Binomial Polynomials Mimicking Riemann's Zeta Function

Home , Chebyshev polynomials

arXiv:1703.09251v2 [math.NT] 14 Jan 2020 n h ioilcecet adtu aclstinl) A triangle). Pascal’s thus (and coefficients binomial the and polynomials hteitbtenbnma offiins ucin hc on which line functions understan the coefficients, further on binomial to between is exist work present that this for motivation The Introduction 1 elntasom,hpremti functions. hypergeometric transforms, Mellin e od n phrases and words Key (secondary). 2010 suggestions. and ioilPlnmasMmcigReansZt Function Zeta Riemann’s Mimicking Polynomials Binomial h uhr ol iet hn rJ .HnmrhadPo.MN M Prof. and Hindmarsh L. J. Dr thank to like would authors The Too h otuiutu bet nmteaisaetese the are mathematics in objects [17] ubiquitous Stolarsky, most B. the of K. “Two and Dilcher K. by stated As ahmtc ujc Classification Subject Mathematics ihol d rm atr.Terslstuho nltcnme th number analytic on touch results The combinatorics. and factors. theory, prime function odd only with aiiso h eaiera xs oevras Moreover axis. real negative the on larities with q oil) h rnfrsaeietfidi em fcmiaoilsu of ‘critical terms combinatorial Their in forms. of identified coefficient terms then binomial are in S:3/1 factors and identified S:4/2 are S:4/3, transforms Gould’s The nomials). factors eetn hs eut oa1prmtrfml fplnmaswt z relation with functional polynomials the of possess family that 1-parameter line a critical to results these extend we n λ ( s omlsto ilstertoa function rational the yields Normalisation o h hbse oyoil eoti h ipe :/ ioilfor binomial S:2/1 simpler the obtain we polynomials Chebyshev the For h gnrlsd elntasom fGgnae oyoil have polynomials Gegenbauer of transforms Mellin (generalised) The ) C → n ,adpienumbers. prime and ), ℜ p the n λ rmbelow. from 1 s ( s n ,woezrsalleo h ciia line’ ‘critical the on lie all zeros whose ), 1 = hCtlnnme,w eue4 deduce we number, Catalan th rtclplnmas ioilcecet,Gudcombin Gould coefficients, binomial polynomials, critical : / rzrso h elln,adhneot eerdt as to referred henceforth and line, real the on zeros or 2 .W Coffey W. M. 16 piay,0A0 32,3C5 20,4A0 30D05 44A20, 42C05, 33C45, 33C20, 05A10, (primary), 11B65 : aur 0 2020 20, January a Abstract 3 n .C Lettington C. M. and F 2 1 yegoercfntos Furthermore, functions. hypergeometric (1) 1 C n − q p s 1 n λ n p ( ( ∞ → 2 s s n hs eoiao a singu- has denominator whose ) ; ( β s ℜ = ) and ) s ln h oiiera axis, real positive the along 1 = ± h rageo connections of triangle the d ulyfrterhlflcomments helpful their for Huxley p yhv rtclzrs(those zeros critical have ly C oncinbtentetwo the between connection n / n (1 cle rtclpoly- critical (called 2 p b 2 uneo rm numbers prime of quence n − +1 s ( ; rsol nthe on only eros s β il integers yield ) ). srltdto related ms tra summations, atorial oy special eory, polynomial’ polynomial ,and m, critical is given by a well-known characterisation of the prime numbers: Consider the entries in the kth row of Pascal’s triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime”. By considering a modified form of Pascal’s triangle, whose kth row consists of the integers

(2k 1)(2k + 1) k + j a(k, j) := − , k N, 0 j k 1, (1) 2j + 3 2j + 1 ∈ ≤ ≤ − Dilcher and Stolarsky obtained an analogous characterisation of pairs of twin prime numbers (2k 1, 2k + 1). This says that the entries in the kth row of the a(k,s) number − triangle are divisible by 2k 1 with exactly one exception, and are divisible by 2k +1 with − exactly one exception, if and only if (2k 1, 2k + 1) is a pair of twin prime numbers. − The corresponding sequence of polynomials Ak(x) obtained from the kth row of the number triangle generated by the integers a(k, j) is given by

k 1 k 1 − − (2k 1)(2k + 1) k + j A (x)= a(k, j)xj = − xj. k 2j + 3 2j + 1 Xj=0 Xj=0 It was shown in [17] that this polynomial family satisﬁes the four-term recurrence relation

