Factorial polynomials

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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

x k  :=    k  − 1 ∏   i  = 0    (x − i), k ∈ ℕ,

and the raising factorial is defined as

x k  :=    k  − 1 ∏   i  = 0    (x + i), k ∈ ℕ,

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

x (k )  :=  x k, k ∈ ℕ,

and for negative

k

we have the definition

x ( − k )  :=  [x k  ]  − 1, k ∈ ℕ,

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

P (x)  :=    n ∑   k  = 0    ak x (k ), n ∈ ℕ,

where for

n = 0

we get the constant polynomial

a 0

.

Generalized factorial polynomials

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

L (x)  :=    n ∑   k  = m    ak x (k ), m ∈ ℤ, n ∈ ℤ, m ≤ n.

Finite difference operator

With the above definitions, the [finite] difference operator

Δ f  (x)  :=  (E − I  ) f  (x)  =  f  (x + 1) − f  (x)

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
Δ  x (0)  =  0	D  x 0  =  0
Δ  x (k )  =  k x (k  − 1), k ∈ ℕ+	D  x k  =  k x k  − 1, k ∈ ℕ+
Δ  x ( − k )  =  ( − k ) x ( − k  − 1), k ∈ ℕ+	D  x  − k  =  ( − k ) x  − k  − 1, k ∈ ℕ+
Δ{ ∑ n k  = m ak x (k )} = ∑ n k  = m ak Δ  x (k ), m ∈ ℤ, n ∈ ℤ, m   ≤   n	D{ ∑ n k  = m ak x k} = ∑ n k  = m ak D  x k, m ∈ ℤ, n ∈ ℤ, m   ≤   n

Converting from polynomial representation to factorial polynomial representation and vice versa

From polynomial representation to factorial polynomial representation, we have

(b0, b1, ..., bn  )T  =  Fn  + 1 × (a0, a1, ..., an  )T,

and from factorial polynomial representation to polynomial representation, we have

(a0, a1, ..., an  )T  =  (Fn  + 1)  − 1 × (b0, b1, ..., bn  )T,

where

(a0, a1, ..., an  )

are the polynomial coefficients,

(b0, b1, ..., bn  )

are the factorial polynomial coefficients (in both cases the degree being

n

), and where

Fn  + 1

is a

(n + 1)  ×  (n + 1)

transformation (from polynomial to factorial polynomial) upper triangular matrix while

(Fn  + 1)  − 1

is the matrix inverse.

     F1  = 

1

,  F2  = 

1		0
0		1

,  F3  = 

1		0		0
0		1		1
0		0		1

,  F4  = 

1		0		0		0
0		1		1		1
0		0		1		3
0		0		0		1

,  F5  = 

1		0		0		0		0
0		1		1		1		1
0		0		1		3		7
0		0		0		1		6
0		0		0		0		1

,  F6  = 

1		0		0		0		0		0
0		1		1		1		1		1
0		0		1		3		7		15
0		0		0		1		6		25
0		0		0		0		1		10
0		0		0		0		0		1

,  F7  = 

1		0		0		0		0		0		0
0		1		1		1		1		1		1
0		0		1		3		7		15		31
0		0		0		1		6		25		90
0		0		0		0		1		10		65
0		0		0		0		0		1		15
0		0		0		0		0		0		1

,  ..., 

S (0, 0)

S (1, 0)

S (2, 0)

S (3, 0)

S (4, 0)

S (5, 0)

S (6, 0)

⋯

0

S (1, 1)

S (2, 1)

S (3, 1)

S (4, 1)

S (5, 1)

S (6, 1)

⋯

0

0

S (2, 2)

S (3, 2)

S (4, 2)

S (5, 2)

S (6, 2)

⋯

0

0

0

S (3, 3)

S (4, 3)

S (5, 3)

S (6, 3)

⋯

0

0

0

0

S (4, 4)

S (5, 4)

S (6, 4)

⋯

0

0

0

0

0

S (5, 5)

S (6, 5)

⋯

0

0

0

0

0

0

S (6, 6)

⋯

⋮

⋮

⋮

⋮

⋮

⋮

⋮

⋱

,
where

S  (n, k )

are Stirling numbers of the second kind.

Fn  +1

is obtained recursively from

Fn

:
     

F1  =  (1); Fn  +1(i,  j)  =  Fn (i,  j), n ≥ 1, 0 ≤ i ≤ n − 1, 0 ≤  j ≤ n − 1; Fn  +1(n,  j)  =  0, n ≥ 1, 0 ≤  j ≤ n − 1; Fn  +1(0, n)  =  0, n ≥ 1; Fn  +1(i, n)  =  i Fn (i, n − 1) + Fn (i − 1, n − 1), n ≥ 1, 1 ≤ i ≤ n.

