Stirling Numbers of the Second Kind: new formulae ...

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where the symbol (), introduced by John Riordan, stands for th the Bell .... R. Sitgreaves, Some Properties of Stirling Numbers of the Second Kind, Fibonacci ...
Stirling Numbers of the Second Kind: new formulae via unorthodox method Amrik Singh Nimbran Abstract: This paper discusses Stirling numbers. The author gives a few results of his own and presents some new formulae relating to Stirling numbers of the second kind derived through an unorthodox method. His method enables one to derive rest of the closed formulae. AMS Subject Classification: 11B73. Key words and phrases: Bell numbers, Stirling numbers of the first and second kind, arithmetical progression. 1.

The Scottish mathematician James Stirling (1692-1770) is known for his work on factorials, particularly the formula: ! ∼ √2

( / ) which appears as Example 2 to Proposition 28 of his

Methodus Differentialis sive Tractatus de Summatione et Interpolatione Serierum Infinitarum (1730). He prepared tables of the coefficients for conversion of factorials into powers and showed that the columns of the tables gave the coefficients for the inverse expressions, of powers in factorials. These coefficients were subsequently named as Stirling numbers of the first and second kind. Stirling numbers of the first kind, denoted by ( , ) with small , appear as the coefficients of

in the polynomial ( ) = ( − 1)( − 2) … ( −

( ) =∑

( , )

+ 1):

.

(1)

Thus (1) relates the factorial power with the normal power of . We further find in [2]: ( , )=0 for > ; (1, )=(−1)

( − 1)!; ( , − 1)=–

, ( , )=1; and

( + 1, )= ( , − 1) − ( , ). The absolute value | ( , )|=(−1)

( , ), denoted by

, stands for the ways of permuting

numbers into cycles. They satisfy the recurrence relation: = ( − 1)

−1

+

−1 , ∈ ℕ ; and ∑ −1

=

+1 . +1

The Stirling numbers of the second kind, denoted by S( , ) with capital S or the symbol is the number of ways of partitioning an –element set into .A

,

nonempty disjoint subsets, called

of a set is defined to be a mutually disjoint class of its non-empty sub-sets of

whose union is the whole set itself. We have

=0 for > ;

We have this explicit formula for S( , ): 1

1

=

=1;

=

−1 + −1

−1

.

(

( , )=∑

)

(

)

!

.

(2)

This identity involves Stirling numbers of the second kind and the binomial coefficients: −

=∑



.

The next formula, inverse to (1), relates the normal power with the factorial power of =∑

( , )( ) .

:

(3)

This result is given in [2]: ∑

= (

!

− 1) .

!

This relation is quite well known: ∑

( , )= ( ),

(4)

where the symbol

( ), introduced by John Riordan, stands for

the Bell

th

number after Eric Temple Bell (1883-1960), who studied them in mid 1930s, though Ramanujan is reported to have derived several of their properties in his Notebooks almost 25-30 years earlier.[1] ( ) is the number of ways of partitioning an –element set into disjoint, non-empty subsets. Bell

( )=∑



. I have constructed this

(



)!



( )

= ∑

numbers are defined by the exponential generating function: ℎ



!

which yields first

ten Bell numbers: 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975: 1 1 1 1 1 1 1 1 1 1

31

127 255

511

9330

6

90

966

10 65

350

1701 7770

34105

1

25

301

3025

1

7 15

63

1 3

1 15

140

1050

266

6951 2646

42525

1 21

1 28

462

22827 5880

1 36 750

1 45

1

I have discovered some results involving S( , ) and the prime numbers. 2

and

2. – prime>2 ⇒ S( , )≡0 (mod ), 1< < .

Theorem 1:

This theorem is known already and can be proved by using formula (2) along with Fermat’s little theorem, that is, if g.c.d.(a, )=1, then

≡ (mod ).

I discovered theorem that completely identifies all such numbers divisible by odd primes: > 3 ⇒ S( +

Theorem 2: – 1, ℎ

+ 2, . . . , − 1





≥ 2,

=







− 1, ) ≡ 0 (

),

= 1, 2, 3, . . . , − 1; = 1,

=

+

+ 1, + 2, . . . , .





. ℎ





(

)(

)

.

We deduce these particular formulae from (2): S( , 2)=

.

!

S( , 3)=

.

!

.

S( , 4)=

.

! .

S( , 5)=

.

.

!

S( , 6)=



.



.

.

.

! .

S( , 7)=

.

.

.

.

!

The above formulae give





have discovered corresponding (

S( , − 1) =

=

S( , − 2) =

(

)

S( , − 3) =

(

)(

)

(

)( (

=

)( ) (

)



)(

)(









ℎ .I

:

) (

(6) )

.

(7) (

= (



.

)

(

=

)

(

)(

3

)(

) (

(

S( , − 6)= =





(5)

S( , − 4) = S( , − 5) =





.

= )





) (

)(

)(

)(

)(

)(

)(

)

)(

)

)(

.

(8)

.

)(

(9)

)

.

(10)

We find that ( , of ( ,





+ ) is always divisible by

. This never happens in the case

). However, I do not see discern any general pattern in the formulae.

