# Generalized Bell Numbers and Peirce Matrix via Pascal Matrix

Authors: Eunmi Choi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

With the Stirling matrixS and the Pascal matrix T, we show that TkS (k≥0) satisfies a type of generalized Stirling recurrence. Then, by expressing the sum of components of each row of TkS as k-Bell number, we investigate properties of k-Bell numbers as well as k-Peirce matrix.


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## Body

## 1. Introduction

The Stirling numbersm,n of the second kind is the number of ways to partition an m elements set into n nonempty subsets for any m,n≥0. sm,n can be expressed by n!sm,n=∑i=0n(-1)ini(n-i)m, and the Stirling matrix S=[sm,n](m,n≥0) satisfies the Stirling recurrence sm,n=sm-1,n-1+nsm-1,n [1]. Some researches including [2–4] were devoted to investigating the Stirling matrix S with the Pascal matrix T and binomial expressions. On the other hand, the sum Bm=∑k=0msm,k of numbers of the mth row of S is called a Bell number, so {Bm∣m≥0}={1,1,2,5,15,⋯} counts the number of partitions of an m elements set. One effective way for generating Bell numbers is to use the Peirce matrix (often called Bell table or Aitken’s array), on which Bell numbers appear along both borders [5–7].A main purpose of work is to studyTkS for k≥1. We show that TkS satisfies a kind of generalized Stirling recurrence, and then by expressing the sum of components of each row of TkS as a k-Bell number we investigate k-Bell numbers as well as k-Peirce matrix. We discuss recurrence rules of a k-Peirce matrix and then interrelationships between each k-Peirce matrices.


