Moments of Orthogonal Polynomials and Exponential Generating

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So H1 = 1, H3 = 1, H5 = 2, H7 = 8, H7 = 56, H11 = 608, H13 = 9440; see [38, 16, 18, 20]. In his combinatorial approach to orthogonal polynomials [39, p.V-10], Viennot brieﬂy alludes to the monic orthogonal polynomials whose moments are the Genocchi numbers (i.e., the nth moment is G2n+2) and the median Genocchi numbers H2n+1, respectively, but he does not give any explicit formula for them. Many years ago, one of the authors (IG) was learning about continued fractions and their connection to orthogonal polynomials, and saw in Viennot’s work [39] the simple continued fraction for the Genocchi numbers. He wondered what the corresponding orthogonal polynomials were and wrote to Richard Askey, asking if he knew. Askey immediately wrote back to say that they were a special case of the continuous dual Hahn polynomials, and this paper grew out of an attempt to understand his reply. A comprehensive discussion of the combinatorial properties of Genocchi numbers has been given by Viennot [38]. In particular, he showed that the Genocchi numbers and median Genocchi numbers H2n+1 have the S-fraction expansions

∞ G t2n = S(t2;12, 1 2, 22, 2 3, 32, 3 4, 42,...), (2) 2n+2 · · · n=0 X∞ 2n 2 2 2 2 2 2 2 H2n+1 t = S(t ; 1 , 1 , 2 , 2 , 3 , 3 ,...). (3) n=0 X

They are somewhat more recondite than the other sequences mentioned above, and unlike the closely related tangent numbers, there is no straightforward derivation of any of their combinatorial interpretations directly from the exponential generating function, and in fact it is nontrivial to prove that they are integers. Some recent papers (see [4, 21, 29]) have shown there are renewed interest on Genocchi numbers and median Genocchi numbers. As the Wilson polynomials are the most general (q = 1) classical orthogonal polynomials, we shall ﬁrst consider the moments of the Wilson polynomials. In the general case, there are four parameters a,b,c,d and the ordinary the generating function of these moments can be expressed as a hypergeometric series. We show that these moments have a simple exponential generating function when when a =0 or a =1/2. We then consider the continuous dual Hahn polynomials and the (continuous) Hahn polynomials along with their rescaled versions. In particular, we show that the moments of a rescaled version of the continuous dual Hahn polynomials are the generalized Dumont-Foata polynomials, which are a reﬁnement of both Genocchi numbers and median Genocchi numbers. It would be possible to derive the moment generating function for Hahn polynomials from that for Wilson polynomials, but we will give a separate derivation and use our method for variety. MOMENTS OF ORTHOGONAL POLYNOMIALS 3

2. Wilson polynomials

The monic Wilson polynomials Wn(x) (see [2, 40, 25]) are deﬁned by (a + b) (a + c) (a + d) W (x2)=( 1)n n n n W (x2; a,b,c,d) (4) n − (a + b + c + d + n 1) − n where f n, n + a + b + c + d 1, a + ix, a ix W (x2; a,b,c,d)= F ;1 . 4 3 − a + b, a + c,− a + d − If a,b,c,df are positive or a = ¯b and/or c = d¯ and the real parts are positive, then the orthogonality reads as follows:

1 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2 W (x2)W (x2) dx 2π Γ(2ix) m n Z0 n!Γ(n + a + b) Γ(n + c + d) = ··· δ . (5a) (n + a + b + c + d 1) Γ(2n + a + b + c + d) mn − n Consider the corresponding moment sequence wn(a) := wn(a,b,c,d) of the Wilson polynomials deﬁned by

1 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2 x2ndx 2π Γ(2ix) Z0 Γ(a + b)Γ(a + c)Γ(a + d )Γ(b + c)Γ(b + d)Γ(c + d) = w (a) . (5b) n Γ(a + b + c + d) It follows that w (a)=1 and for n 0, 0 ≥ (a + b)(a + c)(a + d) w (a)= w (a + 1) a2w (a). (6) n+1 a + b + c + d n − n Here is the formal approach to the Wilson polynomials. We deﬁne a linear functional 2n L on even polynomials by L (x )= wn(a). Lemma 1. For k 0, ≥ 2n (a + b)k (a + c)k (a + d)k L (x (a + ix)k (a ix)k)= wn(a + k). (7) − (a + b + c + d)k Proof. We prove this by induction on k. It is clear for k = 0. For k = 1 we derive from (6) that (a + b)(a + c)(a + d) L (x2n(a2 + x2)) = a2w (a)+ w (a)= w (a + 1). n n+1 a + b + c + d n MOMENTS OF ORTHOGONAL POLYNOMIALS 4

Suppose this is true for k. Then L (x2n(a + ix) (a ix) ) k+1 − k+1 = (a + k)2 + x2) L (x2n(a + ix) (a ix) k − k (a + b)k (a + c)k (a + d)k 2 = (a + k) wn(a + k)+ wn+1(a + k) . (a + b + c + d)k The result for k + 1 follows then from (6).

Lemma 2. For integers m, n 0, ≥ L ((a + ix) (a ix) (b + ix) (b ix) ) m − m n − n (a + b) (a + c) (a + d) (b + c) (b + d) = m+n m m n n . (a + b + c + d)m+n Proof. This can be proved directly from the case m = n = 0 of (5a): replacing a with a + m and b with b + n in 1 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix) 2 dx 2π Γ(2ix) Z0 Γ(a + b)Γ(a + c)Γ(a + d)Γ(b + c)Γ(b + d)Γ(c + d) = . (8) Γ(a + b + c + d) together with the deﬁnition of L and (8) gives Lemma 2.

Theorem 3. We have (a + b) (a + c) (a + d) (b + c) (b + d) (c + d) L (W (x2)W (x2)) = n! n n n n n n δ . n m (a + b + c + d + n 1) (a + b + c + d) mn − n 2n Proof. We evaluate L W˜ (x2; a,b,c,d)(b + ix) (b ix) n m − m (a + b)m(b + c)m(b + d)m n, n + a + b + c + d 1, a + b + m = 3F2 − − ;1 (a + b + c + d) a + b, a + b + c + d + m m (a + b) (b + c) (b + d) (1 n c d) ( m) = m m m − − − n − n (a + b + c + d) (a + b) (1 n a b c d m) m n − − − − − − n (by the Pfaﬀ-Saalsch¨utz theorem) (a + b) (b + c) (b + d) (c + d) ( m) = m m m n − n . (a + b + c + d)m+n(a + b)n This is 0 for m < n, which implies orthogonality. MOMENTS OF ORTHOGONAL POLYNOMIALS 5

By Favard’s theorem [8] there are complex numbers bn and λn such that

xWn(x)= Wn+1(x)+ bnWn(x)+ λnWn−1(x). (9) Corollary 4. We have λ = A C , b = A + C a2, n n−1 n n n n − where (n + a + b + c + d 1)(n + a + b)(n + a + c)(n + a + d) A = − , n (2n + a + b + c + d 1)(2n + a + b + c + d)  −  n(n + b + c 1)(n + b + d 1)(n + c + d 1) C = − − − n (2n + a + b + c + d 2)(2n + a + b + c + d 1). − − Proof.  From Theorem 3 we derive that 2 2 2 2 2 L (x Wn−1(x )Wn(x )) L (Wn(x ) ) λn = 2 2 = 2 2 = An−1Cn. L (Wn−1(x ) ) L (Wn−1(x ) ) Extracting the coeﬃcient of xn in (9) we have b = [xn−1]W (x) [xn]W (x) (10) n n − n+1 k k where [x ]Wn(x) is the coeﬃcient of x in Wn(x). As k−1 (a + i√x) (a i√x) = ((a + l)2 + x), k − k Yl=0 we derive from (4) that n(a + b + n 1)(a + c + n 1)(a + d + n 1) n−1 [xn−1]W (x)= − − − + (a + l)2, n − a + b + c + d +2n 2 − Xl=0 which yields b = A + C a2 by (10). n n n − According to the general theory of orthogonal polynomials [8], the above recurrence relation is equivalent to the following J-fraction expansion of the generating function for the moment wn(a): ∞ w (a)tn = J(t; A + C a2, A C ,..., n 0 0 − 0 1 n=0 2 X An + Cn a , AnCn+1,...). (11) − Proposition 5. There holds (a + b) (a + c) (a + d) tn w (a)tn = n n n n (a + b + c + d) n (1+(a + l)2t) n≥0 n≥0 n l=0 X X 1 Q a + b, a + c, a + d, 1 = F ;1 . 1+ a2t 4 3 a + b + c + d, a +1+ i/√t, a +1 i/√t − MOMENTS OF ORTHOGONAL POLYNOMIALS 6

n Proof. Set F (a, t)= n≥0 wn(a)t . The recurrence (6) implies that (a + b)(a + c)(a + d) F (a, t)=1+P tF (a +1, t) a2tF (a, t). (12) (a + b + c + d) − Hence 1 (a + b)(a + c)(a + d)t F (a, t)= + F (a +1, t). 1+ a2t (a + b + c + d)(1 + a2t) The formula follows then by iterating the above functional equation.

3. Exponential generating functions 1 If a = 0or a = 2 then there is a nice exponential generating function for the moments of the Wilson polynomials. To derive exponential generating functions from ordinary generating functions we use the following lemma, which we will also apply to other orthgonal polynomials. Let ε : Q[t] Q[t] be the linear transformation deﬁned by −→ tn ε u tn = u . (13) n n n! n ! n X X Lemma 6. For any nonnegative integers m and n we have tm (et 1)m ε = eαt − . (1 αt)(1 (α + 1)t) (1 (α + m)t) m! − − ··· − First proof. Expanding the left side by partial fractions we get tm 1 m m ( 1)k m = − . k=0(1 (α + k)t) m! k 1 (α + m k) t − Xk=0 − − 1 β t The result followsQ from the fact that ε( 1−β t )= e and the binomial theorem.

Second proof. Expanding eαt(et 1)m/m! by the binomial theorem we obtain a linear combination of terms e(α+k)t, k−= 0,...,m, and ǫ−1(e(α+k)t)=1/(1 (α + k)t). So ǫ−1 eαt(et 1)m/m! must be a proper rational function of the form − − N(t) , (1 αt)(1 (α + 1)t) (1 (α + m)t) − − ··· − where the degree of N(t) is at most m. But since the first nonzero coefficient of eαt(et 1)m/m! is tm/m!, N(t) must be tm. − Remark 1. We may also reduce the first proof to α =0 and apply the formula ∞ tn 1 ∞ t n ǫ−1 eαt u = u . (14) n n! 1 αt n 1 αt n=0 ! n=0 X − X − MOMENTS OF ORTHOGONAL POLYNOMIALS 7

If α = m/2 then we have − m t m t m t m (e 1) (2 sinh ) ε = e− 2 t − = 2 . (15a) (1 + m t) (1 m t) m! m! 2 ··· − 2 If m =2n, this is

t2n (2 sinh t )2n ε = 2 , (15b) (1 t2)(1 22 t2) (1 n2 t2) (2n)! − − ··· − and if m =2n + 1 this is t2n+1 (2 sinh t )2n+1 ε = 2 . (15c) (1 ( 1 )2 t2)(1 ( 3 )2 t2) (1 (n + 1 )2 t2) (2n + 1)! − 2 − 2 ··· − 2 Returning to the moments of the Wilson polynomials, for a = 0 we have

∞ ∞ (b) (c) (d) t2n ( 1)nw (0) t2n = ( 1)n n n n . − n − (b + c + d) n (1 l2t2) n=0 n=0 n l=0 X X − So by (15b), Q

∞ t2n ∞ (b) (c) (d) (2 sinh t )2n ( 1)nw (0) = ( 1)n n n n 2 − n (2n)! − (b + c + d) (2n)! n=0 n=0 n X X b, c, d t = F ; sinh2 . 3 2 b + c + d, 1 − 2 2 Replacing t with it, we get

∞ t2n b, c, d t w (0) = F ; sin2 . (16) n (2n)! 3 2 b + c + d, 1 2 n=0 2 X Recall the following contraction formulae [39] transforming an S-fraction to a J- fraction:

S(t; α1,...,αn,...)= J(t; γ0, β1,γ0, β1,...,γn, βn+1,...) (17a) ′ ′ ′ ′ =1+ γ0tJ(t; γ0, β1,...,γn, βn,...) (17b) with γ = α ,γ′ = α + α and for n 1 0 1 0 1 2 ≥

γn = α2n + α2n+1, βn = α2n−1α2n; ′ ′ γn = α2n−1 + α2n, βn = α2nα2n+1. MOMENTS OF ORTHOGONAL POLYNOMIALS 8

Thus, when a = 0 we can transform (11) to the S-continued fraction

∞ bcd (b + c)(c + d)(b + d) w (0)tn = S t; , , , n b + c + d (b + c + d)(b + c + d + 1) ··· n=0 X (b + n)(c + n)(d + n)(b + c + d + n 1) − , (b + c + d +2n 1)(b + c + d +2n) − (n + 1)(b + c + n)(c + d + n)(b + d + n) ,... . (18) (b + c + d +2n)(b + c + d +2n + 1) Similarly, for a =1/2 we have ∞ ∞ (b +1/2) (c +1/2) (d +1/2) ( 1)nt2n w (1/2)tn = n n n − . n (b + c + d +1/2) n (1 (l +1/2)2t2n) n=0 n=0 n l=0 X X − So Q ∞ ∞ (b +1/2) (c +1/2) (d +1/2) t2n+1 ( 1)nw (1/2)t2n+1 = ( 1)n n n n . − n − (b + c + d +1/2) n (1 (l +1/2)2t2n) n=0 n=0 n l=0 X X − Therefore by (15c), Q ∞ ( 1)nt2n+1 ∞ (b +1/2) (c +1/2) (d +1/2) (2 sinh t )2n+1 w (1/2) − = ( 1)n n n n 2 . n (2n + 1)! − (b + c + d +1/2) (2n + 1)! n=0 n=0 n X X Replacing t with it, we may write this as ∞ t2n+1 t b + 1 , c + 1 , d + 1 t w (1/2) = 2 sin F 2 2 2 ; sin2 . (19) n (2n + 1)! 2 3 2 b + c + d + 1 , 3 2 n=0 2 2 X Remark 2. We can also get the exponential generating functions for a =0 and a =1/2 as the following. We have z ∞ z2n cosh 2x arcsin = x2(x2 +12) (x2 +(n 1)2) . 2 ··· − (2n)! n=0 X So with a =0, we have z ∞ (b) (c) (d) z2n L cosh 2x arcsin = n n n . 2 (b + c + d) (2n)! n=0 n X t Setting z = 2 sin 2 , we get ∞ t2n L (cosh(xt)) = L (x2n) (2n)! n=0 X b, c, d t = F ; sin2 . (20) 3 2 b + c + d, 1 2 2 MOMENTS OF ORTHOGONAL POLYNOMIALS 9

Similarly, z sinh 2x arcsin 2 ∞ 2n+1 2 2 2 z = x x2 + 1 x2 + 3 x2 + n 1 . 2 2 ··· − 2 (2n + 1)! n=0 X 1 So with a = 2 we have ∞ t2n+1 L x−1 sinh(xt) = L (x2n) (2n + 1)! n=0 X t b + 1 , c + 1 , d + 1 t =2 sin F 2 2 2 ; sin2 . 2 3 2 b + c + d, 1 2 2 4. Continuous dual Hahn polynomials If we take the limit as d in the Wilson polynomials we get the continuous dual →∞ Hahn polynomials pn(x) := pn(x; a, b, c) deﬁned by

2 n n, a + ix, a ix pn(x )=( 1) (a + b)n(a + c)n 3F2 − − ;1 . (21) − a + b, a + c 2 The ﬁrst two values of pn(x ; a, b, c) are the following: p (x2)= x2 (ab + bc + ca) 1 − p (x2)= x4 [1 + 2(a + b + c)+2(ab + ac + bc)]x2 2 − + a2b2 +2 a2bc + a2c2 +2 b2ac +2 c2ab + b2c2 + a2b + a2c + b2a +4 abc + ac2 + b2c + bc2 + ab + ac + bc. When either a, b, and c are all positive or one is positive and the other two are complex conjugates with positive real parts, Wilson’s result [40] reduces to

1 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix) 2 p (x2)p (x2) dx 2π Γ(2ix) m n Z0

= Γ(n + a + b)Γ( n + a + c)Γ(n + b + c)n! δmn. (22)

The corresponding moments µn(a, b, c) are given by by 1 ∞ Γ(a + ix)Γ(b + ix)Γ(c + ix) 2 x2ndx 2π Γ(2ix) Z0

= µn( a, b, c)Γ(a + b)Γ(a + c)Γ(b + c). (23)

De Branges [12] ([2, p. 152] and [1]) also proved that µ0(a, b, c) = 1. The counterpart of (6) reads as follows µ (a, b, c)=(a + b)(a + c)µ (a +1,b,c) a2µ (a, b, c), (24) n+1 n − n MOMENTS OF ORTHOGONAL POLYNOMIALS 10 which is equivalent to the generating function (a + b) (a + c) tn µ (a, b, c)tn = n n n n (1+(a + l)2t) n≥0 n≥0 l=0 X X 1Q a + b, a + c, 1 = F ;1 . (25) 1+ a2t 3 2 a +1+ i/√t, a +1 i/√t − We remark that the above recurrence is nothing but the deﬁnition of a sequence of polynomials introduced by Dumont and Foata [13, 7, 43] as an extension of Genocchi numbers. In this context the following result was conjectured by Gandhi [19] in 1970 and ﬁrst proved by Carlitz [6], and Riordan and Stein [35]. Here we provide a direct proof starting from (23). Proposition 7. For n 0 we have ≥ µn(1, 1, 1) = G2n+4. (26) Proof: Since Γ(ix) 2 = π and Γ(1 + ix) 2 = πx , we have | | x sinh(πx) | | sinh(πx) 1 ∞ (Γ(1 + ix))3 2 µ (1, 1, 1) = x2n dx n 2π Γ(2ix) Z0 ∞ 2n+4 cosh(πx) =2π x 2 dx 0 sinh (πx) Z ∞ = 2 x2n+4d(1/(sinh(πx))). − Z0 Integrating by parts yields ∞ x2n+3dx µ (1, 1, 1) = 4(n + 2) . n sinh(πx) Z0 Equation (26) follows then from the known integral expression of Bernoulli numbers [17, p. 39]: 2n ∞ x2n−1dx B = , | 2n| 22n 1 sinh(πx) − Z0 and the formula G = 2(22n 1) B . 2n − | 2n| From the formula for Wilson polynomials we derive the recurrence for the continuous dual Hahn polynomials pn(x) p (x)=(x b )p (x) λ p (x) (n 0), (27a) n+1 − n n − n n−1 ≥ where b =(n + a)(n + b)+(n + a)(n + c)+(n + b)(n + c) n(n + 1), n − (27b) λ = n(n 1+ a + b)(n 1+ a + c)(n 1+ b + c). ( n − − − MOMENTS OF ORTHOGONAL POLYNOMIALS 11

Proposition 8. Let ϕ : K[x] K be the linear functional such that ϕ(xn) = µ (a, b, c) for n 0. Then −→ n ≥ ϕ(pm(x)pn(x)) = n!(a + b)n(a + c)n(b + c)n δmn. (27c) From the known facts of orthogonal polynomials we derive the following J-fraction expansion for the generating function of the moments:

n µn(a, b, c)t = J(t;(ab + bc + ca), (a + b)(b + c)(c + a),... n≥0 X (a + n)(b + n)+(b + n)(c + n)+(c + n)(a + n) n(n + 1), (n + 1)(a + b + n)(b + c + n)(c + a + n),...). (27d) − By (18) the S-fraction for µn(0,b,c) is ∞ n µn(0,b,c)t = S(t; bc, b + c, (b + 1)(c + 1), 2(b + c + 1),...), (28) n=0 X as is well-known, and b = c = 1 gives the Genocchi numbers. In other words, the Genocchi numbers G2n+2 are the moments of the continuous dual Hahn polynomials pn(x, 0, 1, 1). As we recall the exponential generating function was given by Carlitz [7]. By (16) the corresponding generating functions for a = 0 is ∞ t2n b, c t µ (0,b,c) = F ; sin2 . (29) n (2n)! 2 1 1 2 n=0 2 X In particular, the exponential generating function of Genocchi numbers G2n+2 = µn(0, 1, 1) also has the hypergeometric series representations: ∞ t2n 1, 1 t 1+ G = F ; sin2 2n+2 (2n)! 2 1 1 2 n=1 2 X t 1, 1 t = sec2 F − 2 ; tan2 , 2 2 1 1 − 2 2 where the last formula follows from Pfaﬀ’s transformation (see [2, p. 68]). 1 The J-fraction for µn( 2 ,b,c) is

∞ 1 µ ( 1 ,b,c)tn = J(t; bc + (b + c), (b + c)(b +1/2)(c +1/2), n 2 2 n=0 X 5 bc + (b + c)+2 12, (b + c + 1)(b +3/2)(c +3/2), 2 · 9 bc + (b + c)+2 22, (b + c + 2)(b +5/2)(c +5/2),...). (30) 2 · MOMENTS OF ORTHOGONAL POLYNOMIALS 12

1 By (19) the corresponding generating functions for a = 2 is ∞ t2n+1 t b + 1 , c + 1 t µ ( 1 ,b,c) = 2 sin F 2 2 ; sin2 . (31) n 2 (2n + 1)! 2 2 1 3 2 n=0 2 X The following two nice special cases of (29) and (31) are known (see[17, p. 101]):

cos(at/2) (1 a)/2, (1 + a)/2 2 t = 2F1 − ; sin , (32) cos t/2 1/2 2 sin(at/2) (1 a)/2, (1 + a)/2 2 t = 2F1 − ; sin . (33) a sin t/2 3/2 2 To ﬁnd the orthogonal polynomials whose moments are median Genocchi numbers we consider a general rescaled version of the continuous dual Hahn polynomials:

Zn(x) := Zn(x; α, β, γ, α,¯ β,¯ γ¯)= pn(x + d; a, b, c), (34) where α, β, γ, α,¯ β,¯ γ¯ are six parameters such that a = 1 (α +α ¯ + β β¯ +¯γ γ), 2 − − b = 1 (¯α α + β + β¯ + γ γ¯),  2 − −  1 (35) c = (α α¯ β + β¯ + γ +¯γ),  2 − − d = αα¯ + α(β β¯) α¯(γ γ¯) a2.  − − − −  From the corresponding results for the continuous dual Hahn polynomials we derive n n Z (x2)= ( 1)n−k (¯a + β + k) (α +¯γ + k) n − k n−k n−k Xk=0 k−1 x2 +(α + l)(¯α + l)+(α + l)(β β¯) (¯α + l)(γ γ¯) . (36) × − − − l=0 Y Proposition 9. The polynomials Zn(x) have the recurrence relation Z (x)=(x b )Z (x) λ Z (x) (n 0), (37a) n+1 − n n − n n−1 ≥ with

b =(n + α)(n + β¯)+(n +α ¯)(n + γ)+(n + β)(n +¯γ) n(n + 1), n − (37b) λ = n(n 1+¯α + β)(n 1+ α +¯γ)(n 1+ β¯ + γ). ( n − − −

Let Γn+1(α, α¯) := Γn+1(α, α,¯ β, β,¯ γ, γ¯) be the moments of the polynomials Zn(x)= pn(x + d; a, b, c). Then the above recurrence implies that the generating function of MOMENTS OF ORTHOGONAL POLYNOMIALS 13

Γn(α, α¯) has the following J-fraction expansion

n Γn+1(α, α¯)t = J(t; αβ¯ + βγ¯ + γα,¯ (¯α + β)(β¯ + γ)(¯γ + α),..., n≥0 X (α + n)(β¯ + n)+(β + n)(¯γ + n)+(γ + n)(¯α + n) n(n + 1), − (n + 1)(¯α + β + n)(β¯ + γ + n)(¯γ + α + n),...). (38)

n Moreover, let ψ : K[x] K be a linear functional such that ψ(x )=Γn+1(α, α¯) for n 0. Then −→ ≥ ψ(Zm(x)Zn(x)) = n!(¯α + β)n(α +¯γ)n(β¯ + γ)nδmn. (39)

On the other hand, it is easy to see that the moments of pn(x + d; a, b, c) are related to those of pn(x; a, b, c) as follows: n n Γ (α, α¯)= ϕ((x d)n)= ( d)n−kµ (a, b, c). (40) n+1 − k − k Xk=0 Therefore n n Γ (α, α¯)tn = ( d)n−kµ (a, b, c) tn n+1 k − k n≥0 n≥0 X X Xk=0 n + k = µ (a, b, c)tk ( d t)n k k − n≥0 Xk≥0 X k −k−1 = µk(a, b, c)t (1 + dt) . Xk≥0 Invoking (35) we derive from (25) the generating function

n Γn+1(α, α¯)t n≥0 X (α +¯γ) (β +α ¯) tn = n n . (41) n 1 [(α + k)(β¯ β) (¯α + k)(¯γ γ) (α + k)(¯α + k)]t n≥0 k=0 X − − − − − Clearly, the aboveQ equation is equivalent to the following recurrence:

Γn+1(α, α¯)=(α +¯γ)(β +α ¯)Γn(α +1, α¯ + 1) + [α(β¯ β) α¯(¯γ γ) αα¯]Γ (α, α¯) (42) − − − − n with Γ1(α, α¯) = 1. Remark 3. As a refinement of the Dumont-Foata polynomials, Dumont [15] conjectured a combinatorial interpretation of Γn(α, α¯) n in (38). This conjecture was first proved by Randrianarivony [34] and Zeng{ [43] independently.} Another proof of Du- mont’s conjecture was later given by Josuat-Vergès [22]. MOMENTS OF ORTHOGONAL POLYNOMIALS 14

By contracting the S-fraction (3) using (17a) we obtain the following J-fraction ∞ H tn = J(t;2 12, (1 2)2, 2 22, (2 3)2,..., 2 n2, (n (n + 1))2,...). (43) 2n+1 · · · · · · n=0 X Comparing with (38) we see that H2n+1 = Γn+1(1, 1, 1, 0, 1, 1). Hence, the median Genocchi numbers H are the moments of the rescaled continuous dual Hahn { 2n+1} polynomials pn(x 1/4, 1/2, 1/2, 1/2). Our method does not produce an exponential generaring function− for the median Genocchi numbers here since the denominator in (41) with α =1andα ¯ = 0 does not factorize nicely.

5. Hahn polynomials In this section we introduce another method for determining generating functions for moments of orthogonal polynomials. We apply it only to the Hahn polynomials, but it can also be used for the Wilson polynomials.

Lemma 10. Let a0, a1, a2,... be arbitrary. Then ∞ tn 1 (x + a )(x + a ) (x + a ) = . (44) 0 1 ··· n−1 (1 + a t) (1 + a t) 1 xt n=0 0 n X ··· − Proof. We have the indefinite sum m tn (x + a ) (x + a ) 0 ··· n−1 (1 + a t) (1 + a t) n=0 0 n X ··· 1 tm+1 = 1 (x + a ) (x + a ) , (45) 1 xt − 0 ··· m (1 + a t) (1 + a t) − 0 ··· m which is easily proved by induction. The lemma follows by taking m . →∞ Remark 4. As pointed out by Knuth [24, equation (2.16)], a formula equivalent to Lemma 10 was discovered by Fran¸cois Nicole [31] in 1727. Nicole’s formula was also used by Apéry to derive the formula 5 ∞ ( 1)n−1 ζ(3) = − ; 2 3 2n n=1 n n X see van der Poorten [33, section 3]. The identity in the proof is equivalent to a special case of Newton’s interpolation formula (see [30]). Moreover, equating coefficients of tn in (44) gives the expansion of the polynomial xn in the basis (x + a ) (x + { 0 ··· ak) 0≤k≤n−1, in which the coefficents can be computed by Newton’s interpolation formula.} We give such an example for the moments of Askey-Wilson polynomials at the end of this paper. The following result is an immediate consequence of Lemma 10. MOMENTS OF ORTHOGONAL POLYNOMIALS 15

Lemma 11. Let L be a linear functional defined on polynomials in x and let a0, a1,... be complex numbers or indeterminates that do not involve x. Let v = L (x + a )(x + a ) (x + a ) . n 0 1 ··· n−1 Then ∞ ∞ tn L(xn)tn = v . n (1 + a t) (1 + a t) n=0 n=0 0 n X X ··· The Hahn polynomials are defined in [25] by n, n + α + β +1, x Q (x; α,β,N)= F ;1 . (46) n 3 2 − α +1, N − − (Some authors define Q (x; α,β,N) with N + 1 replacing N in (46).) If N is n − − a nonnegative integer then the polynomials Qn(x; α,β,N) are defined only for n = 0, 1,...,N and it is known that they are orthogonal with respect to the linear functional L0 given by N α + x β + N x L (p(x)) = − p(x). (47) 0 x N x x=0 − Applying Vandermonde’s theoremX we find that

(α + β + 2)N (α + 1)n(α + β + N + 2)n L0 (x + α + 1)n = . N! · (α + β + 2)n This suggests that in studying moments of Hahn polynomials we reparametrize them by setting α = A 1, β = C A 1, and N = B C so that α+1 = A, α+β+N +2 = B, and α + β +2=−C. − − − Thus we deﬁne polynomials Rn(x; A,B,C) by R (x; A,B,C)= Q (x; A 1,C A 1, B C) n n − − − − n, n + C 1, x = F ;1 3 2 − A, C −B − − where A, B, and C are indeterminates. We will show that these polynomials are orthogonal with respect to the linear functional L on polynomials in x deﬁned by

(A)m(B)m L (x + A)m = . (48) (C)m The polynomials Rn(x; A,B,C) are closely related to the continuous Hahn polynomials, deﬁned in [25] by

n (a + c)n(a + d)n n, n + a + b + c + d 1, a + ix pn(x; a,b,c,d)= i 3F2 − − ;1 . n! a + c, a + d Thus n! R (x; A,B,C)=( i)n p (ix;0, B A, A, C B) n − (A) (C B) n − − n − n MOMENTS OF ORTHOGONAL POLYNOMIALS 16 and

(a + c) (a + d) p (x; a,b,c,d)= in n n R ( a ix; a + c, b + c, a + b + c + d). (49) n n! n − −

Lemma 12. For nonnegative integers m and n we have

(A)m+n(B)m(C B)n L (x + A)m( x)n = − . (50) − (C)m+n Proof. By Vandermonde’s theorem we have

n n ( x) = ( 1)i (x + A + m) (A + m + i) . − n − i i n−i i=0 X Thus

n n L (x + A) ( x) = L ( 1)i (x + A) (A + m + i) m − n − i m+i n−i i=0 X n n (A) (B) = ( 1)i m+i m+i (A + m + i) − i (C) n−i i=0 m+i X (A) (B) (C B) = m+n m − n (C)m+n by Vandermonde’s theorem again.

Proposition 13. The polynomials Rn(x; A,B,C) are orthogonal with respect to the linear functional L, and

(B) (C A) L R (x; A,B,C)2 =( 1)nn! n − n . (51) n − (A) (C B) (C) (C +2n 1) n − n n−1 − Proof. To show that the polynomials Rn are orthogonal, it suﬃces to show that for 0 m < n, ≤

L(Rn(x; A,B,C)(x + A)m)=0. MOMENTS OF ORTHOGONAL POLYNOMIALS 17

We have n ( n)k(n + C 1)k L(Rn(x; A,B,C)(x + A)m)= L − − ( x)k(x + A)m k!(A)k(C B)k − ! Xk=0 − n ( n) (n + C 1) (A) (B) (C B) = − k − k m+k m − k k!(A)k(C B)k (C)m+k Xk=0 − (A)m(B)m n, n + C 1, A + m = 3F2 − − ;1 (C) A, C + m m (A) (B) ( m) (A n C + 1) = m m − n − − n (C) (A) ( m n C + 1) m n − − − n (A) (B) (C A) ( m) = m m − n − n , (52) (C)m+n(A)n where we have used Saalsch¨utz’s theorem to evaluate the F . So if 0 m < n this 3 2 ≤ is 0. Since Rn(x; A,B,C) has leading coeﬃcient (C 1)2n/(A)n(C B)n(C 1)n, the case m = n of (52) gives (51). − − −

n Proposition 14. Let Mn(A,B,C)= L(x ). Then ∞ ∞ (A) (B) tn M (A,B,C)tn = n n , (53) n (C) n (1+(A + l)t) n=0 n=0 n l=0 X X and Q ∞ tn A, B M (A,B,C) = e−At F ;1 e−t . (54) n n! 2 1 C − n=0 X Proof. Equation (53) follows from Proposition 11 and equation (48). By Lemma 6 we have tn (et 1)n (1 e−t)n ε = e−(A+n)t − = e−At − n (1+(A + l)t) n! n! l=0 so applying ε to (53) gives (54). Q

Theorem 15. The following S-fraction holds:

∞ A(B C) B(C A) M (A,B,C)tn = S t; − , − , n C C(C + 1) n=0 X (A + 1)(B C 1)C 2(B + 1)(C A + 1) − − , − , (C + 1)(C + 2) (C + 2)(C + 3) (A + 2)(B C 2)(C + 1) 3(B + 2)(C A + 2) − − , − ,... . (55) (C + 3)(C + 4) (C + 4)(C + 5) MOMENTS OF ORTHOGONAL POLYNOMIALS 18

∞ n Proof. Let F (t; A,B,C)= n=0 Mn(A,B,C)t . By (53) we have 1 t F (t; A,B,CP )= G ; A,B,C (56a) 1+ At 1+ At with ∞ (A) (B) tn G(t; A,B,C)= n n . (56b) (C) n (1 + l t) n=0 n l=0 X We ﬁrst expand G(t; A,B,C) as an S-fraction Q G(t; A,B,C)= S(t; C1,C2,...,Cn,...), which can be written as a J-fraction

G(t; A,B,C)= J(t; C1,C1C2,C2 + C3,C3C4,...), (57) where Cn := Cn(A,B,C) (n 1) are to be determined. Rewriting (56b) as ≥ AB t t G(t; A,B,C)=1+ G ; A +1, B +1,C +1 C · 1+ t · 1+ t and using the J-fraction (57) to write 1 t G ; A +1, B +1,C +1 1+ t · 1+ t = J(t; C+ 1,C+ C+,C+ + C+ 1,C+ C+,...) 1 − 1 2 2 3 − 3 4 + with Cn = Cn(A +1, B +1,C + 1), we obtain AB G(t; A,B,C)=1+ t J(t; C+ 1,C+ C+,C+ + C+ 1,C+ C+,...). (58) C · 1 − 1 2 2 3 − 3 4 On the other hand, contracting the S-fraction by (17b) we have G(t; A,B,C)=1+ C t J(t; C + C ,C C ,C + C ,C C ,...). (59) 1 · 1 2 2 3 3 4 4 5 Comparing the above two J-fractions we derive AB C = , C + C = C+ 1, C C = C+ C+, 1 C 1 2 1 − 2 3 1 2 and for n 2 ≥ C + C = C+ + C+ 1, 2n−1 2n 2n−2 2n−3 − + + C2n C2n+1 = C2n−1 C2n+2. This yields immediately (A + n 1)(B + n 1)(C + n 2) C = − − − ; 2n−1 (C +2n 3)(C +2n 2)  n(B C n +− 1)(A C n−+ 1) (60) C2n = − − − − .  (C +2n 2)(C +2n 1) − −   MOMENTS OF ORTHOGONAL POLYNOMIALS 19

Next, by (56a) and (57) we have the J-fraction for F (t; A,B,C): F (t; A,B,C)= J(t; C A, C C ,C + C A, C C ,...). (61a) 1 − 1 2 2 3 − 3 4 ′ It remains to determine a sequence αns such that

F (t; A,B,C)= S(t; α1,α2,...,αn,...) (61b)

= J(t; α1,α1α2,α2 + α3,α3α4,...). (61c) From (60), (61c) and (61a) we derive A(B C) α = C A = − 1 1 − C C1C2 B(C A) α2 = = − . α1 C(C + 1) and for n 2, ≥ α = C + C A 2n−1 2n−1 2n−2 − (A + n)(B C n)(C + n 1) = − − − , (C + n)(C + n + 1)

C2nC2n−1 α2n = α2n−1 n(B + n 1)(C A + n 1) = − − − . (C +2n 1)(C +2n 2) − − Plugging these values in (61b) yields the S-fraction in Theorem 15.

