$\Lambda $-Analogues of R-Stirling Numbers of the First Kind

$$\Lambda $-Analogues of R-Stirling Numbers of the First Kind$

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The r-Stirling numbers of the first kind are defined by the generating function ∞ n 1 k (r) t log(1 + t) (1 + t)r = S (n, k) , (see [3, 20 − 23]). (1.5) k! 1 n! n=k X where k ∈ N ∪{0} and r ∈ R. The unsigned r-Stirling numbers of the first kind are defined as n (x + r)(x + r + 1) ··· (x + r + n − 1) = n+r xk, (see [1, 17, 22]). k+r r (1.6) k=0 X Thus, by (1.5), we get n (r) k (x + r)n = (x + r)(x + r − 1) ··· (x + r − n +1)= S1 (n, k)x , (see [1]). k=0 X (1.7) From (1.5) and (1.7), we note that − − S( r)(n, k) = (−1)n k n+r . (1.8) 1 k+r r The higher-order Daehee polynomials are defined by ∞ log(1 + t) k tn (1 + t)x = D(k)(x) , (see [5, 18, 19, 24]). (1.9) t n n! n=0 X (k) (k) When x = 0, Dn = Dn (0) are called the higher-order Daehee numbers. In (1) particular, for k = 1, Dn(x)= Dn (x), (n ≥ 0), are called the ordinary Daehee polynomials. In this paper, we consider λ-analogues of r-Stirling numbers of the first kind which are derived from the λ-analogues of the falling factorial sequence and investigate some properties for these numbers. Specifically, we give some identities and recurrence relations for the λ-analogues of r-Stirling numbers of the first kind and show their connections with the λ-Stirling numbers of the first kind and higher-order Daehee polynomials.

2. λ-analogues of r-Stirling numbers of the ﬁrst kind From (1.3) and (1.4), we have ∞ ∞ k k k x t n t (1 + λt) λ = (x) = S (k,n)x k,λ k! 1,λ k! k=0 k=0 n=0 ! X∞ ∞ X X (2.1) tk xn = n! S (k,n) . 1,λ k! n! n=0 k n ! X X= λ-ANALOGUES OF r-STIRLING NUMBERS OF THE FIRST KIND 3

On the other hand, we also have

∞ n n x x log(1+λt) log(1 + λt) x (1 + λt) λ = e λ = . (2.2) λ n! n=0 X

Therefore, by (2.1) and (2.2), we get the generating function for S1,λ(n, k), (n, k ≥ 0), which is given by

∞ 1 log(1 + λt) n tk = S1,λ(k,n) . (2.3) n! λ k! k n X= Now, we deﬁne λ-analogues of r-Stirling numbers of the ﬁrst kind as

k ∞ n 1 log(1 + λt) r (r) t (1 + λt) λ = S (n, k) , (2.4) k! λ 1,λ n! n k X= where k ∈ N ∪{0}, and r ∈ R. (0) From (2.3) and (2.4), we note that S1,λ(n, k)= S1,λ(n, k), (n ≥ k ≥ 0). Also, it is easy to show that

∞ n x r t (1 + λt) λ (1 + λt) λ = (x + r)n,λ . (2.5) n! n=0 X By (2.5), we get

∞ n ∞ t x + r n r x log(1+λt) (x + r) = t = (1+ λt) λ e λ n,λ n! n n=0 n=0 λ X X∞ k k 1 log(1 + λt r = x (1 + λt) λ (2.6) k! λ k=0 X∞ ∞ ∞ tn n tn = xk S(r)(n, k) = S(r)(n, k)xk . 1,λ n! 1,λ n! k n k n=0 k ! X=0 X= X X=0 Therefore, by comparing the coeﬃcients on both sides of (2.6), we obtain the following theorem.