A (x) = (2x + 4) (A (x)+ A (x)) (4x2 + 4x + 6)A (x) A (x), k+4 k+3 k+1 − k+2 − k as opposed to a three-term recurrence relation required for orthogonality (e.g. see [28]), and so they do not constitute an orthogonal polynomial system. However it is also shown in [17] that the polynomials Ak(x) are closely linked to the orthogonal polynomial system of Gegenbauer polynomials Cλ(x), deﬁned for λ > 1/2, n − λ = 0 (e.g., [2]), by the hypergeometric series representation [1] (p. 773-802) 6 (2λ) 1 1 x Cλ(x)= n F 2λ + n, n; λ + ; − , (2) n n! 2 1 − 2 2 λ λ (so that Cn (1) = (2λ)n/n!). The hypergeometric series for Cn (x) can be re-written in terms of binomial coeﬃcients and powers of 2 such that

n/2 ⌊ ⌋ λ r n r n r 1+ λ n 2r C (x)= ( 1) − − − (2x) − , (3) n − r n r r=0 X − and taking λ = 2, one ﬁnds that the relation to the Ak(x) polynomial system is given by

x + 2 x + 2 x + 2 A (x)= C2 + (x + 6)C2 + C2 . k k 1 2 k 2 2 k 3 2 − − −

2 The Legendre Polynomials Pn(x) are the case λ = 1/2 of the Gegenbauer polynomials 1/2 Cn (x), and a close connection between these polynomials, the prime numbers and the 1 absolute value of the Riemann zeta function, ζ(s) = ∞ s , deﬁned for (s) > 1, was n=1 n ℜ established in [10], where ζ(s) is expressed as an inﬁnite sum over products of Legendre | | P polynomials and functions derived from prime numbers. The location of the zeros of the Riemann zeta function is famously known as the Riemann Hypothesis (1859), which states that all of the non-trivial zeros of ζ(s) (the trivial zeros lie at the negative even integers) lie on the critical line s = 1/2. In 1901 ℜ von Koch reinforced the connection between ζ(s) and the prime numbers, demonstrating that the Riemann Hypothesis is equivalent to the statement that the error term for π(x), the number of primes up to x, is of order of magnitude O (√x log(x)) [23]. Riemann had originally shown that

∞ µ(n) x du π(x) Li(x)+ Li(x1/n), where Li= , ∼ n log u n=2 2 X Z and where µ(n) is the Möbius function, which returns 0 if n is divisible by a prime squared and ( 1)k if n is the product of k distinct primes. − The Báez-Duarte equivalence to the Riemann Hypothesis [4, 24] links the the Riemann Hypothesis (and so the prime numbers) to binomial coefficients, via the infinite sequence of real numbers ct, defined such that t t 1 c := ( 1)s , t − s ζ(2s + 2) Xs=0 3/4+ǫ with the assertion that the Riemann hypothesis is true if and only if ct = O(t− ), for integers t 0, and for all ǫ> 0. ≥ In relation to understanding the triangle of connections that exist between the three objects consisting of the prime numbers, the binomial coefficients, and functions which only have critical zeros, it is those between the binomial coefficients and the ‘critical polynomials’ that appears to be the least studied, thus motivating the results contained in this paper. Before elaborating further, we mention some standard notation in which 2F1 denotes the Gauss hypergeometric function, pFq the generalized hypergeometric function, Γ(1 a) (a) = Γ(a + n)/Γ(a) = ( 1)n − n − Γ(1 a n) − − the Pochhammer symbol, with Γ the Gamma function [2, 5, 19]. We also define ε = 0 for n even and ε = 1 for n odd. Our starting point is a special case of the more general class of integrals