As an example of Proposition 14 let us consider the case A =1, B = 1,C = 2. We have F (1, 1; 2; z)= z−1 log(1 z), so 2 1 − − ∞ 1, 1 t tn e−t F ;1 e−t = = B , 2 1 2 − et 1 n n! n=0 − X where Bn is the nth Bernoulli number. So Mn(1, 1, 2) = Bn. The orthogonal polynomials whose moments are Bernoulli numbers were considered by Touchard [37], but he was not able to ﬁnd an explicit formula for them. An explicit formula (diﬀerent from ours, but equivalent) was found by Wyman and Moser [41] and these polynomials were further studied by Carlitz [5]. One can show similarly that Mn(1, 2, 3) = 2Bn+1 and Mn(1, 1, 3) = 2(Bn + Bn+1). Krattenthaler [26, Section 2.7] proved a result− equivalent to et ∞ tn ∞ tn M (2, 2, 4) = B . 6 n n! n+2 n! n=0 n=0 X X MOMENTS OF ORTHOGONAL POLYNOMIALS 20

Fulmek and Krattenthaler [27, equation (5.14)] expressed the moments of general continuous Hahn polynomials (with some integrality and nonnegativity restrictions on the parameters) in terms of Bernoulli numbers, generalizing all of these formulas. Chapoton [9] studied some special cases of Racah polynomials whose moments are somehow the median Bernouli numbers since they are, up to an easy power of 2, the main diagonal of Seidel’s diﬀerence tableau of Bernoulli numbers [36, p.181]. For the moments µn of the original Hahn polynomials

Qn(x; α,β,N)= Rn(x; α +1,α + β +2,α + β + N + 2), equation (54) gives ∞ tn α +1, α + β + N +2 µ = e−(α+1)t F ;1 e−t . n n! 2 1 α + β +2 − n=0 X Applying Pfaﬀ’s transformation gives ∞ n t α +1, N t µn = 2F1 − ;1 e . n! α + β +2 − n=0 X If N is a nonnegative integer, we may apply the terminating 2F1 transformation

a, N (b a)N a, N 2F1 − ; z = − 2F1 − ;1 z b (b) 1+ a b N − N − − to obtain ∞ n t (β + 1)N α +1, N t µn = 2F1 − ; e . n! (α + β + 2) β N n=0 N X − − Equating coeﬃcients of xn/n! gives

N (β + 1)N (α + 1)k( N)k n µn = − k (α + β + 2)N k!( β N)k Xk=0 − − N! n α + k β + N k = − kn, (α + β + 2)N k N k Xk=0 − so we see that, as expected, in this case the linear functional L is a constant multiple of the linear functional L0 deﬁned by (47).

6. Concluding remarks The Askey-Wilson polynomials [3] are an important q-analogue of the Wilson polynomials. In the last decade much work have been done to extend Viennot’s results for moments of classical orthogonal polynomials to the moments of the Askey-Wilson polynomials; see [10, 11, 23, 28]. It would be interesting to see to what extent the methods of this paper can be q-generalized. Let us just pointing out that Lemma 11 MOMENTS OF ORTHOGONAL POLYNOMIALS 21 works also for the moments of the Askey-Wilson polynomials. In what follows the standard q-notations (see [2, 25]) will be used. For 0

n 1 (ab, ac, ad, bc, bd, cd; q)∞ Lq x = 2π (abcd; q)∞ π (cos θ)n (e2iθ, e−2iθ; q) dθ ∞ . (aeiθ, ae−iθ, beiθ, be−iθ, ceiθ, ce−iθ, deiθ, de−iθ; q) Z0 ∞ iθ −iθ Thus, the Askey-Wilson integral reads Lq(1)=1. As (az, a/z; q)n = (ae , ae ; q)n, n the value Lq (az, a/z; q)n amounts to shifting a to aq in the integral, so

(ab, ac, ad; q)n Lq (az, a/z; q)n = . (62) (abcd; q)n −j j As x =(z +1/z)/2, choose aj = t +(q /a + aq )/2 for j N and t C, then ∈n ∈ (t + x a ) (t + x a )=( 1)n(2a)−nq−(2)(az, a/z; q) . − 0 ··· − n−1 − n It follows from Lemma 11 that ∞ ∞ n (ab, ac, ad; q) ( 1)n(2a)−nq−(2)un L (t + x)n un = n − . (63) q (abcd; q) (1 a u) ... (1 a u) n=0 n=0 n − 0 − n X X It is interesting to note that computing the coeﬃcent of un in the right side of (63) by partial fraction decomposition or using Newton’s interpolation formula [28] yields the double sum formula for the Askey-Wilson moments in [11, Theorem 1.13].

References [1] Askey (R.). Beta integrals and the associated orthogonal polynomials. In: Number Theory (ed. K. Alladi). Lecture Notes in Mathematics 1395, Springer-Verlag, New York, 1989, 84–121. [2] Andrews (G. E.); Askey (R.); Roy (R.). Special functions. Encyclopedia of Mathematics and its Applications, 71. Cambridge University Press, Cambridge, 1999. xvi+664 pp. [3] Askey (R.) and Wilson (J.). Some basic hypergeometric orthogonal polynomials that generalize Jacobi polynomials, Memoirs of AMS. 54 (1985). [4] Bigeni (A.). A generalization of the Kreweras triangle through the universal sl2 weight system, J. Comb. Theory Ser. A 161 (2019), 309–326. [5] Carlitz (L.). Some polynomials of Touchard connected with the Bernoulli numbers, Canadian J. Math. 9 (1957), 188–190. [6] Carlitz (L.). A conjecture concerning Genocchi numbers, Koninkl. norske Vidensk. Selsk. Sk. 9 (1972), 1–4. [7] Carlitz (L.). Explicit formulas for the Dumont-Foata polynomials, Discrete Math. 30 (1980), 211–225. [8] Chihara (T.S.). An introduction to orthogonal polynomials, Gordon and Breach, New- York/London/Paris, 1978. MOMENTS OF ORTHOGONAL POLYNOMIALS 22

