ON GENERALIZED STIRLING NUMBERS AND POLYNOMIALS

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which is the result for the generalized polynomials T n(α) (x, r, −p) from [17] and[13].Remark 1. From (3) we get directly relation (13), because (α + rj) n = r n ( α r + j)n .Remark 2. The fact T n(α) (x, r, −p) = r n A n(px r , α ) Toscano also found in [19], in a different way,rusing differential operators.3. ON A RESULT OF R. C. SINGH CHANDELThe main results from the paper [15] are the consequence of the results fromA.M.Chak [6].Starting with the following notation by Singh Chandel( ) αα (k−1.n) = (k − 1) n =k − 1 nα(α + k − 1) · · · (α + (k − 1)(n − 1))we see that(α + rj) (k−1,n) = r n ( αr + j ) (k−1r ,n) ,and so, from (5), we haveS (α,λ) (n, k, r) = (−1)kk!( )k∑ k(−1) j r n ( α j rj=0+ j)(λ−1r ,n) = r n A ( α r , λ r )n,kwhere A (α,λ)n.kare numbers defined by A.M.Chak [6] in the following way(x λ D) n = x (n−1)λ n ∑i=0A (α,λ)n,i x i+α D i x −α .So, using proved equations and the recurrence from [6]we get a recurrence from [15]A (α,λ)n+1,k = A(α,λ) n,k−1+ (nλ − n + α + k)A(α,λ)n,kS (α,λ) (n + 1, k, r) = rS (α,λ) (n, k − 1, r) + (nλ − n + α + rk)S (α,λ) (n, k, r)For polynomials treated in [15] we get from (6) easilyT (α,λ)n (x, r, −p) = r n x n(λ−1) G ( α r )(−px r )where G (α)n,λ(x) are a class of polynomials from [6].n, λ r4
4. EXPLICIT EXPRESSIONS FORS (α) (n, k, r) AND S (α,λ) (n, k, r)In this section we use the results from Cakić [1](see Theorem 4. and commentby L.Carlitz) and Milovanović and Cakić [10] (see Theorem 3.)Applying Carlitz’s [2] extension of a result of Wang [20], Singh Chandel andDwivedi [16] proved the following explicit formula for S (0,λ) (n, k, r) :S (0,λ) (n, k, r) = n! ∑ k∏ r (λ−1,i l),k!ii 1 +···+i k =n l=1 l !where i s > 0.In this section we prove the theorem about new explicit expressions fornumbers S (α) (n, k, r) and S (α,λ) (n, k, r).Theorem: We have(14)and(15)S (α) (n, k, r) = r n ∑ ( )α/r n−1( )∏ α/r + s − Ss,i 0 is=1 sS (α,λ) (n, k, r) = r n ∑ ( )α/r n−1( )∏ α/r + s(λ−1r+ 1) − S si 0 is=1 swhere S s = ∑ s−1t=0 i t. The summation on the right in above sums is over alli 0 , . . . , i n−1 ∈ {0, 1} such that i 0 + i 1 + · · · + i n−1 = n − k.P r o o f:Introduce the notations( )α/rn−2( )∏ α/r + s − SsA(i 0 , . . . , i n−2 ) =,i 0 is=1 sandB(i 0 , . . . , i n−2 ) =( )α/rn−2( )∏ α/r + s(λ−1r+ 1) − S s.i 0 is=1 sFor i n−1 ∈ {0, 1} we get from (14) and (15), respectively:S (α) (n, k, r) = r n∑S n−1 =(n−1)−(k−1)r n (α/r + n − 1 − (n − k − 1))5A(i 0 , . . . , i n−2 )+∑S n−1 =(n−1)−kA(i 0 , . . . , i n−2 ),
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ON GENERALIZED STIRLING NUMBERS AND POLYNOMIALS

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