## 2. Generalizedk-Stirling Matrix and k-Bell Numbers

Throughout the work, we writeT for the Pascal matrix and rm for the mth row of T. And Mtr means a transpose matrix of M. Let the Stirling matrix be (1)S=101011013101761⋯andS∗=111131176111525101⋯.ClearlyS111⋯=B0B1B2⋯.Theorem 1.
We have the following:(1)
S123⋯=B1B2B3⋯=S∗111⋯, S∗123⋯=B2-B1B3-B2B4-B3⋯ and S∗234⋯=B2B3B4⋯.(2)
TB0B1B2⋯=B1B2B3⋯ and TB1B2B3⋯=S∗123⋯.Proof.
The first two identities in (1) are easy to observe. And(2)S∗234⋯=S∗111⋯+S∗123⋯=B1B2B3⋯+B2-B1B3-B2B4-B3⋯=B2B3B4⋯.
Clearly111121112=125=B1B2B3. And Bm=∑i=0m-1m-1iBi in [8] proves rm(B0,⋯,Bm-1)tr=Bm for m≥1, so we have TB0B1B2⋯=B1B2B3⋯. Moreover since(3)rm+1=m-10,⋯,m-1m-1,0+0,m-10,⋯,m-1m-1=rm;0+0;rm,it follows that(4)Bm+1=rm+1B0,B1,⋯,Bmtr=rm;0B0,B1,⋯,Bmtr+0;rmB0,B1,⋯,Bmtr=rmB0,⋯,Bm-1tr+rmB1,⋯,Bmtr=Bm+rmB1,⋯,Bmtr.So rm(B1,⋯,Bm)tr=Bm+1-Bm and TB1B2B3⋯=B2-B1B3-B2B4-B3⋯=S∗123⋯.Hence the next corollary follows immediately.Corollary 2 (TS=S∗).
As a generalization ofS∗=TS, we shall consider TkS for k>0. Write S(k)=TkS=[sm,n(k)] for m,n≥0, and let S(0)=S. Clearly S(1)=S∗.Theorem 3.
S ( k ) satisfies the recurrence sm+1,n+1(k)=sm,n(k)+(n+k)sm,n+1(k).Proof.
S ( 1 ) = [ s m , n ( 1 ) ] = S ∗ in Corollary 2 shows sm+1,n+1(1)=sm,n(1)+(n+1)sm,n+1(1). And S(2)=12145181991166555141, S(3)=13197127371218117597181 give identities 55=19+(4)9 and 97=37+(5)12, and it is easy to observe (5)sm+1,n+1k=sm,nk+n+ksm,n+1kfor 1≤n,m≤5; k=2,3. Now we assume sm+1,n+1(k)=sm,n(k)+(n+k)sm,n+1(k) for some k, m. Since (6)sm,nk+1=Sk+1=Tk+1S=TTkS=TSk,any (m,n)th component sm,n(k+1) comes from the mth row rm of T and the nth column of S(k). So(7)sm,nk+1=m-10,⋯,m-1n-2︸n-1,m-1n-1,m-1n,⋯,m-1m-2,m-1m-1·0,⋯,0︸n-1,sn,nk,sn+1,nk,⋯,sm-1,nk,sm,nktr=m-1n-1sn,nk+m-1nsn+1,nk+⋯+m-1m-2sm-1,nk+m-1m-1sm,nkand(8)sm,n+1k+1=m-10,⋯,m-1n-1︸n,m-1n,m-1n+1,⋯,m-1m-2,m-1m-1·0,⋯,0︸n,sn+1,n+1k,sn+2,n+1k,⋯,sm-1,n+1k,sm,n+1ktr=m-1nsn+1,n+1k+m-1n+1sn+2,n+1k+⋯+m-1m-2sm-1,n+1k+sm,n+1k.
