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# Generalized Binomial Coefficients

Our goal is to define a triangle of binomial coefficients where the swinging factorial is always the middle coefficient, and not only in the case when n is even .

To this end we define the generalized binomial coefficient for integer $n\geq 0$ , $m\geq 1$ as

${\text{C}}_{n,k}^{(m)}=0$ if $k<0$ or $k>mn$ , and otherwise
${\text{C}}_{n,k}^{(m)}={\frac {n!}{(n-\left\lceil k/m\right\rceil )!\left\lfloor k/m\right\rfloor !}}\qquad \left(0\leq k\leq mn\right)\,.$ In the sequel we will look only at the case $m=2$ . For this case we will introduce a more persuasive notation and a different enumeration. We set

${\binom {n}{k}}_{2}={\text{C}}_{n,\,n+k}^{(2)}\qquad \left(n\geq 0,\ -n\leq k\leq n\right)\ .$ This gives the following symmetric representation

${\binom {n}{k}}_{2}={\frac {n!}{\Omega _{n-k}!\ \Omega _{n+k}!}}\ ,\quad {\text{where}}\ \Omega _{x}:={\frac {x-\left[{x}\ {\text{odd}}\right]}{2}}\ .$ We see that the central term of this generalization of the binomial coefficients is the swinging factorial. This means

${\binom {n}{0}}_{2}=n\wr \,.$ We see directly that

${\binom {n}{n}}_{2}=1,\ {\binom {n}{n-1}}_{2}=n.$ Also the symmetry relation is clear

${\binom {n}{k}}_{2}={\binom {n}{-k}}_{2}\quad \left(0\leq k\leq n\right)\ .$ If we set ${\binom {n}{n}}_{2}={\binom {n}{-n}}_{2}=1$ and ${\binom {n}{n-1}}_{2}={\binom {n}{1-n}}_{2}=n$ , then the generalized binomial coefficients can be computed for $n\geq 2$ and $-n+2\leq k\leq n-2$ by the recurrence

${\binom {n}{k}}_{2}={\binom {n-1}{k-1}}_{2}+\left[{n-k}\ {\text{odd}}\right]{\binom {n-1}{k}}_{2}+{\binom {n-1}{k+1}}_{2}\ .$ To show this we introduce the abbreviation

$s_{n,k}=\Omega _{n-k}!\ \Omega _{n+k}!\ .$ Now the recurrence equation can be written

${\frac {n!}{s_{n,k}}}={\frac {(n-1)!}{s_{n-1,k-1}}}+\left[{n-k}\ {\text{odd}}\right]{\frac {(n-1)!}{s_{n-1,k}}}+{\frac {(n-1)!}{s_{n-1,k+1}}}\ .$ If $n\geq 2$ and $-n+2\leq k\leq n-2$ this is equivalent to

${\frac {n!}{\left(n-1\right)!}}={\frac {s_{n,k}}{s_{n-1,k-1}}}+\left[{n-k}\ {\text{odd}}\right]{\frac {s_{n,k}}{s_{n-1,k}}}+{\frac {s_{n,k}}{s_{n-1,k+1}}}\ .$ The left hand side of the equation is $n$ , and the right hand side also, because

${\frac {s_{n,k}}{s_{n-1,k-1}}}={\frac {n}{2}}+{\frac {k}{2}}-{\frac {1}{2}}\left[{n-k}\ {\text{odd}}\right]\ ,$ $\left[{n-k}\ {\text{odd}}\right]{\frac {s_{n,k}}{s_{n-1,k}}}=\left[{n-k}\ {\text{odd}}\right]\ ,$ ${\frac {s_{n,k}}{s_{n-1,k+1}}}={\frac {n}{2}}-{\frac {k}{2}}-{\frac {1}{2}}\left[{n-k}\ {\text{odd}}\right].\quad \diamond$ Looking at table 1 below we see the classical binomial triangle embedded: the pictorial representation of Pascal's triangle ${\binom {n}{k}}$ originates from ${\binom {n}{k}}_{2}$ by deleting those entries, where $n$ and $k$ have a different parity. On formal ground we have

${\binom {n}{k}}={\binom {n}{2k-n}}_{2}\qquad \left(0\leq k\leq n\right)\ .$ # Appendix

 n \ k -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 0 1 1 1 1 1 2 1 2 2 2 1 3 1 3 3 6 3 3 1 4 1 4 4 12 6 12 4 4 1 5 1 5 5 20 10 30 10 20 5 5 1 6 1 6 6 30 15 60 20 60 15 30 6 6 1
 n \ k -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 0 1 1 1 1 1 2 2 1 1 1 2 3 6 2 2 1 2 2 6 4 24 6 6 2 4 2 6 6 24 5 120 24 24 6 12 4 12 6 24 24 120 6 720 120 120 24 48 12 36 12 48 24 120 120 720

# Sequences

 A162246 $\ {\text{C}}_{n,k}^{(2)}\$ A056040 ${n}\wr$ A098361 $n!(n-k)!\quad$ A180274 $\Omega _{n-k}!\ \Omega _{n+k}!$ A180064 $n!\ /{n}\wr$ # Maple

C := proc(m,n,k)
if k < 0 or k > m*n then 0 else
n!/(floor(k/m)!*(n-ceil(k/m))!) fi end:
C2 := proc(n,k) C(2,n,n+k) end;
Omega := proc(n) (n-(n mod 2))/2 end:
C2o := proc(n,k) n!/(Omega(n-k)!*Omega(n+k)!) end: