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# Factorial polynomials

The factorial polynomials are to [finite] difference calculus what polynomials are to [infinitesimal] differential calculus.

## Falling factorial and raising factorial

The falling factorial is defined as[1]

${\displaystyle {\begin{array}{l}\displaystyle {x^{\underline {k}}:=\prod _{i=0}^{k-1}(x-i),\quad k\in \mathbb {N} ,}\end{array}}}$

and the raising factorial is defined as[1]

${\displaystyle {\begin{array}{l}\displaystyle {x^{\overline {k}}:=\prod _{i=0}^{k-1}(x+i),\quad k\in \mathbb {N} ,}\end{array}}}$
where, in either case, for
 k = 0
we get the empty product, i.e. 1.

## Factorial polynomials

A factorial term (Boole, 1970: p. 6) or a factorial polynomial (Elaydi, 2005: p. 60) is defined as

${\displaystyle {\begin{array}{l}\displaystyle {x^{(k)}:=x^{\underline {k}},\quad k\in \mathbb {N} ,}\end{array}}}$
and for negative
 k
we have the definition
${\displaystyle {\begin{array}{l}\displaystyle {x^{(-k)}:=\left[x^{\overline {k}}\right]^{-1},\quad k\in \mathbb {N} ,}\end{array}}}$
where, in either case, for
 k = 0
we get the empty product, i.e. 1.
The factorial polynomials of degree
 n
are defined as the sum of factorial terms
${\displaystyle {\begin{array}{l}\displaystyle {P(x):=\sum _{k=0}^{n}a_{k}\,x^{(k)},\quad n\in \mathbb {N} ,}\end{array}}}$
where for
 n = 0
we get the constant polynomial
 a0
.

In a fashion similar to Laurent polynomials, we also have the definition

${\displaystyle {\begin{array}{l}\displaystyle {L(x):=\sum _{k=m}^{n}a_{k}\,x^{(k)},\quad m\in \mathbb {Z} ,n\in \mathbb {Z} ,m\leq n.}\end{array}}}$

### Finite difference operator

With the above definitions, the [finite] difference operator

${\displaystyle {\begin{array}{l}\displaystyle {\Delta \,f(x):=(E-I)\,f(x)=f(x+1)-f(x)}\end{array}}}$
where
 E
is the shift operator and
 I
is the identity operator, behaves with the [whether ordinary or generalized] factorial polynomials like the [infinitesimal] differential operator does with the [whether ordinary or Laurent] polynomials.
Difference operator Differential operator
${\displaystyle {\begin{array}{l}\displaystyle {\Delta \,x^{(0)}=0}\end{array}}}$ ${\displaystyle {\begin{array}{l}\displaystyle {D\,x^{0}=0}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {\Delta \,x^{(k)}=k\,x^{(k-1)},\quad k\in \mathbb {N} ^{+}}\end{array}}}$ ${\displaystyle {\begin{array}{l}\displaystyle {D\,x^{k}=k\,x^{k-1},\quad k\in \mathbb {N} ^{+}}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {\Delta \,x^{(-k)}=(-k)\,x^{(-k-1)},\quad k\in \mathbb {N} ^{+}}\end{array}}}$ ${\displaystyle {\begin{array}{l}\displaystyle {D\,x^{(-k)}=(-k)\,x^{-k-1},\quad k\in \mathbb {N} ^{+}}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {\Delta \left\{\sum _{k=m}^{n}a_{k}\,x^{(k)}\right\}=\sum _{k=m}^{n}a_{k}\,\Delta \,x^{(k)},\quad m\in \mathbb {Z} ,n\in \mathbb {Z} ,m\leq n}\end{array}}}$ ${\displaystyle {\begin{array}{l}\displaystyle {D\left\{\sum _{k=m}^{n}a_{k}\,x^{k}\right\}=\sum _{k=m}^{n}a_{k}\,D\,x^{k},\quad m\in \mathbb {Z} ,n\in \mathbb {Z} ,m\leq n}\end{array}}}$