a n ǫ α+2β+ǫ 2 α 1 2 2 β 1 λ ( 1) (λ)n+ǫc a − α + ǫ x − (a x ) − C (cx)dx = − B , β − 2n+ǫ 21 ǫn! 2 Z0 − 3 α + ǫ 1 α + ǫ + 2β F n,n + λ + ǫ, ; ǫ + , ; a2c2 , (4) ×3 2 − 2 2 2 where ǫ = (1 ( 1)n)/2; a, β > 0; α> ǫ, and B(x,y)= Γ(x)Γ(y) , is the beta function − − ℜ ℜ − Γ(x+y) (see entry 2.21.2(1) on p. 517 of [19]), and which applies to the following deﬁnition and Theorem 2.1.

Deﬁnition 1.1. For λ > 1/2, we deﬁne the generalised Mellin transform M λ(s), such − n that 1 λ s 1 π/2 λ Cn (x)x − s 1 λ λ 1/2 M (s)= dx = cos − θ C (cos θ)sin − θ dθ, (5) n (1 x2)3/4 λ/2 n Z0 − − Z0 wherein x = cos θ, and we assume that s > 0 for n even and s > 1 for n odd, λ ℜ λ ℜ − denoting by pn(s) the polynomial factor of Mn (s). Then for λ = 1 we have the Mellin transform of the Chebyshev functions [25, 27] of the second kind given by

1 s 1 dx Mn(s) x − Un(x) . (6) ≡ (1 x2)1/4 Z0 − In [7, 12, 13], Mellin transforms were used on [0, ). Here we consider Mellin trans- ∞ formations for functions supported on [0, 1],

1 dx ( f)(s)= f(x)xs . M0 x Z0 For properties of the Mellin transform, we mention [8]. λ Our main results show that the polynomial factors pn(s) of the Mellin transforms λ in (5) of the Gegenbauer (and so Chebyshev) functions Cn (x), yield families of ‘critical polynomials’ pλ(s), n = 0, 1, 2,..., of degree n/2 , satisfying the functional equation n ⌊ ⌋ pλ(s) = ( 1) n/2 pλ(1 s). Additionally we ﬁnd that (up to multiplication by a constant) n − ⌊ ⌋ n − these polynomials can be written explicitly as variants of Gould S:4/1 and S:3/2 binomial sums (see [20]), the latter form being

1 n−r 2r−1 n+r+λ−1 n+r (s−2)+r λ n+λ 1 n+ 1 (s+λ) 3 n ( 1) 2 ( r )( r )( 2 r ) p (s)= n!(2n)! 2 4 − 2 , (7) 2n n− n − r=0 n+r 1 (s+λ)− 3 +r ( r )( 2 r 4 ) P 1 n−r 2r n+r+λ n+r+1 (s−1)+r 1 1 n ( 1) 2 ( r )( r )( 2 r ) λ n+λ n+ 2 (s+λ) 4 − 2 +1 p2n+1(s)= n!(2n + 1)! n+1 n − r=0 n+r+1 1 (s+λ)− 1 +r . (8) ( r )( 2 r 4 ) P In the case of the Chebyshev polynomials (λ = 1), this simpliﬁes to the S : 2/1 form, due to cancellation of binomial factors, and with = 1 2n , the nth Catalan number, s Cn n+1 n an integer, we show that polynomials 4 n 1p2n(s) and np2n+1(s) yield integers with only C − C odd prime factors.

4 λ Example 1.2. The ﬁrst few transformed polynomials pn(s), are given by

λ p0 (s) =1/2, λ p1 (s)=λ, 1 1 1 pλ(s)= λ(2λ + 1)(2s 1) = λ(2λ + 1) s , 2 4 − 2 − 2 1 1 pλ(s)= λ(λ + 1)(2λ + 1)(2s 1) = λ(λ + 1)(2λ + 1) s , 3 2 − − 2 1 pλ(s)= λ(λ + 1)(2λ + 1) 8λs2 8λs + 6λ + 12s2 12s + 15 , 4 8 − − 1 2 2 = λ(λ + 1)(2λ + 1) s 1 i√9+9λ+2λ s 1 + i√9+9λ+2λ , 8 − 2 − 3+2λ − 2 3+2λ 1 pλ(s)= λ(λ + 1)(λ + 2)(2λ + 1) 8λs2 8λs + 14λ + 12s2 12s + 51 5 4 − − 1 2 2 = λ(λ + 1)(λ + 2)(2λ + 1) s 1 i√3√2λ +11λ+12 s + 1 i√3√2λ +11λ+12 4 − 2 − 3+2λ 2 − 3+2λ The ‘critical polynomials’ under consideration here, in a sense motivate the Riemann hypothesis, and have many important applications to analytic number theory. For example, using the Mellin transforms of Hermite functions, Hermite polynomials multiplied by a Gaussian factor, Bump and Ng [7] were able to generalise Riemann’s second proof of the functional equation of the zeta function ζ(s), and to obtain a new representation for it. The polynomial factors of the Mellin transforms of Bump and Ng are realised as certain 2F1(2) Gauss hypergeometric functions [12]. In a different setting, the polynomials p (x)= F ( n, x; 1; 2) = ( 1)n F ( n,x + 1; 1; 2) and q (x)= inn!p ( 1/2 ix/2) n 2 1 − − − 2 1 − n n − − were studied [22], and they directly correspond to the Bump and Ng polynomials with s = x. Kirschenhofer, Pethö, and Tichy considered combinatorial properties of p , and − n developed Diophantine properties of them. Their analytic results for pn include univariate and bivariate generating functions, and that its zeros are simple, lie on the line x = 1/2+it, t R, and that its zeros interlace with those of p on this line. We may observe − ∈ n+1 that these polynomials may as well be written as p (x)= n+x F ( n, x; n x; 1), n n 2 1 − − − − − or ( 1)n2nΓ(n x) 1 p (x)= − − F n, n; x + 1 n; . n n!Γ( x) 2 1 − − − 2 − Previous results obtained by the authors related to this area of research are discussed in [14, 15, 11], where in the former paper families of ‘critical polynomials’ are obtained from generalised Mellin transforms of classical orthogonal Legendre polynomials. In [15] the first author has addressed the Mellin transform of certain generalised Hermite functions, where the resulting critical polynomials possess a reciprocity relation. In the latter paper sequences of ‘critical polynomials’ are considered which can also be obtained by generalised Mellin transforms of families of orthogonal polynomials whose coefficients are the weighted

