nu^{(k)}_{r+1}\leqslant\nu^{(1)}_{r+1}+3(k-2)+2\leqslant
3(\alpha_{l}+l-r+k-2)+2.$

As a result, Proposition 4.7 shows that $\nu^{(k)}$ is not homogeneous for
some $k$ (and hence $\lambda$ is not homogeneous) unless the tuple
$(a_{1},\dots,a_{l+1})$ has the form $(1,2^{t},1,2^{t},\dots,1,2^{t},1)$ for
some $t\geqslant 1$. So assume this is the case. Now the fact that $a_{1}<2$
implies $\alpha_{1}=\alpha_{l}$. By Corollary 6.14 we can then assume that
$\alpha=(1^{2})$ or $\varnothing$. In the first case $\lambda=(9,7)$, which
can be ruled out looking at the known decomposition matrices. So we may assume
that $\alpha=\varnothing$. If there are at least three values of $s$ for which
$a_{s}=1$, then we use a degree argument: specifically, we claim that there is
a partition $\mu$ such that $\mu\vartriangleright\lambda$,
$\mu^{\mathrm{reg}}=\lambda^{\mathrm{reg}}$, $\mu_{1}=\lambda_{1}$ and
$\operatorname{ddim}(\mu)>\operatorname{ddim}(\lambda)$, which will show that
$\lambda$ is not homogeneous. We prove this by induction on the number of $1$s
in the tuple $(a_{1},\dots,a_{l+1})$. If there are at least four $1$s, then we
can use Lemma 5.2 and induction, so assume
$(a_{1},\dots,a_{l+1})=(1,2^{t},1,2^{t},1)$. Then we use Lemma 5.10.

We are left with the case where $(a_{1},\dots,a_{l+1})=(1,2^{l-1},1)$. Suppose
first that $l\geqslant 6$. Then

$\nu^{(k)}=(6l-4\stackrel{{\scriptstyle 3}}{{\dots.}}3l+2,\ 3l,\
3l-4\stackrel{{\scriptstyle 3}}{{\dots.}}2).$

Now by Lemma 5.11 $\nu^{(k)}$ is not homogeneous, and hence neither is
$\lambda$.

So we are left with the cases $2\leqslant l\leqslant 5$. If $l=2$ then
$\lambda=(6,4)$, which can be dealt with using known decomposition matrices.
For $3\leqslant l\leqslant 5$, Proposition 5.16 shows that $\lambda$ is not
homogeneous.

#### Case 6.3

Suppose that $a_{1}<2$ and there is no $r\geqslant 2$ for which $a_{r}=2$ and
$a_{r-1}>0$. Suppose in addition that $l\geqslant 3$ and there is $r\geqslant
4$ such that $a_{r}\neq 1$. Since $a_{1}<2$ we have $\alpha_{1}=\alpha_{l}$.
Since $l\geqslant 3$ we can then assume from Corollary 6.14 that
$\alpha=\varnothing$. Now $\lambda$ is not homogeneous by Proposition 5.14.

#### Case 6.4

If neither Case 6.2 nor Case 6.3 applies, then the tuple
$(a_{1},\dots,a_{l+1})$ has the form $(1,1,\dots,1)$, $(0,2,1,1,\dots,1)$ or
$(1,0,2,1,1,\dots,1)$. Again $\alpha_{1}=\alpha_{l}$, so by Corollary 6.14
$\alpha=(1^{2})$ or $\varnothing$. In the first case
$\lambda\in\\{(9,6),(8,7)\\}$ (as $a_{3}=1$ in this case by assumption) which
can both be ruled out looking at known decomposition matrices. So we may again
assume that $\alpha=\varnothing$. These cases can be dealt with using Lemmas
5.7, 5.8 and 5.9, except for the following small cases.

  * $\diamond$

$(6,3)$, $(5,4)$ and $(8,7,3)$ can be shown to be not homogeneous using known
decomposition matrices.

  * $\diamond$

$(11,10,6,3)$, $(14,13,9,6,3)$ and $(17,16,12,9,6,3)$ are not homogeneous by
Proposition 5.16.

This completes the proof of Theorem 3.5.

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