## Stirling numbers

For
 n   ≥   0
, we have
${\displaystyle {\begin{array}{l}\displaystyle {x^{(n)}=\sum _{k=0}^{n}s(n,k)\,x^{k}=\sum _{k=0}^{n}S_{1}(n,k)\,x^{k}=\sum _{k=0}^{n}(-1)^{n+k}\,\vert S_{1}(n,k)\vert \,x^{k},}\end{array}}}$

and

${\displaystyle {\begin{array}{l}\displaystyle {x^{(-n)}:=\left[x^{\overline {n}}\,\right]^{-1}=\left[\sum _{k=0}^{n}(-1)^{n+k}\,S_{1}(n,k)\,\left[x^{-k}\right]^{-1}\right]^{-1}=\left[\sum _{k=0}^{n}(-1)^{n+k}\,S_{1}(n,k)\,x^{k}\right]^{-1}=\left[\sum _{k=0}^{n}\vert S_{1}(n,k)\vert \,x^{k}\right]^{-1}=\left[\sum _{k=0}^{n}\left[{n \atop k}\right]\,x^{k}\right]^{-1},}\end{array}}}$
where
 s(n, k)
or
 S1(n, k)
are Stirling numbers of the first kind, and ${\displaystyle \textstyle {[{n \atop k}]}}$ are unsigned Stirling numbers of the first kind.[2]

For
 n   ≥   0
, we have
${\displaystyle {\begin{array}{l}\displaystyle {x^{n}=\sum _{k=0}^{n}\left\{{n \atop m}\right\}\,x^{(k)}=\sum _{k=0}^{n}S_{k}^{(n)}\,x^{(k)}=\sum _{k=0}^{n}S(n,k)\,x^{(k)}=\sum _{k=0}^{n}S_{2}(n,k)\,x^{(k)},}\end{array}}}$

and

${\displaystyle {\begin{array}{l}\displaystyle {x^{-n}:=\left[x^{n}\right]^{-1}=\left[\sum _{k=0}^{n}(-1)^{n+k}\,S_{2}(n,k)\,\left[x^{(-k)}\right]^{-1}\right]^{-1}=\left[\sum _{k=0}^{n}(-1)^{n+k}\,S_{2}(n,k)\,x^{\overline {k}}\right]^{-1},}\end{array}}}$
where ${\displaystyle \textstyle {\left\{{n \atop k}\right\}}}$,
 S  (n) k
,
 S(n, k)
or
 S2(n, k)
are Stirling numbers of the second kind, and
 ( − 1) n+k S2(n, k)
are signed Stirling numbers of the second kind.[2]

### Stirling numbers of the first kind

We have

${\displaystyle {\begin{array}{l}\displaystyle {x^{(0)}=+\,\ 1\,x^{0},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(1)}=-\,\ 0\,x^{0}+\,\ 1\,x^{1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(2)}=+\,\ 0\,x^{0}-\,\ 1\,x^{1}+\ 1\,x^{2},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(3)}=-\,\ 0\,x^{0}+\,\ 2\,x^{1}-\ 3\,x^{2}+\ 1\,x^{3},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(4)}=+\,\ 0\,x^{0}-\,\ 6\,x^{1}+11\,x^{2}-\ 6\,x^{3}+\ 1\,x^{4},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(5)}=-\,\ 0\,x^{0}+\,24\,x^{1}-50\,x^{2}+35\,x^{3}-10\,x^{4}+\ 1\,x^{5},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(6)}=+\,\ 0\,x^{0}-\,120\,x^{1}+274\,x^{2}-225\,x^{3}+85\,x^{4}-15\,x^{5}+\ 1\,x^{6},}\end{array}}}$
involving the triangle of Stirling numbers of the first kind, where the coefficients of row
 k
are obtained by multiplying the polynomial of row
 k  −  1
by
 (x  −  k + 1)
.

A048994 Triangle read by rows of Stirling numbers of the first kind,
 S1(n, k), n   ≥   0, 0   ≤   k   ≤   n
.
{1, 0, 1, 0, −1, 1, 0, 2, −3, 1, 0, − 6, 11, − 6, 1, 0, 24, −50, 35, −10, 1, 0, −120, 274, −225, 85, −15, 1, 0, 720, −1764, 1624, −735, 175, −21, 1, 0, −5040, 13068, −13132, 6769, −1960, 322, −28, 1, ...}
A008275 Triangle read by rows of Stirling numbers of the first kind,
 S1(n, k), n   ≥   1, 1   ≤   k   ≤   n
.
{1, −1, 1, 2, −3, 1, − 6, 11, − 6, 1, 24, −50, 35, −10, 1, −120, 274, −225, 85, −15, 1, 720, −1764, 1624, −735, 175, −21, 1, −5040, 13068, −13132, 6769, −1960, 322, −28, 1, ...}

