It's an important and popular fact that things are not always what they seem.
sequence $\alpha(arg)$:
Let $x$ be the vector $(0,1,...,(arg-1))$. Let $i$ cycle through the numbers $1,2,\ldots,arg$, repeating after you get to the end. Each time, yield the $i$th coordinate of $x$, then replace that coordinate by the sum of all coordinates of $x$.
sequence $\beta(arg)$:
If arg is $0$, yield $1$ over and over again, forever and ever. Otherwise, let $x$ take the value $0$ initially. Then let $i$ range through the sequence $\beta(arg-1)$. For each such $i$, add $i$ to $x$, then yield $x$.
sequence $\gamma(arg)$:
If arg is $0$, yield $1$ over and over again, forever and ever. Otherwise, let $x$ take the value $0$ initially, and let $j$ be the all $0$s vector in $arg$ dimensional space. Let $i$ range through all vectors whose coordinates are given by the sequences $\gamma(0),\gamma(1),\ldots,\gamma(arg-1)$ (so the first $i$'s coordinates are the first values in the sequences; the second $i$'s coordinates are the second values; etc.). For each such $i$, add the dot product of $i$ with $j$ to $x$, yield $x$, and then set $j$ to be the vector whose coordinates are the same as $i$'s but in reverse order.
sequence $\delta(arg)$:
Let $i$ range through the sequence $\gamma(arg)$. If $i$ is $0$, set $x$ equal to $1$ and yield it. Otherwise, replace $x$ by $x\cdot i$ and yield it.
sequence $\epsilon(arg)$:
Let $x$ start out as the list consisting of the single element $arg$. Then keep doing the following: yield the last element in $x$, and append the dot product of $x$ with its own reversal to the end of $x$.
sequence $\zeta(arg)$:
Let $x$ be the ordered pair $(-1,arg)$. Let $i$ alternate between $1$ and $2$. For each $i$, yield the absolute value of the $i$th coordinate of $x$; then subtract twice the sum of $x$ from that coordinate.
sequence $\eta(arg)$:
Let $x$ be the ordered pair $(1,arg)$. Let $i$ cycle through the numbers $1,2,\ldots,arg$, repeating after you get to the end. For each $i$, do the following: if $i$ is $1$, yield the first coordinate of $x$, then set the first coordinate equal to the second; otherwise, add the first coordinate of $x$ to the second coordinate of $x$.
A $= \gamma(3)$
B $= \alpha(2)$
C $= \zeta(3)$
D $= \delta(3)$
E $= \epsilon(2)$
F $= \zeta(2)$
G $= \alpha(3)$
H $= \eta(1)$
I $= \beta(2)$
J $= \eta(4)$
K $= \zeta(1)$
L $= \epsilon(4)$
M $= \gamma(2)$
N $= \epsilon(3)$
O $= \beta(3)$
P $= \zeta(4)$
Q $= \delta(1)$
R $= \eta(2)$
S $= \epsilon(1)$
T $= \beta(1)$
U $= \delta(4)$
V $= \eta(3)$
W $= \gamma(1)$
X $= \alpha(1)$
Y $= \delta(2)$
Z $= \gamma(4)$
$\displaystyle\frac{x(1+x)}{(1-x)^3}$, $\displaystyle\frac{x}{1-x-x^2}$, $\displaystyle\frac{x(1+4x+x^2)}{(1-x)^4}$, $\displaystyle\frac{1-\sqrt{1-4x}}{2x}$, $\displaystyle\frac{1}{(1-x)^3}$, $\displaystyle\frac{1+x}{(1-x)^2}$
$\displaystyle\frac{1+2x}{(1-x)^2}$, $\displaystyle\frac{1}{1-2x}$, $\displaystyle\frac{1}{(1-x)^4}$, $\displaystyle\frac{x(1+x)}{1-x-x^2-x^3}$, $\displaystyle\frac{1}{1-2x}$, $\displaystyle\frac{x(1+4x+x^2)}{(1-x)^4}$, $\displaystyle\frac{x(1+x)}{(1-x)^3}$, $\displaystyle\frac{x(1+x)}{(1-x)^3}$, $\displaystyle\frac{1}{(1-x)^3}$, $\displaystyle\frac{6}{1+\sqrt{1-12x}}$, $\displaystyle\frac{x(1+x)}{1-x-x^2-x^3}$