Starting with the central diagonal of the upper triangular matrix, the diagonals are given by:

• A000012 Stirling numbers of the second kind

  S (n + 0, n)

  .
  {1, 1, 1, 1, 1, 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 second kind

  S (n + 1, n)

  .
  {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, ...}
• A001296 Stirling numbers of the second kind

  S (n + 2, n)

  .
  {0, 1, 7, 25, 65, 140, 266, 462, 750, 1155, 1705, 2431, 3367, 4550, 6020, 7820, 9996, ...}
• A001297 Stirling numbers of the second kind

  S (n + 3, n)

  .
  {0, 1, 15, 90, 350, 1050, 2646, 5880, 11880, 22275, 39325, 66066, 106470, 165620, 249900, ...}
• A001298 Stirling numbers of the second kind

  S (n + 4, n)

  .
  {0, 1, 31, 301, 1701, 6951, 22827, 63987, 159027, 359502, 752752, 1479478, 2757118, ...}
• A112494 Stirling numbers of the second kind

  S (n + 5, n)

  .
  {0, 1, 63, 966, 7770, 42525, 179487, 627396, 1899612, 5135130, 12662650, 28936908, 62022324, ...}
• A144969 Stirling numbers of the second kind

  S (n + 6, n)

  .
  {0, 1, 127, 3025, 34105, 246730, 1323652, 5715424, 20912320, 67128490, 193754990, ...}

Stirling numbers

For

n   ≥   0

, we have

x (n)  :=  x n  =    n ∑   k  = 0    s (n, k) x k  =    n ∑   k  = 0    S1(n, k) x k  =    n ∑   k  = 0    (−1) n + k | S1(n, k) | x k,

and

x ( − n )  :=  [x n̅  ]  − 1  =  [   n ∑   k  = 0    (−1) n + k S1(n, k ) [x  − k ]  − 1]   − 1  =  [   n ∑   k  = 0    (−1) n + k S1(n, k ) x k]   − 1  =

[   n ∑   k  = 0    | S1(n, k ) | x k]   − 1  =  [   n ∑   k  = 0    [  nk  ]  x k]   − 1,

where

s (n, k )

or

S1(n, k )

are Stirling numbers of the first kind, and

[  nk  ]

are unsigned Stirling numbers of the first kind.

For

n   ≥   0

, we have

x n  =    n ∑   k  = 0    {  nk  } x (k )  =    n ∑   k  = 0    S  (n) k x (k )  =    n ∑   k  = 0    S (n, k ) x (k )  =    n ∑   k  = 0    S2(n, k ) x (k ),

and

x  − n  :=  [x n ]  − 1  =  [   n ∑   k  = 0    (−1) n + k S2(n, k ) [x ( − k ) ]  − 1]   − 1  =  [   n ∑   k  = 0    (−1) n + k S2(n, k ) x k̅]   − 1,

where

{  nk  }

,

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.

Stirling numbers of the first kind

We have

${\begin{aligned}x^{(0)}&=x^{\underline {0}}=+1\,x^{0},\\x^{(1)}&=x^{\underline {1}}=-0\,x^{0}+\ \ \ \ 1\,x^{1},\\x^{(2)}&=x^{\underline {2}}=+0\,x^{0}-\ \ \ \ 1\,x^{1}+\ \ \ \ 1\,x^{2},\\x^{(3)}&=x^{\underline {3}}=-0\,x^{0}+\ \ \ \ 2\,x^{1}-\ \ \ \ 3\,x^{2}+\ \ \ \ 1\,x^{3},\\x^{(4)}&=x^{\underline {4}}=+0\,x^{0}-\ \ \ \ 6\,x^{1}+\ \ 11\,x^{2}-\ \ \ \ 6\,x^{3}+\ \ 1\,x^{4},\\x^{(5)}&=x^{\underline {5}}=-0\,x^{0}+\ \ 24\,x^{1}-\ \ 50\,x^{2}+\ \ 35\,x^{3}-10\,x^{4}+\ \ 1\,x^{5},\\x^{(6)}&=x^{\underline {6}}=+0\,x^{0}-120\,x^{1}+274\,x^{2}-225\,x^{3}+85\,x^{4}-15\,x^{5}+1\,x^{6},\end{aligned}}$

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)

.