#### Row sums of Stirling numbers of the first kind

The row sums of Stirling numbers of the first kind are

${\displaystyle {\begin{array}{l}\displaystyle {\sum _{k=0}^{n}S_{1}(n,k)=0^{n^{(2)}}=0^{n\,(n-1)},\quad n\geq 0.}\end{array}}}$

### Unsigned Stirling numbers of the first kind

We have

${\displaystyle {\begin{array}{l}\displaystyle {x^{(-0)}=\left[x^{\overline {0}}\right]^{-1}=\left[1\,x^{0}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(-1)}=\left[x^{\overline {1}}\right]^{-1}=\left[0\,x^{0}+\ 1\,x^{1}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(-2)}=\left[x^{\overline {2}}\right]^{-1}=\left[0\,x^{0}+\ 1\,x^{1}+\ 1\,x^{2}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(-3)}=\left[x^{\overline {3}}\right]^{-1}=\left[0\,x^{0}+\ 2\,x^{1}+\ 3\,x^{2}+\ 1\,x^{3}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(-4)}=\left[x^{\overline {4}}\right]^{-1}=\left[0\,x^{0}+\ 6\,x^{1}+11\,x^{2}+\ 6\,x^{3}+\ 1\,x^{4}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(-5)}=\left[x^{\overline {5}}\right]^{-1}=\left[0\,x^{0}+24\,x^{1}+50\,x^{2}+35\,x^{3}+10\,x^{4}+\ 1\,x^{5}\right]^{-1}.}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{(-6)}=\left[x^{\overline {6}}\right]^{-1}=\left[0\,x^{0}+120\,x^{1}+274\,x^{2}+225\,x^{3}+85\,x^{4}+15\,x^{5}+\ 1\,x^{6}\right]^{-1},}\end{array}}}$
involving the triangle of unsigned Stirling numbers of the first kind, also called the factorial triangle,[3] where the coefficients of row
 k
are obtained by multiplying the polynomial of row
 k  −  1
by
 (x + k  −  1)
.

A132393 Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows,
 | S1(n, k) |, n   ≥   0, 0   ≤   k   ≤   n
.
{1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, ...}

#### Row sums of triangle of unsigned Stirling numbers of the first kind

The row sums of unsigned Stirling numbers of the first kind are

${\displaystyle {\begin{array}{l}\displaystyle {\sum _{k=0}^{n}|S_{1}(n,k)|=S_{1}(n+1,1)=n!,\quad n\geq 0.}\end{array}}}$

#### Falling diagonals of triangle of unsigned Stirling numbers of the first kind

The coefficients of the
 k
th falling diagonal, where
 k = 0
refers to the rightmost one, are given by a polynomial of degree
 2k
.

A000012 Stirling numbers of the first kind:
 s(n, n), n   ≥   0.
(The simplest sequence of positive numbers: the all 1's sequence.)

[a_0(n) = 1] G.f.: 1/(1-x).

{1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...}
A000217 Stirling numbers of the first kind:
 s(n + 1, n), n   ≥   0.
(Triangular numbers:
 a(n) = C(n + 1, 2) = n (n + 1) / 2 = 0 + 1 + 2 + ... + n
.)

[a_1(n) = (n + n^2) / 2 = n (1 + n) / 2] G.f.: x / (1-x)^3.

{0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, ...}
A000914 Stirling numbers of the first kind:
 s(n + 2, n), n   ≥   0
.

[a_2(n) = (10 n + 21 n^2 + 14 n^3 + 3 n^4) / 24 = n (1 + n) (2 + n) (5 + 3 n) / 24] G.f.: (2*x + x^2) / (1-x)^5.

{0, 2, 11, 35, 85, 175, 322, 546, 870, 1320, 1925, 2717, 3731, 5005, 6580, 8500, 10812, 13566, 16815, 20615, 25025, 30107, 35926, 42550, 50050, 58500, 67977, 78561, 90335, ...}
A001303 Stirling numbers of first kind,
 s(n + 3, n), n   ≥   1,
negated.

[a_3(n) = (36 n + 96 n^2 + 97 n^3 + 47 n^4 + 11 n^5 + n^6) / 48 = n (1 + n) (2 + n)^2 (3 + n)^2 / 48] G.f.: (6*x + 8*x^2 + x^3) / (1-x)^7.

