1 2 Journal of Integer Sequences, Vol. 5 (2002), 3 Article 02.1.6 47 6 23 11 The function M (s; a; z) and some well-known v m sequences Aleksandar Petojevi¶c Aleja Mar·sala Tita 4/2 24000 Subotica Yugoslavia Email address: [email protected] Abstract In this paper we de¯ne the function vMm(s; a; z), and we study the special cases 1Mm(s; a; z) and nM¡1(1; 1; n + 1). We prove some new equivalents of Kurepa's hypothesis for the left factorial. Also, we present a generalization of the alternating factorial numbers. 1 Introduction Studying the Kurepa function +1 tz ¡ 1 K(z) = !z = e¡t dt (Re z > 0); t ¡ 1 Z0 G. V. Milovanovi¶c gave a generalization of the function +1 tz+m ¡ Q (t; z) M (z) = m e¡tdt (Re z > ¡(m + 1)); m (t ¡ 1)m+1 Z0 where the polynomials Qm(t; z), m = ¡1; 0; 1; 2; ::: are given by m m + z Q (t; z) = 0 Q (t; z) = (t ¡ 1)k: ¡1 m k Xk=0 µ ¶ This work was supported in part by the Serbian Ministry of Science, Technology and Development under Grant # 2002: Applied Orthogonal Systems, Constructive Approximation and Numerical Methods. 1 +1 The function fMm(z)gm=¡1 has the integral representation 1 z(z + 1) ¢ ¢ ¢ (z + m) z¡1 m (1¡»)=» 1 ¡ » Mm(z) = » (1 ¡ ») e ¡ z; d»; m! 0 » Z ³ ´ where ¡(z; x), the incomplete gamma function, is de¯ned by +1 ¡(z; x) = tz¡1e¡t dt: (1) Zx Special cases include M¡1(z) = ¡(z) and M0(z) = K(z); (2) where ¡(z) is the gamma function +1 ¡(z) = tz¡1e¡t dt: Z0 The numbers Mm(n) were introduced by Milovanovi¶c [10] and Milovanovi¶c and Petojevi¶c [11]. For non-negative integers n; m 2 N the following identities hold: ¡ ¡ n 1 (¡1)i n 1 m + n M (0) = 0; M (n) = k! : m m i! k + m + 1 i=0 X Xk=i µ ¶ For the numbers Mm(n) the following relations hold: m Mm(n + 1) = n! + Mº(n); º=0 X M (n) lim º = 1; n!+1 Mº¡1(n) M (n) lim m = 1: n!+1 (n ¡ 1)! +1 The generating function of the numbers fMm(n)gn=0 is given by 1 1 + xn A (x)ex¡1 (Ei (1) ¡ Ei (1 ¡ x) + B (x)ex ¡ C (x)) = M (n) m! m m m m n! n=0 ¡ ¢ X where Am(x), Bm(x), and Cm(x) are polynomials de¯ned as follows: A (x) m m (x ¡ 1)k m = ; m! k k! Xk=0 µ ¶ ¡ ¡ ¡ B (x) m 1 m º m (¡1)k¡1 k 1 (¡1)j (x ¡ 1)º m = ; m! k + º k j! º! º=0 à j=0 ! X Xk=1 µ ¶ X ¡ j C (x) m 1 j m (¡1)k¡1 m (x ¡ 1)j m = (¡1)º ; m! º k ¡ º k j! j=0 ú=0 ! X X µ ¶ kX=j+1 µ ¶ 2 Here Ei (x) is the exponential integral de¯ned by x et Ei (x) = p.v. dt (x > 0): (3) t Z¡1 In this paper, we give a generalization of the function Mm(z) which we denote as vMm(s; a; z). These generalization are of interest because its special cases include: 1M1(1; 1; n + 1) = n!; 1M0(1; 1; n) = !n; nM¡1(1; 1; n + 1) = An: where n!, !n, and An are the right factorial numbers (sequence A000142 in [17]), the left factorial numbers (sequence A003422 in [17]) and the alternating factorial numbers (sequence A005165 in [17]), respectively. They are de¯ned as follows: n¡1 n n¡k 0! = 1; n! = n ¢ (n ¡ 1)!; !0 = 0; !n = k! and An = (¡1) k!: (4) Xk=0 Xk=1 2 De¯nitions We now introduce a generalization of the function Mm(z). De¯nition 1 For m = ¡1; 0; 1; 2; :::; and Re z > v ¡ m ¡ 2 the function vMm(s; a; z) is de¯ned by vMm(s; a; z) = v (¡1)2v+1¡k¡(m + z + 2 ¡ k) = L[s; F (a; k ¡ z; m + 2; 1 ¡ t)]; ¡(z + 1 ¡ k)¡(m + 2) 2 1 Xk=1 where v is a positive integer, and s; a; z are complex variables. The hypergeometric function 