EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 4, 2020, 948-963 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Truncated Tangent Polynomials Nestor G. Acala1,∗, Maribeth B. Montero1 1 Mathematics Department, College of Natural Sciences and Mathematics, Mindanao State University-Main Campus, Marawi City, Lanao Del Sur, Philippines Abstract. In this paper, we introduce a class of truncated tangent polynomials which generalizes tangent numbers and polynomials, and establish various properties and identities. Moreover, we obtain some interesting correlations of truncated tangent polynomials with the Stirling numbers of the second kind and with the hypergeometric Bernoulli polynomials. 2020 Mathematics Subject Classifications: 11B68, 11B73, 11B83, 33C15 Key Words and Phrases: Tangent numbers and polynomials, Truncated tangent polynomials, truncated Bernoulli polynomials, Hypergeometric Bernoulli polynomials, Stirling numbers of the second kind, truncated Stirling numbers 1. Introduction The hypergeometric Bernoulli numbers Bm,n (see [8–11, 13]) are defined by 1 1F1(1;m+ 1; t) = tm m! et − ∑m−1 j=0 tj j! = ∞∑ n=0 Bm,n tn n! , (1) where 1F1(a; b; z) = ∞∑ n=0 (a)(n)zn (b)(n)n! (2) is the confluent hypergeometric function with (x)(n) = x(x+ 1) · · · (x+ n− 1) for n ≥ 1, and (x)(0) = 1. When m = 1, Bn := B1,n are the classical Bernoulli numbers given by t et − 1 = ∞∑ n=0 Bn tn n! . ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3839 Email addresses: nestor.acala@msumain.edu.ph (N. Acala), maribeth.montero@msumain.edu.ph (M. Montero) https://www.ejpam.com 948 c© 2020 EJPAM All rights reserved. N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 949 The hypergeometric Bernoulli polynomials Bm,n(x) were also introduced in [8, 9] and defined by ext 1F1(1;m+ 1; t) = tm m!e xt et − ∑m−1 j=0 tj j! = ∞∑ n=0 Bm,n(x) tn n! . (3) When m = 1, Bn(x) := B1,n(x) are the classical Bernoulli polynomials given by the exponential generating function tex et − 1 = ∞∑ n=0 Bn(x) tn n! . Hypergeometric Bernoulli polynomials (numbers) are also called truncated Bernoulli poly- nomials (numbers). In [15], Komatsu and Pita introduced truncated Euler polynomials via the generating function 2tm m! e xt et + 1− ∑m−1 j=0 tj j! = ∞∑ n=0 Em,n(x) tn n! . (4) These polynomials satisfy recurrence relation En,m(x) = 0, n = 0, 1, 2, · · · ,m− 1, and En,n+m(x) = 2 ( n+m n ) xn − n∑ k=0 ( n+m k ) En,k(x), n ≥ 0. When m = 0 in (4), En(x) := E0,n(x) are