{6, 50, 225, 735, 1960, 4536, 9450, 18150, 32670, 55770, 91091, 143325, 218400, 323680, 468180, 662796, 920550, 1256850, 1689765, 2240315, 2932776, 3795000, 4858750, 6160050, ...}
A000915 Stirling numbers of first kind
 s(n + 4, n), n   ≥   1
.

[a_4(n) = ?] G.f.: (24*x + 58*x^2 + 22*x^3 + x^4) / (1-x)^9.

{24, 274, 1624, 6769, 22449, 63273, 157773, 357423, 749463, 1474473, 2749747, 4899622, 8394022, 13896582, 22323822, 34916946, 53327946, 79721796, 116896626, 16842387, ...}

Triangle of polynomials for the numerator of the o.g.f.s for unsigned Stirling numbers of first kind along the falling diagonals:

${\displaystyle {\begin{array}{l}\displaystyle {1,}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {0+\ \ 1\ x,}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {0+\ \ 2\ x+\ \ \ 1\ x^{2},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {0+\ \ 6\ x+\ \ \ 8\ x^{2}+\ \ \ 1\ x^{3},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {0+\ 24\ x+\ \ 58\ x^{2}+\ \ 22\ x^{3}+\ \ \ 1\ x^{4},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {0+120\ x+\ 444\ x^{2}+\ 328\ x^{3}+\ \ 52\ x^{4}+\ \ 1\ x^{5},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {0+720\ x+3708\ x^{2}+4400\ x^{3}+1452\ x^{4}+114\ x^{5}+\ 1\ x^{6}.}\end{array}}}$

A112007 Coefficient triangle for polynomials used for o.g.f.s for unsigned Stirling1 diagonals, for
 n   ≥   1, 1   ≤   k   ≤   n
.
{1, 2, 1, 6, 8, 1, 24, 58, 22, 1, 120, 444, 328, 52, 1, 720, 3708, 4400, 1452, 114, 1, 5040, 33984, 58140, 32120, 5610, 240, 1, 40320, 341136, 785304, 644020, 195800, 19950, ...}

### Stirling numbers of the second kind

We have

${\displaystyle {\begin{array}{l}\displaystyle {x^{0}=1\,x^{(0)},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{1}=0\,x^{(0)}+1\,x^{(1)},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{2}=0\,x^{(0)}+1\,x^{(1)}+\ 1\,x^{(2)},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{3}=0\,x^{(0)}+1\,x^{(1)}+\ 3\,x^{(2)}+\ 1\,x^{(3)},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{4}=0\,x^{(0)}+1\,x^{(1)}+\ 7\,x^{(2)}+\ 6\,x^{(3)}+\ 1\,x^{(4)},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{5}=0\,x^{(0)}+1\,x^{(1)}+15\,x^{(2)}+25\,x^{(3)}+10\,x^{(4)}+\ 1\,x^{(5)},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{6}=0\,x^{(0)}+1\,x^{(1)}+31\,x^{(2)}+90\,x^{(3)}+65\,x^{(4)}+15\,x^{(5)}+\ 1\,x^{(6)},}\end{array}}}$

involving the triangle of Stirling numbers of the second kind.

A048993 Triangle of Stirling numbers of second kind,
 S2(n, k), n   ≥   0, 0   ≤   k   ≤   n
.
{1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 7, 6, 1, 0, 1, 15, 25, 10, 1, 0, 1, 31, 90, 65, 15, 1, 0, 1, 63, 301, 350, 140, 21, 1, 0, 1, 127, 966, 1701, 1050, 266, 28, 1, 0, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, ...}
A106800 Triangle of Stirling numbers of second kind,
 S2(n, n  −  k), n   ≥   0, 0   ≤   k   ≤   n
.
{1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 7, 1, 0, 1, 10, 25, 15, 1, 0, 1, 15, 65, 90, 31, 1, 0, 1, 21, 140, 350, 301, 63, 1, 0, 1, 28, 266, 1050, 1701, 966, 127, 1, 0, 1, 36, 462, 2646, 6951, 7770, 3025, 255, 1, ...}
A008277 Triangle read by rows of Stirling numbers of the second kind,
 S2(n, k), n   ≥   1, 1   ≤   k   ≤   n
.
{1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, ...}

#### Row sums of triangle of Stirling numbers of the second kind

The row sums of Stirling numbers of the second kind are

${\displaystyle {\begin{array}{l}\displaystyle {\sum _{k=0}^{n}S_{2}(n,k)=B_{n},\quad n\geq 0,}\end{array}}}$
where
 Bn
are the Bell (or exponential) numbers.