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kby the induction hypothesis on S(k). But since si,i(k)=1 for all i, it follows immediately from Pascal’s rule that(10)sm,nk+1+n+k+1sm,n+1k+1=m-1n-1+m-1nsn,nk+m-1n+m-1n+1sn+2,n+1k+⋯+m-1m-2+1sm,n+1k+sm+1,n+1k=mnsn,nk+mn+1sn+2,n+1k+⋯+mm-1sm,n+1k+sm+1,n+1k=mnsn+1,n+1k+mn+1sn+2,n+1k+⋯+mm-1sm,n+1k+sm+1,n+1k=m0,⋯,mn-1︸n,mn,⋯,mm-1,1·0,⋯,0︸n,sn+1,n+1k,⋯,sm+1,n+1ktr=sm+1,n+1k+1.Theorem3 agrees with the Stirling rule if k=1, so we may call TkS=S(k) thegeneralizedk-Stirling matrix for k>0. Like the Bell numbers which are the sum of each row of S, we will take sum of each row of the k-Stirling matrix S(k) and call this the k-Bell number and denote it by B(k)={Bi(k)∣i≥0}. The first few terms of the k-Bell numbers are obtained by examining S(k). (11)
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S ( k ) 1 1 1 ⋯ = T B 0 ( k - 1 ) B 1 ( k - 1 ) B 2 ( k - 1 ) ⋯ and S(k)123⋯=B0(k+1)B1(k+1)B2(k+1)⋯.Proof.
SinceS(k)=TkS=TS(k-1), it is clear that (12)B0kB1kB2k⋯=Sk111⋯=TSk-1111⋯=TB0k-1B1k-1B2k-1⋯.And, due to Theorem 1, we have(13)S0123⋯=B10B20B30⋯=B01B11B21⋯andS1123⋯=B20-B10B30-B20B40-B30⋯=B02B12B22⋯.
So if we assumeS(k-1)123⋯=B0(k)B1(k)B2(k)⋯ for some k then we have (14)Sk123⋯=TSk-1123⋯=TB0kB1kB2k⋯=B0k+1B1k+1B2k+1⋯.Theorem4 says TB0B1B2⋯=B1B2B3⋯, so TB1B2B3⋯=T2B0B1B2⋯. Thus the expression rm(B1,⋯,Bm)tr in the proof of Theorem 1, which is the multiplication of the mth row of T with (B1,⋯,Bm)tr, is equal to multiplication of the mth row of T2 with (B0,⋯,Bm-1)tr. Note that T2 equals the arithmetic table T(2x+1) of the polynomial (2x+1)n, while Pascal’s table T is the arithmetic table of (x+1)n. Hence if we let rm(2x+1) be the mth row of T(2x+1)=12144181261 of (2x+1)n then(15)Bm+1-Bm=∑i=0mm-1iBi+1=m-10,m-11,⋯,m-1m-1B1,⋯,Bmtr=rm2x+1B0,⋯,Bm-1tr,which implies T2B0B2⋯=B2-B1B3-B2⋯=B0(2)B1(2)⋯=S(2)11⋯. Thus, owing to Theorem 4, it is clear to have generalizations to k-Bell numbers (k≥0) that(16)TuB0kB1kB2k⋯=TuSk111⋯=Su+k111⋯=B0u+kB1u+kB2u+k⋯.


