Connections with combinatorial numbers Bernoulli number

1 connections combinatorial numbers

1.1 connection worpitzky numbers
1.2 connection stirling numbers of second kind
1.3 connection stirling numbers of first kind
1.4 connection pascal’s triangle
1.5 connection eulerian numbers

connections combinatorial numbers

the connection of bernoulli number various kinds of combinatorial numbers based on classical theory of finite differences , on combinatorial interpretation of bernoulli numbers instance of fundamental combinatorial principle, inclusion–exclusion principle.

connection worpitzky numbers

the definition proceed developed julius worpitzky in 1883. besides elementary arithmetic factorial function n! , power function k employed. signless worpitzky numbers defined as

w

n
,
k


=

∑

v
=
0


k


(
−
1

)

v
+
k


(
v
+
1

)

n





k
!


v
!
(
k
−
v
)
!



.


{\displaystyle w_{n,k}=\sum _{v=0}^{k}(-1)^{v+k}(v+1)^{n}{\frac {k!}{v!(k-v)!}}.}

they can expressed through stirling numbers of second kind

w

n
,
k


=
k
!

{



n
+
1


k
+
1



}

.


{\displaystyle w_{n,k}=k!\left\{{n+1 \atop k+1}\right\}.}

a bernoulli number introduced inclusion–exclusion sum of worpitzky numbers weighted harmonic sequence 1, 1/2, 1/3, …

b

n


=

∑

k
=
0


n


(
−
1

)

k





w

n
,
k



k
+
1



 
=
 

∑

k
=
0


n




1

k
+
1




∑

v
=
0


k


(
−
1

)

v


(
v
+
1

)

n





(


k
v


)



 
.


{\displaystyle b_{n}=\sum _{k=0}^{n}(-1)^{k}{\frac {w_{n,k}}{k+1}}\ =\ \sum _{k=0}^{n}{\frac {1}{k+1}}\sum _{v=0}^{k}(-1)^{v}(v+1)^{n}{k \choose v}\ .}




b0 = 1
b1 = 1 − 1/2
b2 = 1 − 3/2 + 2/3
b3 = 1 − 7/2 + 12/3 − 6/4
b4 = 1 − 15/2 + 50/3 − 60/4 + 24/5
b5 = 1 − 31/2 + 180/3 − 390/4 + 360/5 − 120/6
b6 = 1 − 63/2 + 602/3 − 2100/4 + 3360/5 − 2520/6 + 720/7

this representation has b+

1 = +1/2.

consider sequence sn, n ≥ 0. worpitzky s numbers  a028246,  a163626 applied s0, s0, s1, s0, s1, s2, s0, s1, s2, s3, … identical akiyama–tanigawa transform applied sn (see connection stirling numbers of first kind). can seen via table:

the first row represents s0, s1, s2, s3, s4.

hence second fractional euler numbers  a198631 (n) /  a006519 (n + 1):

e0 = 1
e1 = 1 − 1/2
e2 = 1 − 3/2 + 2/4
e3 = 1 − 7/2 + 12/4 − 6/8
e4 = 1 − 15/2 + 50/4 − 60/8 + 24/16
e5 = 1 − 31/2 + 180/4 − 390/8 + 360/16 − 120/32
e6 = 1 − 63/2 + 602/4 − 2100/8 + 3360/16 − 2520/32 + 720/64

a second formula representing bernoulli numbers worpitzky numbers n ≥ 1

b

n


=


n


2

n
+
1


−
2




∑

k
=
0


n
−
1


(
−
2

)

−
k




w

n
−
1
,
k


.


{\displaystyle b_{n}={\frac {n}{2^{n+1}-2}}\sum _{k=0}^{n-1}(-2)^{-k}\,w_{n-1,k}.}

the simplified second worpitzky s representation of second bernoulli numbers is:

 a164555 (n + 1) /  a027642(n + 1) = n + 1/2 − 2 ×  a198631(n) /  a006519(n + 1)

which links second bernoulli numbers second fractional euler numbers. beginning is:

1/2, 1/6, 0, −1/30, 0, 1/42, … = (1/2, 1/3, 3/14, 2/15, 5/62, 1/21, …) × (1, 1/2, 0, −1/4, 0, 1/2, …)

the numerators of first parentheses  a111701 (see connection stirling numbers of first kind).

connection stirling numbers of second kind

if s(k,m) denotes stirling numbers of second kind 1 has:

j

k


=

∑

m
=
0


k




j


m
_




s
(
k
,
m
)


{\displaystyle j^{k}=\sum _{m=0}^{k}{j^{\underline {m}}}s(k,m)}

where j denotes falling factorial.

if 1 defines bernoulli polynomials bk(j) as:

b

k


(
j
)
=
k

∑

m
=
0


k
−
1





(


j

m
+
1



)



s
(
k
−
1
,
m
)
m
!
+

b

k




{\displaystyle b_{k}(j)=k\sum _{m=0}^{k-1}{\binom {j}{m+1}}s(k-1,m)m!+b_{k}}

where bk k = 0, 1, 2,… bernoulli numbers.

then after following property of binomial coefficient:

(


j
m


)



=



(



j
+
1


m
+
1



)



−



(


j

m
+
1



)





{\displaystyle {\binom {j}{m}}={\binom {j+1}{m+1}}-{\binom {j}{m+1}}}

one has,

j

k


=




b

k
+
1


(
j
+
1
)
−

b

k
+
1


(
j
)


k
+
1



.