the classical Euler polynomials given by 2ex et + 1 = ∞∑ n=0 En(x) tn n! . In recent years, extensive researches on various families of truncated exponential poly- nomials have become popular. Truncation of exponential polynomials have played crucial importance to evaluate integrals including products of special functions [4]. Some of the recent works on truncated numbers polynomials include truncated Fubini polynomials [5], truncated-exponential based Apostol-type polynomials [26], truncated-exponential- based Frobenius-Euler polynomials [17], hypergeometric Cauchy numbers [13], truncated Bernoulli-Carlitz and truncated Cauchy-Carlitz numbers [14, 16], truncated exponential- based Appell polynomials [12] and many others. In [6], Duran and Acikgoz introduced degenerate truncated exponential polynomials and obtain truncated degenerate versions of some special polynomials such as Stirling polynomials of the second kind, Bernoulli polynomials, Euler polynomials, and Bell polynomials. In the next section, we introduce truncated tangent numbers and polynomials and explore some of their interesting properties and formula. N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 950 2. Truncated tangent polynomials In this section, we give a generalization of tangent polynomials in terms of the trun- cated exponential function. For nonnegative integer m, we define the truncated tangent polynomials Tm,n(x) through the generating function 2 tm m! e2t + 1− ∑m−1 j=0 2j t j j! ext = ∞∑ n=0 Tm,n(x) tn n! . (5) When m = 0, Tn(x) := T0,n(x) are the tangent polynomials (see [20, 21]) defined by 2 e2t + 1 = ∞∑ n=0 Tn(x) tn n! , and Tn := Tn(0) are called tangent numbers. When x = 0 in (5), Tm,n := Tm,n(0) are called the truncated tangent numbers given by 2 tm m! e2t + 1− ∑m−1 j=0 2j t j j! = ∞∑ n=0 Tm,n tn n! . (6) Several extensions and generalizations of tangent numbers and polynomials can be seen in [1, 22–25]. The following identities follow directly from the generating function (5). Theorem 1. For m,n ≥ 0, Tm,n(x) = n∑ r=0 ( n r ) Tm,r · xn−r (7) Tm,n(x) = n∑ r=0 ( n r ) Tm,r(y)(x− y)n−r (8) Tm,n(x+ y) = n∑ r=0 ( n r ) Tm,r(x)yn−r (9) Tm,n(px) = n∑ r=0 ( n r ) Tm,r(x)(p− 1)n−rxn−r, p 6= 1. (10) The truncated tangent polynomials satisfy the following derivative and integral prop- erties. Theorem 2. For m ≥ 0 and n ≥ 1, d dx Tm,n(x) = nTm,n−1(x). (11) N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 951∫ Tm,n(x)dx = 1 n+ 1 Tm,n+1 (12) Tm,n(x) = Tm,n + n ∫ x 0 Tm,n−1(t)dt. (13) Proof. Taking the derivative of both sides of (7) with respect to x, we obtain d dx Tm,n(x) = n−1∑ r=0 ( n r ) (n− r)Tm,r · x(n−r−1) = n n−1∑ r=0 ( n− 1 r ) Tm,rx (n−1)−r = nTm,n−1(x). Equation (12) follows from (11). From (12), we have∫ x 0 Tm,n−1(t)dt = 1 n (Tm,n(x)− Tm,n(0)) , which gives (13). Theorem 3. For n ≥ 0, T1,n(x) = 2n(x− 2)n−1. (14) Proof. When m = 1 in (5), we have ∞∑ n=0 T1,n(x) tn−1 n! = 2te(x−2)t = 2 ∞∑ n=0 (x− 2)n tn+1 n! = 2 ∞∑ n=1 n(x− 2)n−1 tn n! = 2 ∞∑ n=0 n(x− 2)n−1 tn n! . Comparing the coefficients of tn n! completes the proof. Theorem 4. The truncated tangent polynomials satisfy the following recurrence relation: Tm,n(x) = 0, n = 0, 1, 2, · · · ,m− 1 and Tm,n+m(x) = 2 ( n+m n ) xn − n∑ j=0 2n+m−j ( n+m j ) Tm,j , n ≥ 0. (15) N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 952 Proof. It follows from (5) that 2tm m! ∞∑ n=0 (xt)n n! = ∞∑ n=0 Tm,n(x) tn n! 1 + ∞∑ j=m 2j tj j!  = ∞∑ n=0 Tm,n(x) tn n! + ( ∞∑ n=0 Tm,n(x) tn n! ) ∞∑ j=m 2j tj j!  . Thus, ( ∞∑ n=0 Tm,n(x) tn n! ) ∞∑ j=m 2j tj j!  (16) = ∞∑ n=0 2xntn+m n!m! − ∞∑ n=0 Tm,n+m(x) tn+m (n+m)! − m−1∑ n=0 Tm,n(x) tn n! = ∞∑ n=0 ( 2 ( n+m n ) xn − Tm,n+m(x) ) tn+m (n+m)! − m−1∑ n=0 Tm,n(x) tn n! . (17) Note that expression (16) can be written as( ∞∑ n=0 Tm,n(x) tn n! ) ∞∑ j=m 2j tj j!  = ∞∑ n=0 n∑ j=0 Tm,j(x) tj j! 2n−j+m tn−j+m (n− j +m)! = ∞∑ n=0 n∑ j=0 2n+m−j ( n+m j ) Tm,j(x) tn+m (n+m)! . (18) Comparing (17) and (18) gives the desired result. Example 1. For m = 1, we have T1,0(x) = 0. Using recurrence relation (15), we obtain T1,n+1(x) = 2 ( n+ 1 n ) xn − n∑ j=0 2n+1−j ( n+ 1 j ) T1,j . (19) Computing for n = 0, 1, 2, 3, 4, we get the following polynomials: T1,1 = 2 T1,2 = 2 ( 2 1 ) x− 2 ( 2 1 ) T1,1 = 4x− 8 = 4(x− 2) T1,3 = 2 ( 3 2 ) x2 − 22 ( 3 1 ) T1,1(x)− 2 ( 3 2 ) T1,2(x) N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 953 = 6x2 − 12(2)− 2(3)4(x− 2) = 6(x− 2)2 T1,4 = 2 ( 4 3 ) x3 − 23 ( 4 1 ) T1,1(x)− 22 ( 4 2 ) T1,2(x)− 2 ( 4 3 ) T1,3(x) = 8x3 − 8(4)(2)− 24(4)(x− 2)− 8(6)(x− 2) = 8(x− 2)3. Note that the above computations can be easily done using (14). Furthermore, taking m = 2, we obtain the recurrence relation: T2,n+2(x) = 2 ( n+ 2 n ) xn − n∑ j=0 2n+2−j ( n+ 2 j ) T2,j , (20) which yields the following polynomials: T2,0(x) = T2,1(x) = 0 T2,2(x) = 2 T2,3(x) = 6x T2,4(x) = 12(x2 − 4). Theorem 5. For m ≥ 0 and n > 0, 2m−1 n∑ k=0 ( n k ) Tm,n−k(y)Tm+1,k(x) = n∑ k=0 ( n k ) Tm+1,n−k(x)yk− n m+ 1 n−1∑ k=0 ( n− 1 k ) Tm,n−1−k(y)xk. Proof. It follows from (5) that 2ext tm+1 (m+ 1)! = e2t + 1− m∑ j=0 (2t)j j!  ∞∑ n=0 Tm+1,n(x) tn n! = e2t + 1− m−1∑ j=0 (2t)j j!  ∞∑ n=0 Tm+1,n(x) tn n! − 2mtm m! ∞∑ n=0 Tm+1,n(x) tn n! . Hence, 2ext tm+1 (m+ 1)! ∞∑ n=0 Tm,n(y) tn n! = 2tm m! eyt ∞∑ n=0 Tm+1,n(x) tn n! − 2mtm m! ∞∑ n=0 Tm,n(y) tn n! ∞∑ n=0 Tm+1,n(x) tn n! . Consequently, ∞∑ n=0 n∑ k=0 ( n k ) Tm,n−k(y)xk tn+1 (m+ 1)n! = ∞∑ n=0 n∑ k=0 ( n k ) Tm+1,n−k(x)yk tn n! − 2m−1 ∞∑ n=0 n∑ k=0 ( n k ) Tm,n−k(y)Tm+1,k(x) tn n! , which provides the desired result. N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 954 Theorem 6. For m,n ≥ 0, Tn,m(x) = m!n! (n+m)! n∑ l=0 ( n+m l ) l∑ k=0 ( l k ) Tm,kBm,l−k(x) tn n! . (21) Proof. Applying (5), we have ∞∑ n=0 Tm,n(x, y) tn n! = 2 tm m!( e2t + 1− ∑m−1 j=0 2j t j j! )ext tm m!( et − ∑m−1 j=0 tj j! ) · ( et − ∑m−1 j=0 tj j! ) tm m! = m! tm ( ∞∑ n=0 Tm,n tn n! ∞∑ n=0 Bm,n(x) tn n! ) ∞∑ j=m tj j! = m! ∞∑ n=0 n∑ k=0 ( n k ) Tm,kBm,n−k(x) tn n! ∞∑ j=0 tj (j +m)! = ∞∑ n=0 n∑ l=0 m!n! ( n+m l ) (n+m)! l∑ k=0 ( l k ) Tm,kBm,l−k(x) tn n! . Comparing the coefficients of tn n! completes the proof. 3. Relations with Stirling numbers of the second kind and its associated truncated Stirling numbers The Stirling numbers of the second kind are given by the generating function (et − 1)k k! = ∞∑ n=0 S2(n, k) tn n! , (22) or by the recurrence relation for a fixed nonnegative integer n, xn = n∑ k=0 S2(n, k)(x)k, (23) where (x)(k) is the falling factorial defined as (x)k = x(x− 1) · · · (x− k + 1) for k ≥ 0, and (x)0 = 1. For the detailed discussion of Stirling numbers of the second kind, see [3, 7]. In [5], Duran and Acikgoz introduced the truncated Stirling numbers of the second kind: ( et − 1− ∑m−1 j=0 )k k! = ∞∑ n=0 S2,m(n, k) tn n! , (24) N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 955 which reduce to the Stirling numbers of the second kind when m = 0. In this section, we derive some relationships between the truncated tangent polynomials and Stirling numbers of the second kind and its associated truncated version. Theorem 7. For m,n ≥ 0, Tm,n(x) = n∑ k=0 n∑ r=k ( n r ) S2(r, k)Tm,n−r · (x)k, (25) Tm,n(x) = n∑ k=0 n∑ r=k ( n r ) S2(r, k)Tm,n−r(−k)x(k). (26) Proof. Applying relation (23) to (7), we get Tm,n(x) = n∑ r=0 ( n r ) Tm,n−r r∑ k=0 S2(r, k)(x)k (27) = n∑ k=0 n∑ r=k ( n r ) S2(r, k)Tm,n−r · (x)k. (28) To do (26), we note that ext = (1− (1− e−t)))−x. Hence, equation (5) becomes ∞∑ n=0 Tm,n(x) tn n! = 2 tm m! e2t + 1− ∑m−1 j=0 2j t j j! (1− (1− e−t)))−x = 2 tm m! e2t + 1− ∑m−1 j=0 2j t j j! ∞∑ k=0 ( x+ k − 1 k ) (1− e−t)k = 2 tm m! e2t + 1− ∑m−1 j=0 2j t j j! ∞∑ k=0 x(k) (et − 1)k k! e−kt = ∞∑ k=0 x(k) ( ∞∑ n=0 S2(n, k) tn n! )( ∞∑ n=0 Tm,n(−k) tn n! ) = ∞∑ k=0 x(k) ∞∑ n=0 n∑ r=0 ( n r ) S2(r, k)Tm,n−r(−k) tn n! = ∞∑ n=0 n∑ k=0 n∑ r=k ( n r ) S2(r, k)Tm,n−r(−k)x(k) tn n! . Comparing the coefficients of tn n! gives (26). Theorem 8. For m,n ≥ 0, Tm,n+m(x) = n∑ k=0 ( n+m m ) (−1)kk!2n−kS2,m(n, k). (29) N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 956 Proof. Note that (5) can be expressed as ∞∑ n=0 Tm,n(x) tn n! = tm m!( 1 + 1 2 ( e2t − 1− ∑m−1 j=0 (2t)j j! )) (30) = tm m! ∞∑ k=0 ( −1 2 )k e2t − 1− m−1∑ j=0 (2t)j j! k (31) = tm m! ∞∑ k=0 (−1)kk!2−k ∞∑ n=0 S2,m(n, k) (2t)n n! (32) = ∞∑ n=0 ∞∑ k=0 (−1)kk!2n−kS2,m(n, k) tn+m m!n! (33) = ∞∑ n=0 n∑ k=0 ( n+m m ) (−1)kk!2n−kS2,m(n, k) tn+m (n+m)! , (34) which gives the desired result. The case when m = 1, we obtain an interesting identity involving Apostol-type Stir- ling numbers of the second kind. These numbers are defined (see [18] by means of the generating function ( λet − 1 )k k! = ∞∑ n=0 S2(n, k;λ) tn n! . (35) Theorem 9. For n ≥ 0, T1,n+1(x) = (n+ 1)2n ∞∑ k=0 (−1)kk!S2 ( n, k; 1 2 ) . (36) Proof. When m = 1 in (5), we have ∞∑ n=0 T1,n(x) tn n! = t( 1 2e 2t − 1 ) + 1 = t ∞∑ k=0 (−1)k ( 1 2 e2t − 1 )k = t ∞∑ k=0 (−1)kk! ( 1 2e 2t − 1 )k k! = t ∞∑ k=0 (−1)kk! ∞∑ n=0 S2 ( n, k; 1 2 ) (2t)n n! = ∑ n=0 (n+ 1)2n ∞∑ k=0 (−1)kk!S2 ( n, k; 1 2 ) tn+1 (n+ 1)! , which gives the desired result. N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 957 4. Relations with hypergeometric Bernoulli polynomials In this section, we show several relations of truncated tangent polynomials with hy- pergeometric Bernoulli polynomials and hypergeometric Bernoulli numbers. Theorem 10. For m,n ≥ 0,( n+m n ) n∑ k=0 ( n k )( 2k−(n+m)Bm,k(x)yn−k − 2−(k+1)Tm,k(y)xn−k ) (37) = n+m∑ k=0 ( n+m k ) 2−(k+1)Tm,k(y)Bm,n+m−k(x). (38) Proof. Applying (5), we obtain 2tm m! eyt = ( ∞∑ n=0 Tm,n(y) tn n! )e2t + 1− m−1∑ j=0 (2t)j j!  = ( ∞∑ n=0 Tm,n(y) tn n! )e2t − m−1∑ j=0 (2t)j j! + ( ∞∑ n=0 Tm,n(y) tn n! ) . Thus, 2tm m! eyt ∞∑ n=0 Bm,n(x) (2t)n n! = (2t)m m! e2xt ( ∞∑ n=0 Tm,n(y) tn n! ) + ( ∞∑ n=0 Tm,n(y) tn n! ) ∞∑ n=0 Bm,n(x) (2t)n n! . Expanding the exponential functions into series and applying Cauchy product, we have tm m! ∞∑ n=0 ( n∑ k=0 ( n k )( 2k+1Bm,k(x)yn−k − 2m+n−kTm,k(y)xn−k )) tn n! = ∞∑ n=0 n∑ k=0 ( n k ) 2n−kTm,k(y)Bm,n−k(x) tn n! . Avoiding the zero-terms on the right-hand side of the above equation leads to 1 m! ∞∑ n=0 ( n∑ k=0 ( n k )[ 2k+1Bm,k(x)yn−k − 2m+n−kTm,k(y)xn−k ]) tn+m n! = ∞∑ n=0 n+m∑ k=0 ( n+m k ) 2n+m−kTm,k(y)Bm,n+m−k(x) tn+m (n+m)! . Comparing the coefficients of both sides gives 2n+m+1 n!m! n∑ k=0 ( n k )[ 2k−(n+m)Bm,k(x)yn−k − 2−(k+1)Tm,j(y)xn−k ] N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 958 = 2n+m+1 (n+m)! n+m∑ k=0 ( n+m k ) 2−(k+1)Tm,k(y)Bm,n+m−k(x), from where the desired result follows. Lemma 1. (see Theorem 1[19]) The polynomial identity n∑ k=0 an,k(x+ α)k = n∑ k=0 bn,k(x+ β)k (39) implies the Bernoulli polynomials identiy n∑ k=0 an,kBk(x+ α) = n∑ k=0 bn,kBk(x+ β). (40) When m = 0 in Theorem 10, we obtain the following corollary. Corollary 1. For n ≥ 0, n∑ k=0 ( n k ) 2−(k+1)Tk(y) ( (x− 1)n−k + xn−k ) = ( x− 1 + y 2 )n , (41) n∑ k=0 ( n k ) 2−(k+1)Tk(y) (Bn−k(x− 1) +Bn−k(x)) = Bn ( x− 1 + y 2 ) . (42) Proof. Note that B0,n(x) = (x− 1)n and T0,n(x) = Tn(x). Setting m = 0 in Theorem 10, we get n∑ k=0 ( n k ) 2−(k+1)Tk(y) ( (x− 1)n−k + xn−k ) = n∑ k=0 ( n k ) 2k−n(x− 1)kyn−k (43) = ( x− 1 + y 2 )n . Applying Lemma 1 in (43), we obtain n∑ k=0 ( n k ) 2−(k+1)Tk(y) (Bn−k(x− 1) +Bn−k(x)) = n∑ k=0 ( n k ) 2k−nBk(x− 1)yn−k = Bn ( x− 1 + y 2 ) . Corollary 2. For n ≥ 0, n∑ k=0 ( n k )( 2k−n−1Bk(x)yn−k − 2−kk(y − 2)k−1xn−k ) N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 959 = 1 n+ 1 n∑ k=0 ( n+ 1 n− k ) 2−k−1(k + 1)(y − 2)kBn−k(x). (44) n∑ k=0 ( n k )( 2k−n−1Bk(x)Bn−k(y)− 2−kkBk−1(y − 2)xn−k ) = 1 n+ 1 n∑ k=0 ( n+ 1 n− k ) 2−k−1(k + 1)Bk(y − 2)Bn−k(x). (45) Proof. Using B1,n(x) = Bn(x) and T1,n(x) = 2n(x− 2)n−1, and applying Theorem 10 when m = 1, we obtain (n+ 1) n∑ k=0 ( n k )( 2k−n−1Bk(x)yn−k − 2−kk(y − 2)k−1xn−k ) (46) = n+1∑ k=0 ( n+ 1 k ) 2−kk(y − 2)k−1Bn+1−k(x) (47) = n∑ k=0 ( n+ 1 n− k ) 2−(k+1)(k + 1)(y − 2)kBn−k(x), (48) which gives (44). Applying Lemma 1 in (44), we obtain (45). Theorem 11. n∑ k=0 ( n k ) Tm+1,n−k(x)yk− n m+ 1 n−1∑ k=0 ( n− 1 k ) Tm,n−k−1(y)xk = 2m−1 n∑ k=0 ( n k ) Tm+1,n−k(x)Tm,k(y). Proof. From (5), 2tm+1 (m+ 1)! ext = e2t + 1− m−1∑ j=0 (2t)j j! − 2mtm m!  ∞∑ n=0 Tm+1,n(x) tn n! = e2t + 1− m−1∑ j=0 (2t)j j!  ∞∑ n=0 Tm+1,n(x) tn n! − 2mtm m! ∞∑ n=0 Tm+1,n(x) tn n! . Consequently, 2tm+1 (m+ 1)! ext ∞∑ n=0 Tm,n(y) tn n! = 2tm m! eyt ∞∑ n=0 Tm+1,n(x) tn n! N. Acala, M. Montero / Eur. J. Pure Appl. Math, 13 (4) (2020), 948-963 960 − 2mtm m! ∞∑ n=0 Tm+1,n(x) tn n! ∞∑ n=0 Tm,n(y) tn n! . Expanding the exponential functions into series and applying Cauchy product, we get n m+ 1 ∞∑ n=1 ( n−1∑ k=0 ( n− 1 k ) Tm,n−1−k(y)xk ) tn n! = ∞∑ n=1 n∑ k=0 ( n k ) Tm+1,n−k(x)yk tn n! − 2m−1 ∞∑ n=1 n∑ k=0 ( n k ) Tm+1,n−k(x)Tm,k(y) tn n! . Comparing the coeeficients of tn n! completes the proof. Corollary 3. For n ≥ 0, n∑ k=0 ( n k )( 2(x− 2)n−kyk − Tn−k(y)xk ) = Tn(x− 2 + y) (49) n∑ k=0 ( n k )( 2Bn−k(x− 2)yk − Tn−k(y)Bk(x) ) = n∑ k=0 ( n k ) Bn−k(x− 2)Tk(y). (50) Proof. Setting m = 0 in Theorem 11, we have n−1∑ k=0 ( n k ) 2(n− k)(x− 2)n−k−1yk − n n−1∑ k=0 ( n− 1 k ) Tn−k−1(y)xk = = n−1∑ k=0 ( n k ) (n− k)(x− 2)n−k−1Tk(y). Using the identity ( n k ) (n− k) = ( n−1 k ) n, the above equation simplifies to n−1∑ k=0 ( n− 1 k )( 2(x− 2)n−k−1yk − Tn−k−1(y)xk ) = n−1∑ k=0 ( n− 1 k ) (x− 2)n−k−1Tk(y), which is equivalent to n∑ k=0 ( n k )( 2(x− 2)n−kyk − Tn−k(y)xk ) = n∑ k=0 ( n k ) (x− 2)n−kTk(y) (51) = Tn(x− 2 + y). Applying Lemma 1 to (51), we obtain (50). Lastly, we obtain a relation of truncated tangent polynomials with the Frobenius-Euler polynomials. REFERENCES 961 Theorem 12. For m,n ≥ 0, Tm,n(x) = n∑ k=0 ( n k ) (1− λ)r r∑ j=0 ( r j ) (−λ)r−jTm,n−k(j)H (r) k (x|λ), (52) where ( 1− λ et − λ )s ext = ∞∑ n=0 H(s) n (x;λ) tn n! are the Frobenius-Euler polynomials (see [2]). Proof. We express (5) as ∞∑ n=0 Tm,n(x) tn n! = ( 1− λ et − λ )r ext ( et − λ 1− λ ) 2 tm m! e2t + 1− ∑m−1 j=0 2j t j j! = 1 (1− λ)r ( ∞∑ n=0 H(r) n (x|λ) tn n! ) r∑ j=0 ( r j ) (−λ)r−j  2 tm m!e jt e2t + 1− ∑m−1 j=0 2j t j j! = 1 (1− λ)r r∑ j=0 ( r j ) (−λ)r−j ( ∞∑ n=0 H(r) n (x|λ) tn n! )( ∞∑ n=0 Tm,n(j) tn n! ) = 1 (1− λ)r r∑ j=0 ( r j ) (−λ)r−j ∞∑ n=0 ( n∑ k=0 ( n k ) H (r) k (x|λ)Tm,n−k(j) ) tn n! = ∞∑ n=0  n∑ k=0 ( n k ) (1− λ)r r∑ j=0 ( r j ) (−λ)r−jTm,n−k(j)H (r) k (x|λ)  tn n! . Comparing the coefficients of tn n! completes the proof. References [1] R P Agarawal, J Y Kang, and C S Ryoo. Some properties of (p, q)-tangent polyno- mials. J. Comp. Anal. Appl., 24(8):1439–1454, 2018. [2] S Araci and M Acikgoz. A note on the Frobenius-Euler numbers and polynomials associated with Bernstein polynomials. Adv. Stud. Contemp. Math., 22(3):399–406, 2012. [3] L Comtet. Advanced Combinatorics. Reidel Publishing Company, 1974. [4] G Dattoli, C Cesarano, and D Sacchetti. 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