Triangle of Stirling numbers of the first kind

s (n, k )

n																										∑ n k  = 1 s (n, k )
0													1													1
1												0		1												1
2											0		−1		1											0
3										0		2		−3		1										0
4									0		−6		11		−6		1									0
5								0		24		−50		35		−10		1								0
6							0		−120		274		−225		85		−15		1							0
7						0		720		−1764		1624		−735		175		−21		1						0
8					0		−5040		13068		−13132		6769		−1960		322		−28		1					0
9				0		40320		−109584		118124		−67284		22449		−4536		546		−36		1				0
10			0		−362880		1026576		−1172700		723680		−269325		63273		−9450		870		−45		1			0
			k = 0		1		2		3		4		5		6		7		8		9		10

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

n ∑   k  = 0    S1(n, k)  =  0 n (2)  =  0 n  (n − 1), n ≥ 0.

Unsigned Stirling numbers of the first kind

We have

${\begin{aligned}x^{(-0)}&=\left[x^{\overline {0}}\right]^{-1}=\left[1\,x^{0}\right]^{-1},\\x^{(-1)}&=\left[x^{\overline {1}}\right]^{-1}=\left[0\,x^{0}+\ \ \ \ 1\,x^{1}\right]^{-1},\\x^{(-2)}&=\left[x^{\overline {2}}\right]^{-1}=\left[0\,x^{0}+\ \ \ \ 1\,x^{1}+\ \ \ \ 1\,x^{2}\right]^{-1},\\x^{(-3)}&=\left[x^{\overline {3}}\right]^{-1}=\left[0\,x^{0}+\ \ \ \ 2\,x^{1}+\ \ \ \ 3\,x^{2}+\ \ \ \ 1\,x^{3}\right]^{-1},\\x^{(-4)}&=\left[x^{\overline {4}}\right]^{-1}=\left[0\,x^{0}+\ \ \ \ 6\,x^{1}+\ \ 11\,x^{2}+\ \ \ \ 6\,x^{3}+\ \ 1\,x^{4}\right]^{-1},\\x^{(-5)}&=\left[x^{\overline {5}}\right]^{-1}=\left[0\,x^{0}+\ \ 24\,x^{1}+\ \ 50\,x^{2}+\ \ 35\,x^{3}+10\,x^{4}+\ \ 1\,x^{5}\right]^{-1},\\x^{(-6)}&=\left[x^{\overline {6}}\right]^{-1}=\left[0\,x^{0}+120\,x^{1}+274\,x^{2}+225\,x^{3}+85\,x^{4}+15\,x^{5}+1\,x^{6}\right]^{-1}.\end{aligned}}$

involving the triangle of unsigned Stirling numbers of the first kind, also called the factorial triangle, 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,

| S (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

n ∑   k  = 0    | S1(n, k) |  =  | S1(n + 1, 1) |  =  n!, n ≥ 0.

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

2 k

.
A000012 Stirling numbers of the first kind:

s (n, n), n   ≥   0.

(The simplest sequence of positive numbers: the all 1’s sequence.) [

a0(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) = (  n + 12  )  = n (n + 1) / 2 = 0 + 1 + 2 + ... + n

.) [

a1(n) = (n + n 2 ) / 2 = n  (1 + n) / 2

] G.f.:

1 / (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

. [

a2(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. [

a3(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

. [

a4(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:

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

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

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

involving the triangle of Stirling numbers of the second kind.

Triangle of Stirling numbers of the second kind

S (n, k )

n																										∑ n k  = 1 S (n, k )
0													1													1
1												0		1												1
2											0		1		1											2
3										0		1		3		1										5
4									0		1		7		6		1									15
5								0		1		15		25		10		1								52
6							0		1		31		90		65		15		1							203
7						0		1		63		301		350		140		21		1						877
8					0		1		127		966		1701		1050		266		28		1					4140
9				0		1		255		3025		7770		6951		2646		462		36		1				21147
10			0		1		511		9330		34105		42525		22827		5880		750		45		1			115975
			k = 0		1		2		3		4		5		6		7		8		9		10

A048993 Triangle of Stirling numbers of second kind,

S 2(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,

S 2(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,

S 2(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

n ∑   k  = 0    S2(n, k )  =  Bn , n ≥ 0,

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

x n  =    n ∑   k  = 0    S2(n, k ) x (k), n ≥ 0,

one obtains the Bell polynomials

Bn(x)  :=    n ∑   k  = 0    S2(n, k ) x k, n ≥ 0,

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

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

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

n ∑   k  = 0    (−1) n + k S2(n, k )  =  (−1) n   n ∑   k  = 0    (−1) k S2(n, k )  =  (−1) n B̃n , n ≥ 0,

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, ...}

Lah numbers

The Lah numbers (also called “Stirling numbers of the third kind”), discovered by Ivo Lah in 1954, are defined as

L (n, k )  :=    n ∑   m  = k    | s (n, m) | S (m, k ) ,

where

| s (n, k ) |

and

S (n, k )

are unsigned Stirling numbers of the first kind and Stirling numbers of the second kind, respectively.