## 3. Matrix of General Bell Numbers

With all thej-Bell numbers B(j)(j≥1), we make a table GB^ in which each jth column is composed of j-Bell numbers. We call it amatrix of general Bell numbers and denote it by GB^=[bi,j] (i,j≥1). (17)
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1Theorem 5.
The entries of the(j+1)th column in GB^=[bi,j] satisfy bi,j+1=bi,j-jbi-1,j. So Bi(j+1)=Bi+1(j)-jBi(j).Proof.
Observe that eachjth column (2≤j≤5) satisfies the following: (18)
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Supposebi,j+1=bi,j-jbi-1,j for some i,j>1. By looking at (19)bi,j=b1,1b2,1b2,2b3,1b3,2b3,3b4,1b4,2b4,3b4,4⋯=B01B11B02B21B12B03B31B22B13B04⋯it is enough to prove the following two cases.
(i) For anyi≥0, assume Bi(j)=Bi+1(j-1)-(j-1)Bi(j-1) in some jth column and show Bi(j+1)=Bi+1(j)-jBi(j) is true at (j+1)th column.
(ii) For anyj≥1, assume Bt-1(j)=Bt(j-1)-(j-1)Bt-1(j-1) at every tth row (1≤t≤i) and show Bi(j)=Bi+1(j-1)-(j-1)Bi(j-1) is true at (i+1)th row.
Using theith row ri of T, the induction hypothesis in (i) with (16) implies(20)
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.On the other hand, we also have(21)Bi+1j-jBij=ri+2B0j-1,B1j-1,⋯,Bi+1j-1tr-jri+1B0j-1,B1j-1,⋯,Bij-1tr=i+10B0j-1+i+11B1j-1+i+12B2j-1+⋯+i+1i+1Bi+1j-1-ji0B0j-1+i1B1j-1+⋯+iiBij-1=i+10-ji0,i+11-ji1,i+12-ji2,⋯,i+1i-jii,i+1i+1B0j-1,⋯,Bij-1,Bi+1j-1tr=-j-1i0,i0-j-1i1,⋯,ii-1-j-1ii,iiB0j-1,B1j-1,⋯,Bi+1j-1tr,for i+1u-jiu=iu-1+iu-jiu=iu-1-(j-1)+iu for all 0≤u≤i. Hence identities (20) and (21) prove (i).
Now, for (ii), the induction hypothesisBt(j-1)+(j-2)Bt(j-2)=Bt+1(j-2) (1≤t≤i) in (ii) with again (16) implies(22)Bij+j-1Bij-1=Bij+j-2Bij-1+Bij-1=ri+1B0j-1,⋯,Bij-1tr+j-2ri+1B0j-2,⋯,Bij-2tr+ri+1B0j-2,⋯,Bij-2tr=ri-1B0j-1+j-2B0j-2,⋯,Bij-1+j-2Bij-2tr+ri+1B0j-2,⋯,Bij-2tr=ri+1B1j-2,⋯,Bi+1j-2tr+ri+1B0j-2,⋯,Bij-2trMoreover(23)Bi+1j-1=ri+2B0j-2,B1j-2,⋯,Bi+1j-2tr=0;ri+1+ri+1;0B0j-2,B1j-2,⋯,Bi+1j-2tr=ri+1B1j-2,⋯,Bi+1j-2tr+ri+1B0j-2,⋯,Bij-2trThen we complete the proof by comparing (22) and (23).Therefore Theorem5 with (16) shows (24)TuB0kB1kB2k⋯=B0u+kB1u+kB2u+k⋯=B1u+k-1-k+u-1B0u+k-1B2u+k-1-k+u-1B1u+k-1B3u+k-1-k+u-1B2u+k-1⋯.As an example, the6-Bell numbers B(6)={Bi(6)}={1,7,50,365,2727,⋯} and then 7-Bell numbers B(7)={Bi(7)}={1,8,65,537,4516,⋯} are obtained sequentially from the 5-Bell numbers. (25)
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537The next theorem gives another way to havek-Bell numbers B(k). Let [0]j denote the j-tuple {0,⋯,0}.Theorem 6.
LetXj=([0]j-1;j,j+1,⋯,n)tr and ρn=(bn,1,⋯,bn,n) be the nth row of GB^. Then ρnXj+1=bn+1,j for any 1≤j≤n.Proof.
ClearlyGB^X1+1=GB^123⋯+1=2515⋯ and GB^X2+1=GB^023⋯+1=1310⋯. Similarly GB^Xj+1 (j=3,4) equals 114⋯ and 111⋯. Thus, for instance, when j=4, with nth row ρn of GB^, we have ρ3X4+1=d4,4=1, ρ4X4+1=d5,4=5, ρ5X4+1=d6,4=26, ρ6X4+1=d7,4=141, and so on. Hence we can say that ρnX4+1=bn+1,4 for n≥3.
Now assumeρnXj+1=bn+1,j-1 for n≥j-1. Then since Xj+1=Xj-([0]j-1;j,0,⋯,0)tr, Theorem 5 implies(26)ρnXj+1=ρnXj-ρn0j-1;j,0,⋯,0tr=bn+1,j-1-jbn,j=bn+1,j+1-1.Theorem6 can be restated in terms of k-Bell numbers.Corollary 7.
( B n ( 1 ) , B n - 1 ( 2 ) , ⋯ , B 1 ( n ) ) ( [ 0 ] j - 1 ; j , j + 1 , ⋯ , n ) t r = B n - j + 2 ( j ) - 1.HenceGB^(1,2,⋯,n)tr=(Bn+1(1),⋯,Bn+1(n))tr and GB^([0]j-1;j,⋯,n)tr=( [ 0 ] j - 1 ; B n + 1 ( j ) , ⋯ , B n + 1 ( n ) ) t r. Precisely, with ρ6=(203,151,77,26,6,1), we have ρ6(0,0,0,0,0,6)tr+1=7, ρ6(0,0,0,0,5,6)tr+1=37, ρ6(0,0,0,4,5,6)tr+1=141, ρ6(0,0,3,4,5,6)tr+1=372, ρ6(0,2,3,4,5,6)tr+1=674, and ρ6(1,2,3,4,5,6)tr+1=877, where these yield the 7th row ρ7=(877,674,372,141,37,7,1).


