Finite colour focussing extends equally coloured progressions to a longer monochromatic arithmetic progression

For $r=1$ and $m=2$, the singleton progression $\{1\}$ with chosen difference $1$ is focused at $2$. For $m\ge 3$, apply $V(m-1,k)$ inside the first half of an interval twice as long. Its monochromatic $(m-1)$-term progression has positive difference at most the length of that half, so its next term still lies in the full interval. In either case there is one focused progression.

Assume $r>1$. Take $m=2$ first, where the block construction below has nothing to work with: a $1$-term progression of block indices carries no difference. It is not needed. Among any $k+1$ points two share a colour, and two points $a<b$ of one colour are a monochromatic $2$-term progression with difference $b-a>0$, so $F(2,k,r)=k+1$ and the first alternative always holds. Assume from here that $m\ge 3$, and let $n=F(m,k,r-1)$. Partition a sufficiently long interval into consecutive blocks of length $2n$. By [L1] there are $k^{2n}$ possible block colour vectors. Use $V(m-1,k^{2n})$ on the sequence of block vectors to obtain identically coloured blocks whose indices are $b,b+t,\dots ,b+(m-2)t$.

Apply the induction hypothesis to the first half of the first selected block, an interval of length $n=F(m,k,r-1)$. It gives either a monochromatic $m$-term progression, which finishes, or $r-1$ colour-focused progressions $A_i=\{a_i+j{d}_i:0\le j<m-1\}$ of pairwise distinct colours focused at $f$. Each $A_i$ lies in that first half, so $a_i+(m-2)d_i\le n$ measured from the block start, and $m\ge 3$ makes both $a_i$ and $d_i$ at most $n$; hence $f=a_i+(m-1)d_i\le 2n$ and the focus lies in the block. That is what the block length $2n$ is for, exactly as in step 1.1. For the second alternative define $A_i^{\mathrm{\prime }}=\{a_i+j(d_i+2nt):0\le j<m-1\}$. Its $j$th term occupies the same relative position in block $b+jt$ as the $j$th term of $A_i$ in block $b$, so identical block vectors preserve its colour.

The progressions $A_i^{\mathrm{\prime }}$ are focused at $f+(m-1)2nt$. Since $f$ lies in block $b$ by step 2.1, the point $f+j2nt$ occupies the same relative position in block $b+jt$ as $f$ does in block $b$, so identical block vectors make $f,f+2nt,\dots ,f+(m-2)2nt$ a monochromatic $(m-1)$-term progression, focused at the same point and coloured as $f$ is. If that colour equals the colour of some $A_i$, then $A_i\cup \{f\}=\{a_i+j{d}_i:0\le j\le m-1\}$ is a monochromatic $m$-term progression and the first alternative holds. Otherwise the new progression differs in colour from all $r-1$ of the $A_i^{\mathrm{\prime }}$, which already have pairwise distinct colours, and the second alternative contains $r$ focused progressions of distinct colours.

The base and step prove the focusing assertion for every $1\le r\le k$.