The Lah numbers express the “rising factorials” in terms of the “falling factorials” and vice versa, i.e.

x n̅  =    n ∑   k  = 1    L (n, k ) x k ,

x n  =    n ∑   k  = 1    (−1) n − k L (n, k ) x k.

Triangle of Lah numbers

L (n, k )

n																										∑ n k  = 1 L (n, k )
1													−1													−1
2												2		1												3
3											−6		−6		−1											−13
4										24		36		12		1										73
5									−120		−240		−120		−20		−1									−501
6								720		1800		1200		300		30		1								4051
7							−5040		−15120		−12600		−4200		−630		−42		−1							−37633
8						40320		141120		141120		58800		11760		1176		56		1						394353
9					−362880		−1451520		−1693440		−846720		−211680		−28224		−2016		−72		−1					−4596553
10				3628800		16329600		21772800		12700800		3810240		635040		60480		3240		90		1				58941091
11			−39916800		−199584000		−299376000		−199584000		−69854400		−13970880		−1663200		−118800		−4950		−110		−1			−824073141
			k = 1		2		3		4		5		6		7		8		9		10		11

A008297 Triangle of Lah numbers (concatenated rows).

{−1, 2, 1, − 6, − 6, −1, 24, 36, 12, 1, −120, −240, −120, −20, −1, 720, 1800, 1200, 300, 30, 1, −5040, −15120, −12600, − 4200, − 630, − 42, −1, 40320, 141120, 141120, 58800, 11760, ...}

Row sums of triangle of Lah numbers

A293125 Expansion of e.g.f.:

exp ( − x / (1 + x))

: for

n   ≥   1

, gives row sums of Lah triangle.

{−1, 3, −13, 73, −501, 4051, −37633, 394353, − 4596553, 58941091, −824073141, 12470162233, −202976401213, 3535017524403, − 65573803186921, 1290434218669921, ...}

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
     

(x + 1) n  =    n ∑   k  = 0    ( n k ) x k, n ≥ 0,

we get

     

Pn (x)  :=    n ∑   k  = 0    ( n k ) x (k ), n ≥ 0,

which is given by the recurrence relation

     

P0 (x)  =  1; P1(x)  =  1 + x; Pn (x)  =  (x − (n − 2)) Pn  − 1(x) + (n − 1) Pn  − 2(x), n ≥ 2.

Examples:

${\begin{aligned}P_{0}(x)&=\sum _{k=0}^{0}{\binom {0}{k}}\,x^{(k)}=1\,x^{0},\\P_{1}(x)&=\sum _{k=0}^{1}{\binom {1}{k}}\,x^{(k)}=1\,x^{0}+\ \ 1\,x^{1},\\P_{2}(x)&=\sum _{k=0}^{2}{\binom {2}{k}}\,x^{(k)}=1\,x^{0}+\ \ 1\,x^{1}+\ \ 1\,x^{2},\\P_{3}(x)&=\sum _{k=0}^{3}{\binom {3}{k}}\,x^{(k)}=1\,x^{0}+\ \ 2\,x^{1}+\ \ 0\,x^{2}+\ \ 1\,x^{3},\\P_{4}(x)&=\sum _{k=0}^{4}{\binom {4}{k}}\,x^{(k)}=1\,x^{0}+\ \ 0\,x^{1}+\ \ 5\,x^{2}-\ \ 2\,x^{3}+\ \ 1\,x^{4},\\P_{5}(x)&=\sum _{k=0}^{5}{\binom {5}{k}}\,x^{(k)}=1\,x^{0}+\ \ 9\,x^{1}-15\,x^{2}+15\,x^{3}-\ \ 5\,x^{4}+1\,x^{5},\\P_{6}(x)&=\sum _{k=0}^{6}{\binom {6}{k}}\,x^{(k)}=1\,x^{0}-35\,x^{1}+94\,x^{2}-85\,x^{3}+40\,x^{4}-9\,x^{5}+1\,x^{6}.\end{aligned}}$

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, 1, 1, 1, 2, 0, 1, 1, 0, 5, −2, 1, 1, 9, −15, 15, −5, 1, 1, −35, 94, −85, 40, − 9, 1, ...}

Related to (although the signs differ):

A269953 Triangle read by rows,

T  (n, k ) = ∑ n j  = 0 (   −  j  −  1 −  n  −  1  )  S1(  j, k )

where

S1

are the Stirling cycle numbers A132393, for

n   ≥   0

and

0   ≤   k   ≤   n

.

