counting the number of solution How many solutions of the equation $x_0+x_1+⋯+x_k=s$ such that $0\leq x_i\leq s$, such that one solution is not the reverse of another. That is suppose $k=3$ and $s=4$ then the solution $(0,1,0,3)$ is the reverse of $(3,0,1,0)$.
 A: Ordinarily (ignoring the reversing condition) the answer would be $\binom{k+s}{s}$ from a stars and bars argument.
However, we now need to correct for the reversing condition.  Most of the time, this operation "pairs up" a sequence with a distinct sequence.  We can calculate the number that are paired with themselves as follows:
We want to find the number of solutions to the given equation for which $x_0=x_k$, and $x_1=x_{k-1}$, and so on.  This is given by


*

*the number of solutions to $2x_0+2x_1+\cdots+2x_{\lfloor k/2 \rfloor}=s$ if $k$ is odd, or

*the number of solutions to $2x_0+2x_1+\cdots+2x_{k/2-1}+x_{k/2}=s$ if $k$ is even.


In the odd $k$ case, the number of solutions is given by $\binom{s/2+\lfloor k/2 \rfloor}{s/2}$ if $s$ is even, or $0$ if $s$ is odd, by the stars and bars method (diving by $2$ both sides).  In the even $k$ case, it's a bit trickier, since $x_{k/2}$ (i.e., the middle number) can take any value $i \in \{0,1,\ldots,s\}$ for which $s-i \equiv 0 \pmod 2$.  So we have $\sum_{\substack{i \in \{0,1,\ldots,s\} \\ s \equiv i \pmod 2}} \binom{s-i+k/2}{(s-i)/2}$.
So, the number of sequences that are partnered with themselves is
$$c(k,s):=\begin{cases} 0 & \text{if $k$ is odd and $s$ is odd} \\ \binom{s/2+\lfloor k/2 \rfloor}{s/2} & \text{if $k$ is odd and $s$ is even} \\ \sum_{\substack{i \in \{0,1,\ldots,s\} \\ s \equiv i \pmod 2}} \binom{(s-i)/2+k/2-1}{(s-i)/2} & \text{if $k$ is even.}\end{cases}$$
Thus the number is $$\frac{1}{2}\left(\binom{k+s}{s}-c(k,s)\right)+c(k,s)=\frac{1}{2}\binom{k+s}{s}+\frac{c(k,s)}{2}.$$

There were some bugs in earlier versions of this answer (but the overall idea didn't change).  So here's some GAP code which verifies it's correct for $k,s \in \{1,2,\ldots,7\}$:
f:=function(k,s)
  local n,c;
  n:=Binomial(k+s,s);
  if(k mod 2=1) then
    if(s mod 2=1) then
      c:=0;
    else
      c:=Binomial(s/2+Int(k/2),s/2);
    fi;
  else
    c:=Sum(Filtered([0..s],i->(s-i) mod 2=0),i->Binomial((s-i)/2+(k/2-1),(s-i)/2));
  fi;
  return (n-c)/2+c;
end;;

g:=function(k,s)
  return Size(Filtered(Tuples([0..s],k+1),T->Sum(T)=s and T<=Reversed(T)));
end;;

for k in [1..7] do for s in [1..7] do Print([k,s]," ",f(k,s)," ",g(k,s),"\n"); od; od;

A: OEIS says the sequence is related to Alkane numbers, which are defined as:
\begin{align*}
  L(c, r) &= \frac{1}{2}\left(\binom{c+r-3}{r} + D(c, r)\right)
\end{align*}
where
$$
   D(c, r) = \left\{\begin{matrix}
\dbinom{(c + r - 3)/2}{ r/2} &    \text{$c$ is odd and $r$ is even} \\\\
0   &  \text{$c$ is even and $r$ is odd} \\\\
\dbinom{(c + r - 4)/2}{ r/2}  & \text{both $c$ and $r$ are even} \\\\  
\dbinom{(c + r - 4)/2}{ (r - 1)/2} & \text{both $c$ and $r$ are odd} 
\end{matrix}\right.
$$
and the required answer is $$L\left(k+3,s\right)$$
For $k=3, s=4$, $$L\left(6,4\right)=19$$
This may be of interest : https://cs.uwaterloo.ca/journals/JIS/cayley.html