Theorem 2.1. For n ≥ 0, we have

n (r) k (x + r)n,λ = S1,λ(n, k)x . k X=0 4 TAEKYUNKIMANDDAESANKIM

Now, we observe that ∞ k k 1 log(1 + λt) r x (1 + λt) λ k! λ k=0 X ∞ ∞ ∞ tm tl = xk S (m, k) (r) 1,λ m! l,λ l! k m k ! l ! X=0 X= X=0 ∞ m m ∞ l k t t = S1,λ(m, k)x (r)l,λ (2.7) m! l! m=0 k=0 ! l=0 ! ∞X X X n m n n k t = S (m, k)(r) − x m 1,λ n m,λ n! n=0 m=0 k ! X X X=0 ∞ n n n n k t = S (m, k)(r) − x . m 1,λ n m,λ n! n=0 k m k ! X X=0 X= Thus, by (2.6) and (2.7), we get n n n (r) k n k S (n, k)x = S (m, k)(r) − x . (2.8) 1,λ m 1,λ n m,λ k k m k ! X=0 X=0 X= Therefore, by comparing the coefficients on both sides of (2.8), we obtain the following theorem. Theorem 2.2. For n ≥ 0, we have n (r) n S (n, k)= S (m, k)(r) − . 1,λ m 1,λ n m,λ m k X= Now, we define λ-analogues of the unsigned r-Stirling numbers of the first kind as follows:

n (x + r)(x + r + λ)(x + r +2λ)+ ··· (x + r + (n − 1)λ)= n+r xk. k+r r,λ (2.9) k X=0 n+r n+r Note that lim → = , (n ≥ k ≥ 0). λ 1 k+r r,λ k+r r By Theorem 2.1 and (2.9), we get n (−r) k (x − r)n,λ = S1,λ (n, k)x , (2.10) k X=0 and n − (x − r) = (−1)n k n+r xk. n,λ k+r r,λ (2.11) k=0 X λ-ANALOGUES OF r-STIRLING NUMBERS OF THE FIRST KIND 5

From (2.10) and (2.11), we can easily derive the following equation (2.12). − − S( r)(n, k) = (−1)n k n+r , (n ≥ k ≥ 0). 1,λ k+r r,λ (2.12) For n ≥ 1, by Theorem 2.1, we get n+1 n+1 (r) k (r) k (x + r)n+1,λ = S1,λ(n +1, k)x = S1,λ(n +1, k)x + (r)n+1,λ. (2.13) k k X=0 X=1 On the other hand, by (1.2), we get

(x + r)n+1,λ = (x + r)n,λ(x + r − nλ) n n (r) k (r) k = x S1,λ(n, k)x − (nλ − r) S1,λ(n, k)x k=0 k=0 nX n X (r) k (r) k n+1 = S1,λ(n, k − 1)x − (nλ − r)S1,λ(n, k)x + (r − nλ)(r)n,λ + x k=1 k=1 Xn X (r) (r) k n+1 = S1,λ(n, k − 1) − (nλ − r)S1,λ(n, k) x + (r)n+1,λ + x . k=1 n o X (2.14) Therefore, by Theorem 2.1 and (2.14), we obtain the following theorem. Theorem 2.3. For 1 ≤ k ≤ n, we have

(r) (r) (r) S1,λ(n +1, k)= S1,λ(n, k − 1) − (nλ − r)S1,λ(n, k). From (2.4), we note that

k k ∞ l l 1 log(1 + λt) r 1 log(1 + λt) r log(1 + λt) (1 + λt) λ = k! λ k! λ l! λ l=0 ∞ X k + l 1 log(1 + λt) k+l = rl l (k + l)! λ l=0 X∞ ∞ k + l tn = rl S (n, k + l) l 1,λ n! l=0 n=k+l X∞ ∞X k + l tn+k = rl S (n + k, k + l) l 1,λ (n + k)! l=0 n=l X∞ Xn n!tk k + l tn = rl S (n + k, k + l) . (n + k)! l 1,λ n! n=0 l=0 ! X X (2.15) 6 TAEKYUNKIMANDDAESANKIM

On the other hand, we have k k k 1 log(1 + λt) r t log(1 + λt) r (1 + λt) λ = (1 + λt) λ k! λ k! λt ∞ ∞ λltl tm tk = D(k) (r) l l! m,λ m! k! (2.16) l=0 ! m=0 ! X∞ X n n k n (k) l t t = D λ (r) − . l l n l,λ n! k! n=0 l ! X X=0 Thus, by (2.15) and (2.16), we get n k+l n l l n (k) l r S1,λ(n + k, k + l)= D λ (r)n−l,λ. (2.17) n+k l l l=0 n l=0 X X Therefore, by (2.17), we obtain the following theorem. Theorem 2.4. For n ≥ 0, we have n n k+l n (k) l l l D λ (r) − = r S (n + k, k + l). l l n l,λ n+k 1,λ l=0 l=0 n X X Now, we observe that ∞ k l k 1 log(1 + λt) r t 1 log(1 + λt) (1 + λt) λ = (r) k! λ l,λ l! k! λ l=0 ! ∞X n n tn = S (m, k)(r) − . m 1,λ n m,λ n! n=k m=k ! X X (2.18) Therefore, by (2.4) and (2.18), we obtain the following theorem. Theorem 2.5. For n, k ≥ 0, with n ≥ k, we have n (r) n S (n, k)= (r) − S (m, k). 1,λ m n m,λ 1,λ m k X= From (2.4), we note that m k 1 log(1 + λt) 1 log(1 + λt) r (1 + λt) λ m! λ k! λ m+k (m + k)! 1 log(1 + λt) r = (1 + λt) λ (2.19) m!k! (m + k)! λ ∞ m + k tn = S(r)(n,m + k) . m 1,λ n! n m k =X+ λ-ANALOGUES OF r-STIRLING NUMBERS OF THE FIRST KIND 7