5 binomial coeﬃcients deﬁned by

k k 2k + 1 k + j B (x)= b(k, j)xj = xj. k 2j + 1 2j Xj=0 Xj=0 Here we find that for s> 1/4, the Mellin transforms ℜ − 0 s 3/4 1 B Bn(x)x − s+5/4 s n 1 Γ s + 4 Mn (s)= 3/4 dx = ( 1) 4 4− − Γ(1/4)pn(s) 2n+1 , 4 (4 + x) − Γ s + Z− 2 yield critical polynomial factors pn(s), which obey the perfect reflection functional equation p (s)= p ( 1 s ). n ± n − The ‘perfect-reflection’ functional equation pλ(s) = ( 1) n/2 pλ(1 s), is similar to n − ⌊ ⌋ n − that for Riemann’s xi function ξ(s), defined by ξ(s) = 1 s(s 1)π s/2Γ 1 s ζ(s), and 2 − − 2 which satisfies ξ(s)= ξ(1 s), so that for t R, the zeros of ξ(1/2+ it) and ζ(1/2+ it) − ∈ λ are identical. Drawing upon this analogy, one interpretation is that the polynomials pn(s) are normalised (from a functional equation perspective) polynomial forms of the rational functions qλ(s), defined for n N by n ∈ 2 pλ (s) 22n+1pλ (s) qλ (s)= 2n = 2n , (9) 2n 2n+2λ 1 n+ 1 (s+λ) 3 n 2 4 (2λ)2n (2s + 2λ + 4j 3) λ(n 1)!(2n)! 2n 1− n − j=1 − − − λ Q2n+1 λ λ p2n+1(s) 2 p2n+1(s) q2n+1(s)= 1 1 = n , (10) 2n+2λ n+ 2 (s+λ) 4 (2λ)2n+1 (2s + 2λ + 4j 1) λ(n)!(2n)! 2n n − j=1 − where both numerator and denominator polynomials of qλQ(s) are of degree n/2 . n ⌊ ⌋ For λ > 1/2, λ = 0, and s > 0, the n/2 linear factors of the denominator − λ 6 ℜ ⌊ ⌋ λ polynomials of qn(s), are each non-zero, so that for these values of λ, we have qn(s) has no singularities with s > 0. Hence the rational function qλ(s) has the same ‘critical ℜ n zeros’ as the polynomial pλ(s), and for t R, the roots of pλ(1/2+ it) and qλ(1/2+ it) n ∈ n n are identical. λ λ The [n/2] poles of qn(s) (so zeros of the denominator polynomial of qn(s)) occur on the negative real axis when 2s = 3 λ 4j or 2s = 1 2λ 4j (depending on the parity − − − λ − of n) and in Theorem 2.8, for s (1, ), we find that qn(s) takes values on (0, 1), with λ ∈ ∞ lims qn(s) = 1 (from below), and obeys the functional equation →∞ 1 1 s+λ+ǫ 3 s+λ+ǫ 3 − λ n/2 n/2 + − 2 4 n/2 + 2 4 λ q (s) = ( 1)⌊ ⌋ ⌊ ⌋ − ⌊ ⌋ − q (1 s). (11) n − n/2 n/2 n − ⌊ ⌋ ⌊ ⌋ R λ It follows that on >1, the behaviour of qn(s) has similarities to that of 1/ζ(s), albeit with a rate of convergence to the limit point 1, considerably slower than for 1/ζ(s). To give an overview, the present work is split into four sections, with the main results concerning the critical polynomials arising from Mellin transforms of Gegenbauer

6 polynomials appearing after this introduction in the second section. In the third section we prove these results, utilising continuous Hahn polynomials to locate the ‘critical zeros’. The fourth and concluding section then considers further possible extensions to these results.

2 Critical Polynomial Results

Theorem 2.1. The Mellin transforms (5) may be written as 3F2(1) hypergeometric functions, such that

λ + n 1 s 1 λ s 1 M λ (s) = ( 1)nM λ(s) − F n, λ + n, ; , + + ; 1 , (12) 2n − 0 n 3 2 − 2 2 2 2 4 and λ + n s 1 3 λ + s 3 M λ (s) = ( 1)n2M λ(s+1)(n+1) F n, λ + n + 1, + ; , + ; 1 , 2n+1 − 0 n + 1 3 2 − 2 2 2 2 4 (13) where λ 1 s λ Γ 2 + 4 Γ 2 λ λ M0 (s)= s+λ 1 , and M1 (s) = 2λM0 (s + 1), (14) 2Γ 2 + 4 and the latter identities for odd and even cases give the equivalent hypergeometric form, which can be written for general n as

λ 1 s+n λ Γ 2 + 4 Γ 2 2λ + n 1 λ 1 1 n n 1 (n + s) M (s)= − 3F2 + , − , ; + λ, 1 ; 1 . n 2Γ s+n+λ + 1 n 2 4 2 − 2 2 − 2 2 4 (15) λ Theorem 2.2. Let Mn (s) be deﬁned as in (5). Then we have (a) the mixed recurrence relation

λ λ λ nMn (s) = 2(λ + n 1)Mn 1(s + 1) (2λ + n 2)Mn 2(s), − − − − − (b) the generating function

1 s 1 λ ∞ λ k 1 x − G (s,t)= Mk (s)t = 2 3/4 λ/2 2 λ dx 0 (1 x ) − (1 2tx + t ) Xk=0 Z − − 1 λ s 2 1 Γ 4 + 2 Γ(λ)Γ 2 λ + 1 λ s 1 s + λ 1 4t = 3F2 , , ; , + ; (1 + t2)λ 2 Γ s+λ + 1 2 2 2 2 2 4 (1 + t2)2 " 2 4 s+1 2 2tΓ(λ + 1) Γ 2 λ + 1 λ s + 1 3 s + λ 3 4t + 3F2 , 1+ , ; , + ; , (1 + t2) Γ s+λ + 3 2 2 2 2 2 4 (1 + t2)2 2 4 # 7 (c) the polynomial factors satisfy the functional equation pλ(s) = ( 1) n/2 pλ(1 s), n − ⌊ ⌋ n − (d) the recurrence relation in s