{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, −1, 230, 595, 609, 315, 91, 14, 1, 1, 1624, 4458, 4844, 2779, 924, 182, 20, 1, −1, ...}

Triangle row sums

It appears that the row sums yield:

Pn (1)  =  n + 1, n ≥ 0.

Triangle columns

It appears that we might have for column 1:

T  (n, 1) = ( − 1) [n belongs to some set] A002741 (n), n   ≥   0,

where [⋅] is the Iverson bracket.

{0, 1, 1, 2, 0, 9, −35, ...}

A002741 Logarithmic numbers: expansion of

−  log (1  −  x) exp ( −  x), n   ≥   0

.

{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
     

(x − 1) n  =    n ∑   k  = 0    ( n k ) (−1) n  − k x k, n ≥ 0,

we get

     

Pn (x)  :=    n ∑   k  = 0    ( n k ) (−1) n  − k x (k ), n ≥ 0,

which is given by the recurrence relation

     

P0 (x)  =  1; P1(x)  =  − 1 + x; Pn (x)  =  (x − n) Pn  − 1(x) − (n − 1) Pn  − 2(x), n ≥ 2.

Examples:

${\begin{aligned}P_{0}(x)&=\sum _{k=0}^{0}{\binom {0}{k}}\,(-1)^{n-k}\,x^{(k)}=+1\,x^{0},\\P_{1}(x)&=\sum _{k=0}^{1}{\binom {1}{k}}\,(-1)^{n-k}\,x^{(k)}=-1\,x^{0}+\ \ \ \ 1\,x^{1},\\P_{2}(x)&=\sum _{k=0}^{2}{\binom {2}{k}}\,(-1)^{n-k}\,x^{(k)}=+1\,x^{0}-\ \ \ \ 3\,x^{1}+\ \ \ \ 1\,x^{2},\\P_{3}(x)&=\sum _{k=0}^{3}{\binom {3}{k}}\,(-1)^{n-k}\,x^{(k)}=-1\,x^{0}+\ \ \ \ 8\,x^{1}-\ \ \ \ 6\,x^{2}+\ \ \ \ 1\,x^{3},\\P_{4}(x)&=\sum _{k=0}^{4}{\binom {4}{k}}\,(-1)^{n-k}\,x^{(k)}=+1\,x^{0}-\ \ 24\,x^{1}+\ \ 29\,x^{2}-\ \ 10\,x^{3}+\ \ \ \ 1\,x^{4},\\P_{5}(x)&=\sum _{k=0}^{5}{\binom {5}{k}}\,(-1)^{n-k}\,x^{(k)}=-1\,x^{0}+\ \ 89\,x^{1}-145\,x^{2}+\ \ 75\,x^{3}-\ \ 15\,x^{4}+\ \ 1\,x^{5},\\P_{6}(x)&=\sum _{k=0}^{6}{\binom {6}{k}}\,(-1)^{n-k}\,x^{(k)}=+1\,x^{0}-415\,x^{1}+814\,x^{2}-545\,x^{3}+160\,x^{4}-21\,x^{5}+1\,x^{6}.\end{aligned}}$

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, −1, 89, −145, 75, −15, 1, 1, − 415, 814, −545, 160, −21, 1, ...}

This absolute values are given by:

A094816 Triangle read by rows:

T  (n, k ), 0   ≤   k   ≤   n,

equals 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:

Pn (1)  =  ( − 1) n +1 (n  −  1), n   ≥   0.

{1, 0, −1, 2, −3, 4, −5, ...}

Triangle columns

It appears that we have for column 1:

T  (n, 1)  =  ( − 1) n +1 A002104 (n), n   ≥   0.

{0, 1, −3, 8, −24, 89, − 415, ...}

A002104 Logarithmic numbers: expansion of

−  log (1  −  x) exp (x), n   ≥   0

.

{0, 1, 3, 8, 24, 89, 415, 2372, 16072, 125673, 1112083, 10976184, 119481296, 1421542641, 18348340127, 255323504932, 3809950977008, ...}

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.

External links

• Weisstein, Eric W., Bell Number, from MathWorld—A Wolfram Web Resource.
• Weisstein, Eric W., Complementary Bell Number, from MathWorld—A Wolfram Web Resource.
• Weisstein, Eric W., Bell Polynomial, from MathWorld—A Wolfram Web Resource.