{\displaystyle j^{k}={\frac {b_{k+1}(j+1)-b_{k+1}(j)}{k+1}}.}

one has following bernoulli polynomials,

b

k


(
j
)
=

∑

n
=
0


k





(


k
n


)




b

n



j

k
−
n


.


{\displaystyle b_{k}(j)=\sum _{n=0}^{k}{\binom {k}{n}}b_{n}j^{k-n}.}

the coefficient of j in (j

m + 1) (−1)/m + 1.

comparing coefficient of j in 2 expressions of bernoulli polynomials, 1 has:

b

k


=

∑

m
=
0


k


(
−
1

)

m





m
!


m
+
1



s
(
k
,
m
)


{\displaystyle b_{k}=\sum _{m=0}^{k}(-1)^{m}{\frac {m!}{m+1}}s(k,m)}

(resulting in b1 = +1/2) explicit formula bernoulli numbers , can used prove von-staudt clausen theorem.

connection stirling numbers of first kind

the 2 main formulas relating unsigned stirling numbers of first kind [n

m] bernoulli numbers (with b1 = +1/2) are

1

m
!




∑

k
=
0


m


(
−
1

)

k



[



m
+
1


k
+
1



]


b

k


=


1

m
+
1



,


{\displaystyle {\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{m+1 \atop k+1}\right]b_{k}={\frac {1}{m+1}},}

and inversion of sum (for n ≥ 0, m ≥ 0)

1

m
!




∑

k
=
0


m


(
−
1

)

k



[



m
+
1


k
+
1



]


b

n
+
k


=

a

n
,
m


.


{\displaystyle {\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{m+1 \atop k+1}\right]b_{n+k}=a_{n,m}.}

here number an,m rational akiyama–tanigawa numbers, first few of displayed in following table.

the akiyama–tanigawa numbers satisfy simple recurrence relation can exploited iteratively compute bernoulli numbers. leads algorithm shown in section algorithmic description above. see  a051714/ a051715.

an autosequence sequence has inverse binomial transform equal signed sequence. if main diagonal zeroes =  a000004, autosequence of first kind. example:  a000045, fibonacci numbers. if main diagonal first upper diagonal multiplied 2, of second kind. example:  a164555/ a027642, second bernoulli numbers (see  a190339). akiyama–tanigawa transform applied 2 = 1/ a000079 leads  a198631 (n) /  a06519 (n + 1). hence:

see  a209308 ,  a227577.  a198631 (n) /  a006519 (n + 1) second (fractional) euler numbers , autosequence of second kind.

( a164555 (n + 2)/ a027642 (n + 2) = 1/6, 0, −1/30, 0, 1/42, …) × ( 2 − 2/n + 2 = 3, 14/3, 15/2, 62/5, 21, …) =  a198631 (n + 1)/ a006519 (n + 2) = 1/2, 0, −1/4, 0, 1/2, ….

also valuable  a027641 /  a027642 (see connection worpitzky numbers).

connection pascal’s triangle

there formulas connecting pascal s triangle bernoulli numbers

b

n


+


=




|


a

n



|



(
n
+
1
)
!



 
 
 


{\displaystyle b_{n}^{+}={\frac {|a_{n}|}{(n+1)!}}~~~}

where




|


a

n



|



{\displaystyle |a_{n}|}

determinant of n-by-n square matrix part of pascal’s triangle elements are:




a

i
,
k


=


{



0



if 

k
>
1
+
i







(



i
+
1


k
−
1



)






otherwise









{\displaystyle a_{i,k}={\begin{cases}0&{\text{if }}k>1+i\\{i+1 \choose k-1}&{\text{otherwise}}\end{cases}}}

example:

b

6


+


=



det


(



1


2


0


0


0


0




1


3


3


0


0


0




1


4


6


4


0


0




1


5


10


10


5


0




1


6


15


20


15


6




1


7


21


35


35


21



)




7
!



=


120
5040


=


1
42




{\displaystyle b_{6}^{+}={\frac {\det {\begin{pmatrix}1&2&0&0&0&0\\1&3&3&0&0&0\\1&4&6&4&0&0\\1&5&10&10&5&0\\1&6&15&20&15&6\\1&7&21&35&35&21\end{pmatrix}}}{7!}}={\frac {120}{5040}}={\frac {1}{42}}}



connection eulerian numbers

there formulas connecting eulerian numbers ⟨n

m⟩ bernoulli numbers:

∑

m
=
0


n


(
−
1

)

m



⟨


n
m


⟩




=

2

n
+
1


(

2

n
+
1


−
1
)



b

n
+
1



n
+
1



,





∑

m
=
0


n


(
−
1

)

m



⟨


n
m


⟩





(


n
m


)




−
1





=
(
n
+
1
)

b

n


.






{\displaystyle {\begin{aligned}\sum _{m=0}^{n}(-1)^{m}\left\langle {n \atop m}\right\rangle &=2^{n+1}(2^{n+1}-1){\frac {b_{n+1}}{n+1}},\\\sum _{m=0}^{n}(-1)^{m}\left\langle {n \atop m}\right\rangle {\binom {n}{m}}^{-1}&=(n+1)b_{n}.\end{aligned}}}

both formulae valid n ≥ 0 if b1 set 1/2. if b1 set −1/2 valid n ≥ 1 , n ≥ 2 respectively.