On the other hand,

m k 1 log(1 + λt) 1 log(1 + λt) r (1 + λt) λ m! λ k! λ ∞ ∞ l j t (r) t = S ,λ(l,m) S (j, k) 1 l!  1,λ j!  (2.20) l m ! j k X= X= ∞ n−m   n tn = S(r)(l, k)S (n − l,m) . l 1,λ 1,λ n! n m k l k ! =X+ X= Therefore, by (2.19) and (2.20), we obtain the following theorem.

Theorem 2.6. For m,n,k ≥ 0 with n ≥ m + k, we have

n−m m + k n S(r)(n,m + k)= S (l, k)S (n − l,m). m 1,λ l 1,λ 1,λ l k X= By (2.3), we get

∞ n k t 1 log(1 + λt) r − r S (n, k) = (1 + λt) λ (1 + λt) λ 1,λ n! k! λ n=k X ∞ ∞ tl − r = S(r)(l, k) λ λmtm 1,λ l! m l=k ! m=0 ! X∞ X∞ tl tm = S(r)(l, k) (−1)m(r + (m − 1)λ) 1,λ l! m,λ m! l=k ! m=0 ! ∞X X n n n (r) n−l t = S (l, k)(−1) (r + (n − l − 1)λ) − . l 1,λ n l,λ n! n=k l=k ! X X (2.21)

Comparing the coeﬃcients on both sides of (2.21), we have the following theorem.

Theorem 2.7. For n, k ≥ 0, with n ≥ k, we have

n n (r) n−l S (n, k)= S (l, k)(−1) (r + λ(n − l − 1)) − . 1,λ l 1,λ n l,λ l k X= 8 TAEKYUNKIMANDDAESANKIM

From (1.9), we have

k k k 1 log(1 + λt) r t log(1 + λt) r (1 + λt) λ = (1 + λt) λ k! λ k! λt ∞ ∞ tk tm tl = D(k)λm (r) k! m m! l,λ l! (2.22) m=0 ! l ! X X=0 k ∞ n n t n (k) m t = D λ (r) − . k! m m n m,λ n! n=0 m=0 ! X X On the other hand, by (2.4), we get

k ∞ n 1 log(1 + λt) r (r) t (1 + λt) λ = S (n, k) k! λ 1,λ n! n=k X ∞ (2.23) tk n!k! tn = S(r)(n + k, k) . k! 1,λ (n + k)! n! n=0 X Thus, by comparing the coeﬃcients on both sides of (2.22) and (2.23), we get

n n (k) m 1 (r) D λ (r)n−m,λ = S (n + k, k). (2.24) m m n+k 1,λ m=0 n X Therefore, by (2.24), we obtain the following theorem.

Theorem 2.8. For n, k ≥ 0, we have n (r) n + k n (k) m S (n + k, k)= D λ (r) − . 1,λ n m m n m,λ m=0 X From (1.9), we note that

k k k 1 log(1 + λt) r t log(1 + λt) r (1 + λt) λ = (1 + λt) λ k! λ k! λt ∞ (2.25) tk tn = λnD(k)( r ) . k! n λ n! n=0 X By (2.23) and (2.25), we get (n + k)! n + k S(r)(n + k, k)= λn D(k)( r )= λn D(k)( r ), (n ≥ 0). (2.26) 1,λ n!k! n λ n n λ In particular, for r = 0, from (2.21) and (2.26) we have λ-ANALOGUES OF r-STIRLING NUMBERS OF THE FIRST KIND 9

