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 S_b the base sequence. Then the Smarandache star derived sequence S_d 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

d_n+1 = S a_(m,r) .b_k+1

k=0

where a_(m,r) is given by

r

a_(m,r) = (1/r!) S (-1)^r-t .^rC_t .t^m , Ref. [1]

t=0

e.g. (1) If the base sequence S_b is 1, 1, 1, . . . then the derived sequence S_d is

1, 2, 5, 15, 52, . . . , i.e. the sequence of Bell numbers. T_n = B_n

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

S_d ------- 1, 3, 10, 37, . . ., we have T_n = B_n+1 -B_n . 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 S_d , to retrieve the original base sequence S_b . We get b_k 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.

2. Sum of the numbers of each row is zero.
3. 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 [b_1,k]_1xn be the row matrix of the base sequence.

[d_1,k]_1xn be the row matrix of the derived sequence.

[S_j,k]_nxn be a square matrix of order n in which s_j,k is the k^th number in the j^th row of the star triangle ( array of the Stirling numbers of the second kind , Ref. [2] ). Then we have

[T_j,k]_nxn be a square matrix of order n in which t_j,k is the k^th number in the j^th row of the array of the Stirling numbers of the first kind , Ref. [2] ). Then we have

[b_1,k]_1xn * [S_j,k]^'_nxn = [d_1,k]_1xn

[d_1,k]_1xn * [T_j,k]^'_nxn = [b_1,k]_1xn

Which suggests that [T_j,k]^'_nxn is the transpose of the inverse of the transpose of the Matrix [S_j,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.