A000110 Bell or exponential numbers: number of ways to partition a set of
 n, n   ≥   0,
labeled elements.
{1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 1382958545, 10480142147, 82864869804, 682076806159, 5832742205057, 51724158235372, ...}

#### Bell polynomials

By replacing the
 x (k)
by
 x k
in
${\displaystyle {\begin{array}{l}\displaystyle {x^{n}=\sum _{k=0}^{n}S_{2}(n,k)\,x^{(k)},\quad n\geq 0,}\end{array}}}$

one obtains the Bell polynomials

${\displaystyle {\begin{array}{l}\displaystyle {B_{n}(x):=\sum _{k=0}^{n}S_{2}(n,k)\,x^{k},\quad n\geq 0,}\end{array}}}$
for which
 Bn (1)
yields the Bell numbers and
 Bn ( − 1)
yields the complementary Bell numbers.

### Signed Stirling numbers of the second kind

We have

${\displaystyle {\begin{array}{l}\displaystyle {x^{-0}=\left[+\,1\,x^{\overline {0}}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{-1}=\left[-\,0\,x^{\overline {0}}+\ 1\,x^{\overline {1}}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{-2}=\left[+\,0\,x^{\overline {0}}-\ 1\,x^{\overline {1}}+\ 1\,x^{\overline {2}}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{-3}=\left[-\,0\,x^{\overline {0}}+\ 1\,x^{\overline {1}}-\ 3\,x^{\overline {2}}+\ 1\,x^{\overline {3}}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{-4}=\left[+\,0\,x^{\overline {0}}-\ 1\,x^{\overline {1}}+\ 7\,x^{\overline {2}}-\ 6\,x^{\overline {3}}+\ 1\,x^{\overline {4}}\right]^{-1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{-5}=\left[-\,0\,x^{\overline {0}}+\ 1\,x^{\overline {1}}-15\,x^{\overline {2}}+25\,x^{\overline {3}}-10\,x^{\overline {4}}+\ 1\,x^{\overline {5}}\right]^{-1}.}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {x^{-6}=\left[+\,0\,x^{\overline {0}}-\ 1\,x^{\overline {1}}+31\,x^{\overline {2}}-90\,x^{\overline {3}}+65\,x^{\overline {4}}-15\,x^{\overline {5}}+\ 1\,x^{\overline {6}}\right]^{-1},}\end{array}}}$

involving the triangle of signed Stirling numbers of the second kind.

A?????? Triangle of signed Stirling numbers of second kind,
 ( − 1) n+k S2(n, k), n   ≥   0, 0   ≤   k   ≤   n
.
{1, 0, 1, 0, −1, 1, 0, 1, −3, 1, 0, −1, 7, − 6, 1, 0, 1, −15, 25, −10, 1, 0, −1, 31, − 90, 65, −15, 1, 0, 1, − 63, 301, −350, 140, −21, 1, 0, −1, 127, − 966, 1701, −1050, 266, −28, 1, ...}
A?????? Triangle read by rows of signed Stirling numbers of the second kind,
 ( − 1) n+k S2(n, k), n   ≥   1, 1   ≤   k   ≤   n
.
{?, ...}

#### Row sums of triangle of signed Stirling numbers of the second kind

The row sums of signed Stirling numbers of the second kind

${\displaystyle {\begin{array}{l}\displaystyle {\sum _{k=0}^{n}(-1)^{n+k}\,S_{2}(n,k)=(-1)^{n}\sum _{k=0}^{n}(-1)^{k}\,S_{2}(n,k)=(-1)^{n}\,{\tilde {B}}_{n},\quad n\geq 0,}\end{array}}}$
where
 B̃n
are the Rao Uppuluri-Carpenter numbers (or complementary Bell numbers), yield the sequence
{1, (−) −1, 0, (−) 1, 1, (−) −2, − 9, (−) − 9, 50, (−) 267, 413, (−) −2180, −17731, (−) −50533, 110176, (−) 1966797, 9938669, (−) 8638718, −278475061, (−) −2540956509, − 9816860358, (−) 27172288399, ...}