While going through literature on the subject, we see that (5) is commonly known [2][3] while formulae (6) & (7) are given in [5]. However, formulae (8), (9) and (10) do not figure anywhere else, to the best of my knowledge. I have discovered these formulae by means of an unorthodox method which enables one to discover all the closed formulae. I found empirically: Theorem 3: Let

( )

( )= +

+1 − +1

+

be defined as the first order difference of the ( + 1)th

diagonal from right to left of the Stirling triangle of the second kind. Let

( )

, denoting the

differences of the consecutive differences of the first order, be called the second order difference. Let

( )

symbolize the differences of the second order differences and be termed as the third order

difference, and so on. Then, this chain of differences eventually, after (2k-1) steps, lead to an arithmetical progression with first term=2{5.7...(2k+1)} and common difference=1.3.5... (2k-1). For example take S( , − 2)=1, 7, 25, 65, 140, 266, 462, 750, . . . The first sequence of differences is: 6, 18, 40, 75, 126, 196, 288, . . . The second order sequence of differences is: 12, 22, 35, 51, 70, 92, . . . The third order sequence of differences is: 10, 13, 16, 19, 22, . . . and we thus reach an A.P. with a=10, d=3. Theorem 3 forms the basis of my method. I shall now illustrate my method by an example. Suppose we want to derive the formula for the seventh column from right to left of the Stirling numbers of the second kind. We take the numbers 1, 127, 3025, 34105, 246730, 1323652, 5715424, 20912320, . . . Since it is likely that the formula will have multiples of the binomial coefficients: 1*

, 127*

,

*

,

*

,

*

,

*

,

*

,

*

, . . ., so

instead of the original numbers, we take the multiples of the binomial coefficients with same denominator, the l.c.m. of all denominators. It is found to be 72 here. We now compute successive differences till we find common difference. Note that the stage is reached earlier than as given by theorem 3, because we have not taken the original numbers. This accelerated the process. 72 ( ) ( )

1143 1071

6050 4907

3836

20463 14413

9506

( )

5670

( )

=

33369 18956

9450 3780 d=

53832 66500 33131 14175

4725 945

120332

4

119476 52976

19845 5670

945

239808

6615 945

198912 79436

26460

438720

So we have an arithmetical progression of numbers 3780, 4725, 5670, 6615, and so on. Its sum is given by the formula:

= {2*3780 + (n-1)945}. Since it is preceded by the term 5670 on

the earlier stage of differences, so we have the sum up to + stage

(

+

with the formula:

). We have to begin with (n-1) to get the first number at n=1. So we reach

by 5670 +

(

+ 5 − 6). ∑(

We go back to earlier stage by the formula =3836 + 5670 + need

to

change

n

into

=3836+5670 ( − 1) + reach stage

{

(

(n-1)

) (

{

(

=70+1001n+



(

)

+

+6 (

. formula=1071+1001(n-1)+

+

+5

get )

3836

72+70(n-1)+1001 (

)

= (

)

{(

(

+

+

So

{

+

we +

have }. We

+ 11 }] at n-1 and add 1071 to the result to reach to ) + 6



(

)

+ (

{ )

+

(

(

) (

)

+ 11

+ (6



)

+ 30 −

(

)

+ 30

+

(

∑( )

+6

− 390

+

We repeatedly summed above an A.P. with

=3780,

∑ +( − ) =

(

)

th

}

)

(

+ 7 + 18)

.

+6 (

(

)

+ 11 + 11

− 18 )] at n-1: (

) (

)

+

)



+ 324 )

). +

general formula for the

(

). This takes us to stage

(

)

Hence, the required formula: ( + , )=



+

{( − 1) + +4(2 − 1) + 22}=70+1001n+

}=2+70n+1001 +

n=1.

− 6 − 6}=

We finally reach the multiples by: 72+∑[70 + 1001n +

18

at

by this.

We compute ∑[1001 +

=

)

to

+ 5 − 6). We again

(





=945. I conclude with a useful

order sum of an A.P. with first term and common difference : )(

)…(

){ ( (

) (

)!

*****

5

) }

.

.

(11)

References: 1. 2.

3.

4. 5.

B.C. Berndt, Ramanujan’s Notebooks: Part I, Springer-Verlag, New York, 1985, pp. 11, 44, 85. K. Goldberg, M. Newman & E. Haynsworth, in Milton Abramowitz and Irene A. Stegun (ed.), Handbook of Mathematical Functions, U. S. Department of Commerce, National Bureau of Standards, Applied Mathematics (1964), chapter 24 , pp. 824-825. http://mintaka.sdsu.edu/faculty/wfw/ABRAMOWITZ-STEGUN/page_824.htm Shy-Der Lin, Shih-Tong Tu & H. M. Srivastava, Some generating functions involving the Stirling numbers of the second kind, Rend. Sem. Univ. Pol. Torino, 59, (2001), 199-224, p.199. http://seminariomatematico.dm.unito.it/rendiconti/59-3/199.pdf R. Sitgreaves, Some Properties of Stirling Numbers of the Second Kind, Fibonacci Quarterly, Vol. 8, No.2 (March 1970), pp. 172-181. http://www.fq.math.ca/Scanned/8-2/sitgreaves.pdf J.-M. Sixdeniers; K. A. Penson & A. I. Solomon, Extended Bell and Stirling Numbers From Hypergeometric Exponentiation, Journal of Integer Sequences, Vol. 4 (2001), issue 1, Article 01.1.4, formulae (31) & (33), p.5. http://www.cs.uwaterloo.ca/journals/JIS/VOL4/SIXDENIERS/bell.pdf

Amrik Singh Nimbran, IPS ADG of Police, CID, Bihar, 6, Polo Road, Patna 800002 Email address: [email protected]

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