Take $0\leq r < m$, and let all values be nonnegative and integer. Consider the function on a sequence ${x(n)}$, $\Phi_mx(mn+r)=mx(n)+\frac{r}{m}(x(n+1)-x(n))$, where we consider $x(0)=0$.
As an example, $\Phi_3$ maps $1,1,2,3$ to $1,2,3,3,3,3,4,5,6,7,8,9$. Would anyone be able to walk me through why this is? I end up with different values.
It appears that the definition of $\Phi_m$ was written incorrectly in the original article, and that the correct definition is $$\begin{aligned}(\Phi_mx)(mq+r)&=m\left(x(q)+\frac{r}{m}\left(x(q+1)-x(q)\right)\right)\\ &=mx(q)+r(x(q+1)-x(q)).\end{aligned}$$ Here $x$ is a sequence with initial term $x(0)=0,$ $m$ is a fixed positive integer, and $r$ is an integer satisfying $0\le r<m.$
If $y=\Phi_mx,$ then the $(mq)^\text{th}$ term of $y$ is $m$ times the $q^\text{th}$ term of $x.$ The $(mq)^\text{th}$ through $(mq+m)^\text{th}$ terms of $y$ form an arithmetic progression. So $y$ can be thought of as a piecewise linear function on the nonnegative integers.
To perform some explicit computations, let $x$ be the sequence $0,1,1,2,3.$ Then $(\Phi_3x)(0)=0,$ $(\Phi_3x)(3)=3x(1)=3,$ $(\Phi_3x)(6)=3x(2)=3,$ $(\Phi_3x)(9)=3x(3)=6,$ and $(\Phi_x)(12)=3x(4)=9.$ The full sequence $\Phi_3x$ contains two terms between each two successive terms in the sequence $0, 3, 3, 6, 9,$ and these two terms interpolate linearly between successive terms. Hence, one gets $$\mathbf{0}, 1, 2, \mathbf{3}, 3, 3, \mathbf{3}, 4, 5, \mathbf{6}, 7, 8, \mathbf{9}.$$ Alternatively, we may just apply the formula. For example, $(\Phi_3x)(7)=(\Phi_3x)(3\cdot2+1)=3x(2)+1\cdot(x(3)-x(2))=3\cdot1+1\cdot(2-1)=4,$ which agrees with the result above.
To take another example, $(\Phi_2x)$ is the sequence $$\mathbf{0}, 1, \mathbf{2}, 2, \mathbf{2}, 3, \mathbf{4}, 5, \mathbf{6}.$$ The even indexed terms (in boldface) are simply $2$ times the sequence $x,$ while the remaining terms interpolate linearly between the terms in boldface. An explicit computation of $(\Phi_2x)(3)$ is $$(\Phi_2x)(3)=(\Phi_2x)(2\cdot1+1)=2\cdot x(1)+1\cdot(x(2)-x(1)=2\cdot1+1\cdot(1-1)=2.$$
The functions $\Phi_m$ have the sequence $z(0)=0,$ $z(1)=1,$ $\ldots,$ $z(n)=n,$ $\ldots,$ that is, the sequence of nonnegative integers, as a fixed point. In other words, $(\Phi_mz)(n)=z(n)=n$ for all integers $m>0$ and $n\ge 0$. To see this, simply compute $$(\Phi_mz)(mq+r)=mz(q)+r(z(q+1)-z(q))=mq+r((q+1)-q)=mq+r.$$