2F1(a; b; c; x) is de¯ned by the series 1 (a) (b) xn F (a; b; c; x) = n n (jxj < 1); 2 1 (c) n! n=0 n X and has the integral representation ¡(c) 1 F (a; b; c; x) = tb¡1(1 ¡ t)c¡b¡1(1 ¡ tx)¡adt; 2 1 ¡(b)¡(c ¡ b) Z0 3 in the x plane cut along the real axis from 1 to 1, if Re c > Re b > 0. The symbols (z)n and L[s; F (t)] represent the Pochhammer symbol ¡(z + n) (z) = ; n ¡(z) and Laplace transform 1 L[s; F (t)] = e¡stF (t)dt: Z0 Table 1: The numbers 1Mm(1; a; n) for m = 1; 2; 3; 4 and a = 0; 1; 2 a = 0 in [17] a = 1 in [17] a = 2 in [17] 1M1(1; 0; n) A000217 1M1(1; 1; n) A014144 1M1(1; 2; n) A007489 1M2(1; 0; n) A000292 1M2(1; 1; n) unlisted 1M2(1; 2; n) A014145 1M3(1; 0; n) A000332 1M3(1; 1; n) unlisted 1M3(1; 2; n) unlisted 1M4(1; 0; n) A000389 1M4(1; 1; n) unlisted 1M4(1; 2; n) unlisted The term \unlisted" in Table 1 means that the sequence cannot currently be found in Sloane's on-line encyclopedia of integer sequences[17]. Lemma 1 Let m = ¡1; 0; 1; 2; : : : : Then 1Mm(1; 1; z) = Mm(z): Proof. The proof presented here is due to Professor G. V. Milovanovi¶c. Since (¡1)k¡(z) (k + m + 1)! = (m + 1)!(m + 2) and (1 ¡ z) = ; k k ¡(z ¡ k) we have m + z ¡(m + z + 1) ¡(m + z + 1) (1 ¡ z) (¡1)k = = ¢ k ; k + m + 1 ¡(z ¡ k)(k + m + 1)! ¡(z)(m + 1)! (m + 2) µ ¶ k so that 1 tz+m ¡ Q (t; z) + m + z m = (t ¡ 1)k (jt ¡ 1j < 1) (t ¡ 1)m+1 k + m + 1 Xk=0 µ ¶ 1 ¡(m + z + 1) + (1 ¡ z) (1) (1 ¡ t)k = k k ¢ ¡(z)(m + 1)! (m + 2)k k! Xk=0 ¡(m + z + 1) = F (1; 1 ¡ z; m + 2; 1 ¡ t): ¡(z)(m + 1)! 2 1 4 3 The function 1Mm(s; a; z) +1 +1 +1 3.1 The numbers f1Mm(1; ¡n; r)gr=0 n=0 m=¡1 We have ¡(m + z + 1) M (1; ¡n; z) = L[1; F (¡n; 1 ¡ z; m + 2; 1 ¡ t)]; 1 m ¡(z)¡(m + 2) 2 1 starting with the polynomials 1 m + z n (1 ¡ t)k; n 2 N: k + m + 1 k Xk=0 µ ¶µ ¶ Since n! (¡n) = (¡1)k k (n ¡ k)! we have 1 m + z n ¡(m + z + 1) (1 ¡ t)k = F (¡n; 1 ¡ z; m + 2; 1 ¡ t) (5) k + m + 1 k ¡(z)¡(m + 2) 2 1 Xk=0 µ ¶µ ¶ or, by continuation, 1 n = ¼ cosec ¼z »1¡z(1 ¡ »)m+z+1 1 ¡ (1 ¡ t)» d»: Z0 ¡ ¢ Hence, the following de¯nition is reasonable. De¯nition 2 For n 2 N and m = ¡1; 0; 1; 2; :::; the polynomials z 7! mPn(z) are de¯ned by mPn(z) = L[1; 2F1(¡n; 1 ¡ z; m + 2; 1 ¡ t)]: Table 2: The polynomials mP2(z), mP3(z) and mP4(z) m mP2(z) mP3(z) mP4(z) 1 2 3 1 3 7 2 49 3 4 61 3 193 2 ¡1 2 z ¡ 2 z + 2 ¡ 3 z + 2 z ¡ 6 z + 6 8 z ¡ 12 z + 8 z ¡ 1 2 1 4 1 3 2 29 5 509 0 6 z ¡ 2 z + 3 ¡ 12 z + z ¡ 12 z + 2 ¡ 12 z + 24 1 2 1 7 1 3 9 2 67 17 1 12 z ¡ 4 z + 6 ¡ 30 z + 20 z ¡ 60 z + 10 The polynomials mPn(z) can be expressed in terms of the derangement numbers (sequence A000166 in [17]) k (¡1)º S = k! (k ¸ 0) k º! º=0 X in the form 5 Theorem 1 For m = ¡1; 0; 1; 2; ::: and n 2 N we have 1 n + m + 1 ¡1 n + m + 1 z ¡ 1 P (z) = (¡1)kS : m n n k + m + 1 k k µ ¶ Xk=0 µ ¶µ ¶ Proof. Using the relation (5) we have 1 1 ¡(z)¡(m + 2) ¡t m + z n k mPn(z) = e (1 ¡ t) dt ¡(m + z + 1) 0 k + m + 1 k Z Xk=0 µ ¶µ ¶ 1 ¡(z)(m + 1)! ¡(m + z + 1) n 1 = e¡t(1 ¡ t)kdt ¡(m + z + 1) ¡(z ¡ k)(k + m + 1)! k 0 Xk=0 µ ¶ Z 1 n + m + 1 ¡1 n + m + 1 z ¡ 1 1 = e¡t(1 ¡ t)kdt: n k + m + 1 k 0 µ ¶ Xk=0 µ ¶µ ¶ Z Now use e®s¡(z; ®s) L[s; (t + ®)z¡1] = (Re s > 0); sz to obtain 1 n + m + 1 ¡1 n + m + 1 z ¡ 1 ¡(k + 1; ¡1) P (z) = (¡1)k : m n n k + m + 1 k e µ ¶ Xk=0 µ ¶µ ¶ Here ¡(z; x) is the incomplete gamma function.