SMARANDACHE STAR (STIRLING) DERIVED SEQUENCES

Amarnath Murthy, S.E.(E&T), WLS, Oil and Natural Gas Corporation Ltd., Sabarmati, Ahmedabad,-380005 INDIA.

Let b1, b2, b3, . . . be a sequence say Sb the base sequence. Then the Smarandache star derived sequence Sd using the following star triangle {ref. [1]} is defined

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

. . .

as follows

d1 = b1

d2 = b1 + b2

d3 = b1 + 3b2 + b3

d4 = b1 + 7b2 + 6b3 + b4

. . .

n

dn+1 = S a(m,r) .bk+1

k=0

where a(m,r) is given by

r

a(m,r) = (1/r!) S (-1)r-t .rCt .tm , Ref. [1]

t=0

e.g. (1) If the base sequence Sb is 1, 1, 1, . . . then the derived sequence Sd is

1, 2, 5, 15, 52, . . . , i.e. the sequence of Bell numbers. Tn = Bn

(2) Sb ----- 1, 2, 3, 4, . . . then

Sd ------- 1, 3, 10, 37, . . ., we have Tn = Bn+1 -Bn . Ref [1]

The Significance of the above transformation will be clear when we consider the inverse transformation. It is evident that the star triangle is nothing but the Stirling Numbers of the Second kind ( Ref. [2] ). Consider the inverse Transformation : Given the Smarandache Star Derived Sequence Sd , to retrieve the original base sequence Sb . We get bk for k = 1, 2, 3, 4 etc. as follows ;

b1 = d1

b2 = -d1 + d2

b3 = 2d1 - 3d2 + d3

b4 = -6d1 + 11d2 - 6d3 + d4

b5 = 24d1 - 50d2 + 35d3 - 10d4 + d5

………………

we notice that the triangle of coefficients is

 1 -1 1 2 -3 1 -6 11 -6 1 24 -50 35 -10 1

Which are nothing but the Stirling numbers of the first kind.

Some of the properties are

(1) The first column numbers are (-1) r-1.(r-1)! , where r is the row number.

1. Sum of the numbers of each row is zero.
2. Sum of the absolute values of the terms in the r th row = r! .

More properties can be found in Ref. [2].

This provides us with a relationship between the Stirling numbers of the first kind and that of the second kind, which can be better expressed in the form of a matrix.

Let [b1,k]1xn be the row matrix of the base sequence.

[d1,k]1xn be the row matrix of the derived sequence.

[Sj,k]nxn be a square matrix of order n in which sj,k is the kth number in the jth row of the star triangle ( array of the Stirling numbers of the second kind , Ref. [2] ). Then we have

[Tj,k]nxn be a square matrix of order n in which tj,k is the kth number in the jth row of the array of the Stirling numbers of the first kind , Ref. [2] ). Then we have

[b1,k]1xn * [Sj,k]'nxn = [d1,k]1xn

[d1,k]1xn * [Tj,k]'nxn = [b1,k]1xn

Which suggests that [Tj,k]'nxn is the transpose of the inverse of the transpose of the Matrix [Sj,k]'nxn .

The proof of the above proposition is inherent in theorem 10.1 of ref. [3].

Readers can try proofs by a combinatorial approach or otherwise.

REFERENCES:

[1] "Amarnath Murthy", 'Properties of the Smarandache Star Triangle' , SNJ, Vol. 11, No. 1-2-3, 2000.

[2] "V. Krishnamurthy" , 'COMBINATORICS Theory and applications' ,East West Press Private Limited, 1985.

[3] " Amarnath Murthy", 'Miscellaneous results and theorems on Smarandache Factor Partitions.', SNJ,Vol. 11,No. 1-2-3, 2000.