## 4.k-Peirce Matrix

The Peirce matrix[pi,j](i,j≥0) was designed to generate Bell numbers. (27)P=pi,j=1122355710151520273752where it holds a recurrence pi,j=pi,j-1+pi-1,j-1 with p0,0=B0, pn,0=pn-1,n-1=Bn. So the 6th row of P begins with 52 followed by 67, 87, 114, and 151 and then reach 203, the next Bell number of 52. Thus all Bell numbers are on both borders of P. The matrix is often called the Bell matrix or Aitken’s array named after E. T. Bell and A. Aitken. Here we call it Peirce matrix after C. S. Peirce [10] to avoid confusion with the matrix of general Bell numbers in Section 3.Now, for anyk≥1, let P(k)=[pi,j(k)](i,j≥1) be a matrix satisfying the following two rules.(i) Eachith row begins with the Bell number pi,1(k)=Bi-1(0).(ii)(28)pi,jk=pi,j-1k+kpi-1,j-1kforalli>1,andpi,jk=0forj>i.ObviouslyP(1) is the Peirce matrix, and P(k) (k=2,3) are equal to (29)P2=1132410591737andP3=11425175112677Notice that the left border is always comprised of Bell numbers, while the right diagonals (r.diag.) ofP(2) and P(3) are of 2 and 3-Bell numbers, respectively. And the second right diagonal (2nd r.diag.) of P(2) equals the r.diag. of P(3). The next theorem shows that the r.diag. of P(k)=[pi,j(k)] (k≥1) is composed of k-Bell numbers Bi(k).Theorem 8.
(1) The entries of r.diag. ofP(k) are k-Bell numbers; i.e., pi,i(k)=Bi-1(k), while the entries of left border are Bell numbers.
(2) The 2nd r.diag. is composed of the(k+1)-Bell numbers; i.e., pi,i-1(k)=Bi(k+1).Proof.
Following the rules in (28), P(k)=[pi,j(k)] is equal to (30)11k+12k+2k2+2k+252k+5k2+4k+5k3+3k2+6k+5155k+152k2+10k+15k3+6k2+15k+15k4+4k3+12k2+20k+155215k+525k2+30k+522k3+15k2+45k+52k4+8k3+30k2+60k+52⋯
Whenk=1,2,3, the r.diag. of P(k) yields {1,2,5,15,⋯}, {1,3,10,37,⋯}, and {1,4,17,77,⋯}, where these correspond to B(1), B(2), and B(3).
Assume the r.diag(31)1,k+1,k2+2k+2,k3+3k2+6k+5,k4+4k3+12k2+20k+15,⋯of P(k) equals the k-Bell numbers B(k). Then, due to (16), we have(32)B0k+1B1k+1B2k+1B3k+1B4k+1=TB0kB1kB2kB3kB4k=T1k+1k2+2k+2k3+3k2+6k+5k4+4k3+12k2+20k+15=1111211331146411112215631152012411kk2k3k4=121541151561526030811kk2k3k4
And it is not hard to see that it corresponds to the r.diag. ofP(k+1)(33)11k+22k+3k2+4k+552k+7k2+6k+10k3+6k2+15k+15155k+202k2+14k+27k3+9k2+30k+37k4+8k3+30k2+60k+525215k+675k2+40k+872k3+21k2+81k+114⋯
Moreover, by comparingP(k) with P(k+1), we also notice that the r.diag. of P(k+1) equals the 2nd r.diag. of P(k).Owing to Theorem8, we may call P(k) thePeircek-matrix. Write diag[a,b,c,⋯] for the diagonal matrix having diagonal entries a,b,c,⋯. Every k-Bell number can be obtained by Bell numbers as follows.Theorem 9.
B n ( k ) = r n d i a g [ B 0 , B 1 , ⋯ , B n - 1 ] ( k n - 1 , ⋯ , k , 1 ) t r.Proof.
Whenk=1, it is due to Theorem 1. And, by Theorem 8, we have (34)B1kB2kB3kB4kB5k=11+k2+2k+k25+6k+3k2+k315+20k+12k2+4k3+k4=1112215631152012411kk2k3k4.But we notice (35)111122136514122015=111121133114641112515=TdiagB0,B1,⋯,B4.Hence, with the nth row rn of T, the nth k-Bell number Bn(k) is equal to rndiag[B0,⋯,Bn-1](kn-1,⋯,k,1)tr.For example, withr6=(1,5,10,10,5,1), the 6thk-Bell number B6(k) equals(36)r6diagB0,⋯,B5k5,⋯,1tr=r6diag1,1,2,5,15,52k5,⋯,1tr=k5+5k4+20k3+50k2+75k+52.



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*Source: 9096761-2018-08-08.xml*