A000587 Rao Uppuluri-Carpenter numbers (or complementary Bell numbers),
 n   ≥   0
: e.g.f. =
 exp(1  −  exp(x))
.
{1, −1, 0, 1, 1, −2, − 9, − 9, 50, 267, 413, −2180, −17731, −50533, 110176, 1966797, 9938669, 8638718, −278475061, −2540956509, − 9816860358, 27172288399, 725503033401, 5592543175252, ...}

## Factorial polynomials inspired from ordinary binomial expansions

### Factorial polynomials inspired from ordinary binomial expansion of (x  +  1)n

By replacing the
 x k
by
 x (k)
in the binomial expansion
${\displaystyle {\begin{array}{l}\displaystyle {(x+1)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}\,x^{k},\quad n\geq 0,}\end{array}}}$

we get

${\displaystyle {\begin{array}{l}\displaystyle {P_{n}(x):=\sum _{k=0}^{n}{\binom {n}{k}}\,x^{(k)},\quad n\geq 0.}\end{array}}}$

Examples:

${\displaystyle {\begin{array}{l}\displaystyle {P_{0}(x)=\sum _{k=0}^{0}{\binom {0}{k}}\,x^{(k)}=1\,x^{0},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{1}(x)=\sum _{k=0}^{1}{\binom {1}{k}}\,x^{(k)}=1\,x^{0}+\ 1\,x^{1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{2}(x)=\sum _{k=0}^{2}{\binom {2}{k}}\,x^{(k)}=1\,x^{0}+\ 1\,x^{1}+\ 1\,x^{2},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{3}(x)=\sum _{k=0}^{3}{\binom {3}{k}}\,x^{(k)}=1\,x^{0}+\ 2\,x^{1}+\ 0\,x^{2}+\ 1\,x^{3},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{4}(x)=\sum _{k=0}^{4}{\binom {4}{k}}\,x^{(k)}=1\,x^{0}+\ 0\,x^{1}+\ 5\,x^{2}-\ 2\,x^{3}+\ 1\,x^{4},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{5}(x)=\sum _{k=0}^{5}{\binom {5}{k}}\,x^{(k)}=1\,x^{0}+\ 9\,x^{1}-15\,x^{2}+15\,x^{3}-\ 5\,x^{4}+\ 1\,x^{5},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{6}(x)=\sum _{k=0}^{6}{\binom {6}{k}}\,x^{(k)}=1\,x^{0}-35\,x^{1}+94\,x^{2}-85\,x^{3}+40\,x^{4}-\ 9\,x^{5}+\ 1\,x^{6}.}\end{array}}}$
Triangle read by rows of ordinary polynomial coefficients resulting from replacing x^k by x^(k) in the binomial expansion of (x+1)^n, for
 n   ≥   0
.
{1, 1, 1, 1, 1, 1, 1, 2, 0, 1, 1, 0, 5, −2, 1, 1, 9, −15, 15, −5, 1, 1, −35, 94, −85, 40, − 9, 1, ...}

#### Triangle row sums

It appears that the row sums yield

${\displaystyle {\begin{array}{l}\displaystyle {P_{n}(1)=n+1,\quad n\geq 0.}\end{array}}}$

#### Triangle columns

It appears that we might have for column 1 =
 { 0, 1, 1, 2, 0, 9, -35, ...}
${\displaystyle {\begin{array}{l}\displaystyle {T(n,1)=(-1)^{[n{\text{ belongs to some set}}]}\,A002741(n),\quad n\geq 0,}\end{array}}}$
where
 [ · ]
is the Iverson bracket.

A002741 Logarithmic numbers: expansion of
 −  log(1  −  x) exp( −  x)
.
{0, 1, −1, 2, 0, 9, 35, 230, 1624, 13209, 120287, 1214674, 13469896, 162744945, 2128047987, 29943053062, 451123462672, 7245940789073, ...}

### Factorial polynomials inspired from ordinary binomial expansion of (x  −  1)n

By replacing the
 x k
by
 x (k)
in the binomial expansion
${\displaystyle {\begin{array}{l}\displaystyle {(x-1)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}\,(-1)^{n-k}\,x^{k},\quad n\geq 0,}\end{array}}}$

we get

${\displaystyle {\begin{array}{l}\displaystyle {P_{n}(x):=\sum _{k=0}^{n}{\binom {n}{k}}\,(-1)^{n-k}\,x^{(k)},\quad n\geq 0.}\end{array}}}$