n + k λn D(k) = S (n + k, k) k n 1,λ n+k (2.27) n + k (r) n+k−l = S (l, k)(−1) (r + (n + k − l − 1)λ) − , l 1,λ n+k l,λ l k X= where n, k ≥ 0. Therefore, by (2.27), we obtain the following theorem. Theorem 2.9. For n, k ≥ 0, we have

n+k n n + k (k) n + k (r) n+k−l λ D = S (l, k)(−1) (r + (n + k − l − 1)λ) − . k n l 1,λ n+k l,λ l k X= In addition,

n+k − 1 n + k l 1 n+k l D(k) = n n+k l k λ k l k X= n+k−l (k) r × (r + (n +k − l − 1)λ)n+k−l,λ(−1) Dl−k( λ ). Now, we observe that ∞ n k t 1 log(1 + λt) r − r log(1+λt) S (n, k) = (1 + λt) λ e λ 1,λ n! k! λ n=k X ∞ ∞ tl 1 log(1 + λt) m = S(r)(l, k) (−1)mrm 1,λ l! m! λ l k ! m=0 X= X ∞ ∞ ∞ l j (r) t m m t = S (l, k) (−1) r S ,λ(j, m) 1,λ l!  1 j!  l k ! m=0 j=m X= X X ∞ −   n k j n n m m t = (−1) r S ,λ(j, m)S ,λ(n − j, k) .  j 1 1  n! n k j=0 m=0 X= X X  (2.28) Therefore, by comparing the coeﬃcients on both sides of (2.28), we obtain the following theorem Theorem 2.10. For n, k ≥ 0, with n ≥ k, we have − j n k n S (n, k)= (−1)mrmS (j, m)S (n − j, k). 1,λ j 1,λ 1,λ j=0 m=0 X X 10 TAEKYUNKIMANDDAESANKIM

For m,n ≥ 0, we deﬁne λ-analogues of the Whitney’s type r-Stirling numbers of the ﬁrst kind as

(mx + r)n,λ = (mx + r)(mx + r − λ)(mx + r − 2λ) ··· (mx + r − (n − 1)λ) n (r) k = T1,λ (n, k|m)x . k=0 X (2.29)

By (2.29), we get

∞ ∞ tn n tn (mx + r) = T (r)(n, k|m)xk n,λ n! 1,λ n! n=0 n=0 k=0 ! X X∞ X∞ (2.30) tn = T (r)(n, k|m) xk. 1,λ n! k n k ! X=0 X= On the other hand, by binomial expansion, we get ∞ ∞ tn mx + r (mx + r) = tn n,λ n! n n=0 n=0 λ X X mx+r r mx( log(1+λt) ) =(1+ λt) λ = (1+ λt) λ e λ (2.31) ∞ k k m log(1 + λt) r k = (1 + λt) λ x . k! λ k X=0 Comparing the coeﬃcients on both sides of (2.30) and (2.31), the generating (r) function for T1,λ (n, k|m), (n, k ≥ 0), is given by

∞ k k n m log(1 + λt) r (r) t (1 + λt) λ = T (n, k|m) . (2.32) k! λ 1,λ n! n k X= From (2.4) and (2.32), we note that 1 S(r)(n, k)= T (r)(n, k|m), (n ≥ k ≥ 0). (2.33) 1,λ mk 1,λ It is known that the r-Whitney numbers are deﬁned as

n n k (mx + r) = m Wm,r(n, k)(x)k, (see [3]). (2.34) k X=0 By (1.3), we get λ-ANALOGUES OF r-STIRLING NUMBERS OF THE FIRST KIND 11

n l (mx + r)n,λ = S1,λ(n,l)(mx + r) l X=0 n l j = S1,λ(n,l) m Wm,r(l, j)(x)j l j=0 X=0 X n n S n,l mj W l, j x = 1,λ( ) m,.r( )( )j (2.35) j=0 l j X X= n n j j k = S1,λ(n,l)m Wm,r(l, j) S1(j, k)x j=0 l j k X X= X=0 n n n j k = S ,λ(n,l)S (j, k)m Wm,r(l, j) x .  1 1  k j k l j X=0 X= X= Therefore, by (2.29) and (2.35), we obtain the following theorem. Theorem 2.11. For n, k ≥ 0, with n ≥ k, we have n n (r) j T1,λ (n, k|m)= S1,λ(n,l)S1(j, k)m Wm,r(l, j). j k l j X= X= References

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Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea E-mail address: [email protected]

Department of Mathematics, Sogang University, Seoul 121-742, Republic of Ko- rea E-mail address: [email protected]