[9] Chapoton (F.). Ramanujan-Bernoulli numbers as moments of Racah polynomials, J. Théor. Nombres Bordeaux 32 (2020), no. 1, 205–215. [10] Corteel (S.), Kim (J.S.), Stanton (D.). Moments of orthogonal polynomials and combinatorics, in: Recent Trends in Combinatorics, in: The IMA Volumes in Mathematics and Its Applications, vol. 159, 2016. [11] Corteel (S.), Stanley (R.), Stanton (D.), Williams (L.). Formulae for Askey-Wilson moments and enumeration of staircase tableaux. Trans. Amer. Math. Soc. 364 (2012), no. 11, 6009– 6037. [12] De Branges (L.). Gauss spaces of entire functions, J. Math. Anal. Appl. 37 (1972), 1–41. [13] Dumont (D.) and Foata (D.). Une propriétéde symétrie des nombres de Genocchi, Bull. Soc. Math. France 104 (1976), no. 4, 433–451. [14] Dominici (D.). Polynomial sequences associated with the moments of hypergeometric weights, SIGMA Symmetry Integrability Geom. Methods Appl. 12 (2016), Paper No. 044, 18 pp. [15] Dumont (D.). Conjectures sur des symétries ternaires liées aux nombres de Genocchi, Discrete Math. 139 (1995), 469–472. [16] Dumont (D.) and Zeng (J.). Further results on the Euler and Genocchi numbers, Aequationes Math. 47 (1994), 31–42. [17] Erdely´ et al. (A.). Higher transcendental functions, Vol. 1 of Bateman Manuscript Project. McGraw-Hill Book Company, New-York, 1953. [18] Feigin (E.). The median Genocchi numbers, q-analogues and continued fractions, Europ. J. Combin. 33 (2012), 1913–1918. [19] Gandhi (J. M.). A conjectured representation of Genocchi numbers, Amer. Math. Monthly 77 (1970), 505–506. [20] Gessel (I. M.). Applications of the classical umbral calculus. Algebra Universalis 49 (2003), no. 4, 397–434. [21] Hetyei (G.). Alternation acyclic tournaments. European J. Combin. 81 (2019), 1–21. [22] Josuat-Verges` (M.). Generalized Dumont–Foata polynomials and alternative tableaux. Sém. Lothar. Combin. 64 (2011) 17 pp. Art. B64b. [23] Kim (J.S.), Stanton (D.). Moments of Askey-Wilson polynomials. J. Combin. Theory Ser. A 125 (2014), 113–145. [24] Knuth (D. E.). Two notes on notation, Amer. Math. Monthly 99 (1992), no. 5, 403–422. [25] Koekoek (R.), Lesky (P.) and Swarttouw (R.). Hypergeometric orthogonal polynomials and their q-analogues. With a foreword by Tom H. Koornwinder. Springer Monographs in Mathemat- ics. Springer-Verlag, Berlin, 2010. xx+578 pp. [26] Krattenthaler (C.). Advanced determinant calculus, Sém. Lothar. Combin. 42 (1999), Art. B42q, 67 pp. [27] Fulmek (M.) and Krattenthaler (C.). The number of rhombus tilings of a symmetric hexagon which contain a fixed rhombus on the symmetry axis. II, European J. Combin. 21 (2000), 601–640. [28] Guo (V. J. W.), Ishikawa (M.), Tagawa (T.), Zeng (J.), A quadratic formula for basic hypergeometric series related to Askey-Wilson polynomials. Proc. Amer. Math. Soc. 143 (2015), no. 5, 2003–2015. [29] Lazar (A.) and Wachs (M. L.). On the homogenized Linial arrangement: intersection lattice and Genocchi numbers, Sém. Lothar. Combin. 82B (2020), Art. 93, 12 pp. [30] Milne-Thomson (L. M.). The Calculus of Finite Differences. Macmillan and Co., Ltd., London, 1951. [31] Nicole (F.), Méthode pour sommer une infinitéde Suites nouvelles, dont on ne peut trouver les Sommes par les Méthodes connuës, Mémoires de l’Academie Royale des Sciences, 1727, 257–268. MOMENTS OF ORTHOGONAL POLYNOMIALS 23

[32] Njionou Sadjang (P.), Koepf (W.), Foupouagnigni (M.). On moments of classical orthogonal polynomials, J. Math. Anal. Appl. 424 (2015), no. 1, 122–151. [33] van der Poorten (A. J.). A proof that Euler missed . . . Apéry’s proof of the irrationality of ζ(3), Math. Intelligencer 1 (1978/79), no. 4, 195–203. [34] Randrianarivony (A.). Sur une extension des polynômes de Dumont-Foata, Sém. Lothar. Com- bin. 32 (1994), Art. B32d, approx. 12 pp. [35] Riordan (J.) and Stein (P.). Proof of a conjecture on Genocchi numbers, Discrete Math. 5 (1973), 381–388. [36] Seidel (L.). Uber¨ eine einfache entstehungsweise der Bernoullischen zahlen und einiger verwandten reihen, Sitzungsber. Münch. Akad. 4 (1877), 157–187, http://publikationen.badw.de/de/003384831. [37] Touchard (J.), Nombres exponentiels et nombres de Bernoulli, Canad. J. Math. 8 (1956 ), 305–320. [38] Viennot (G.). Une théorie combinatoire des nombres d’Euler et Genocchi, Séminaire de Théorie des nombres de l’UniversitéBordeaux, Exposéno. 11, 1980–1981, Publications de l’Université Bordeaux I. [39] Viennot (G.). Une théorie combinatoire des polynômes orthogonaux généraux, Lecture Notes, Universitédu Québec àMontréal, 1983. [40] Wilson (J.). Some hypergeometric orthogonal polynomials, SIAM J. Anal. 11 (1980), 690–701. [41] Wyman (M.) and Moser (L.). On some polynomials of Touchard, Canad. J. Math. 8 (1956), 321–322. [42] Zeng (J.). Weighted derangements and the linearization coefficients of orthogonal Sheffer polynomials. Proc. London Math. Soc. (3) 65 (1992), no. 1, 1–22. [43] Zeng (J.). Sur quelques propriétés de symétrie des nombres de Genocchi, Discrete Math. 153 (1996), no. 1-3, 319–333.

(Ira M. Gessel) Department of Mathematics, Brandeis University, Waltham, MA 02453, USA Email address: [email protected]

(Jiang Zeng) Universite´ de Lyon; Universite´ Lyon 1; Institut Camille Jordan; UMR 5208 du CNRS; 43, boulevard du 11 novembre 1918, F-69622 Villeurbanne Cedex, France Email address: [email protected]