Examples:

${\displaystyle {\begin{array}{l}\displaystyle {P_{0}(x)=\sum _{k=0}^{0}{\binom {0}{k}}\,(-1)^{n-k}\,x^{(k)}=+\ 1\,x^{0},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{1}(x)=\sum _{k=0}^{1}{\binom {1}{k}}\,(-1)^{n-k}\,x^{(k)}=-\ 1\,x^{0}+\ 1\,x^{1},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{2}(x)=\sum _{k=0}^{2}{\binom {2}{k}}\,(-1)^{n-k}\,x^{(k)}=+\ 1\,x^{0}-\ 3\,x^{1}+\ 1\,x^{2},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{3}(x)=\sum _{k=0}^{3}{\binom {3}{k}}\,(-1)^{n-k}\,x^{(k)}=-\ 1\,x^{0}+\ 8\,x^{1}-\ 6\,x^{2}+\ 1\,x^{3},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{4}(x)=\sum _{k=0}^{4}{\binom {4}{k}}\,(-1)^{n-k}\,x^{(k)}=+\ 1\,x^{0}-24\,x^{1}+29\,x^{2}-10\,x^{3}+\ 1\,x^{4},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{5}(x)=\sum _{k=0}^{5}{\binom {5}{k}}\,(-1)^{n-k}\,x^{(k)}=-\ 1\,x^{0}+89\,x^{1}-145\,x^{2}+75\,x^{3}-15\,x^{4}+\ 1\,x^{5},}\end{array}}}$
${\displaystyle {\begin{array}{l}\displaystyle {P_{6}(x)=\sum _{k=0}^{6}{\binom {6}{k}}\,(-1)^{n-k}\,x^{(k)}=+\ 1\,x^{0}-415\,x^{1}+814\,x^{2}-545\,x^{3}+160\,x^{4}-21\,x^{5}+\ 1\,x^{6}.}\end{array}}}$
A?????? Triangle read by rows of ordinary polynomial coefficients resulting from replacing x^k by x^(k) in the binomial expansion of (x-1)^n, for
 n   ≥   0
.
{1, −1, 1, 1, −3, 1, −1, 8, − 6, 1, 1, −24, 29, −10, 1, ...}

Related to (although the signs differ)

A094816 Triangle read by rows: T(n,k), 0<=k<=n, = coefficients of Charlier polynomials: A046716 transposed.

{1, 1, 1, 1, 3, 1, 1, 8, 6, 1, 1, 24, 29, 10, 1, 1, 89, 145, 75, 15, 1, 1, 415, 814, 545, 160, 21, 1, 1, 2372, 5243, 4179, 1575, 301, ...}

#### Triangle row sums

It appears that the row sums yield

${\displaystyle {\begin{array}{l}\displaystyle {P_{n}(1)=(-1)^{n+1}\,(n-1),\quad n\geq 0.}\end{array}}}$

#### Triangle columns

It appears that we have for column 1 =
 { 0, 1, -3, 8, -24, 89, -415, ...}
${\displaystyle {\begin{array}{l}\displaystyle {T(n,1)=(-1)^{n+1}\,A002104(n),\quad n\geq 0.}\end{array}}}$
A002104 Logarithmic numbers: expansion of
 −  log(1  −  x) exp(x)
.
{0, 1, 3, 8, 24, 89, 415, 2372, 16072, 125673, 1112083, 10976184, 119481296, 1421542641, 18348340127, 255323504932, 3809950977008, ...}

## Notes

1. Using the
 x k
and
 x k
notation proposed in (Graham, Knuth, and Patashnik, 1994: p. 48).
2. Named after the Scottish mathematician James Stirling.
3. Steven Schwartzman, "The Factorial Triangle and Polynomial Sequences", The College Mathematics Journal, Vol. 15, No. 5 (Nov., 1984), pp. 424-426.

## References

• Boole, G. Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
• Elaydi, S.N. An Introduction to Difference Equations, 3rd ed. Springer, 2005.
• Graham, R. L.; Knuth, D. E.; and Patashnik, O. Concrete Mathematics: A Foundation for Computer Science, 2nd ed. Reading, MA: Addison-Wesley, 1994.