[6 4(λ + 2λn + n2) 16s + 8s(s + 1)]M λ(s) − − n +[ 9 + 4(n + λ)2 + 16(s + 2) 4(s + 2)(s + 3)]M λ(s + 2) − − n 4(s 1)(s 2)M λ(s 2) = 0, − − − n − with λ 1 s λ Γ 2 + 4 Γ 2 λ λ M0 (s)= s+λ 1 , and M1 (s) = 2λM0 (s + 1), 2Γ 2 + 4 λ λ and (e) (location of zeros) the polynomial factors pn(s) of Mn (s), have zeros only on the critical line.

Lemma 2.3. Let λ = 1, so that Mn(s) is the Mellin transform of the Chebyshev function given in (6). Then we have that (a)

3 n+s Γ 4 Γ 2 3 1 n n 3 n + s Mn(s) = (n + 1) 3F2 , − , ; , 1 ; 1 , 2Γ n+s + 3 4 2 − 2 2 − 2 2 4 and (b)

3 n+s n 1 Γ 4 Γ 2 1 n n 1 (n + s) n + s Mn(s) = 2 − 3F2 − , , ; n, 1 ; 1 . Γ n+s + 3 2 − 2 4 − 2 − − 2 2 4 Proof. The interchange of ﬁnite summation and the integration of (6) is used in both instances.

Theorem 2.4. (a) When λ = 1, the polynomials pn, corresponding to the Chebyshev polynomials of the second kind, satisfy the simpliﬁed recursion relation, with p0 = Γ(3/4)/2 and p1 = Γ(3/4). for n even,

1 1 pn(s)= spn 1(s + 1) s + n pn 2(s), (16) − − 2 − 2 − and for n odd, 1 1 pn(s) = 2pn 1(s + 1) s + n pn 2(s). (17) − − 2 − 2 − (b) The polynomials p (s), of degree n/2 , satisfy the functional equation n ⌊ ⌋ n/2 p (s) = ( 1)⌊ ⌋p (1 s). n − n − (c) These polynomials have zeros only on the critical line. Further, all zeros = 1/2 occur 6 in complex conjugate pairs.

8 Theorem 2.5. The Mellin transforms (5) may be written as a constant multiplied by a variant on Gould’s combinatorial S:4/2 and S:3/1 functions, such that

s− s+λ n n r 2r n+r+λ 1 n+r 2 +r n+ 3 2 2 − 4 λ λ ( 1) − 2 n+r− 2r r n r M (s)= M (s) − s λ − 2n 0 n n+ + 3 2 −4 Xr=0 r n s−2 n n r 2r n+r+λ 1 n+r 2 +r λ ( 1) − 2 n+r− 2r r = M (s) − s λ , 0 + 3 +r r=0 2 − 4 X r and s− s+λ 1 n n r 2r+1 n+r+λ n+r+1 1 +r n+ 2 2 − 4 λ λ ( 1) − 2 n+r+1 2r+1 r n r M (s)= M (s + 1) − s λ − 2n+1 0 n n+ + 1 r=0 2 −4 X r n s−1 n n r 2r+1 n+r+λ n+r+1 2 +r λ ( 1) − 2 n+r+1 2r+1 r = M (s + 1) − s λ . 0 + 1 +r r=0 2 − 4 X r Corollary 2.6. When s = 2t is an even positive integer then we have 1 1 λ + 1 + t 1 − M λ(s)= 2 4 − , 0 4t t and when s = 2t + 1 is an odd positive integer

1 λ λ + 1 + t − M λ(s) = 2λM λ(s + 1) = 2λM λ(2t +2) = 2 4 . 1 0 0 2t + 2 t + 1 Hence, when s is a non-negative integer then we have the Gould’s S:4/3 summation variants λ 3 n n r 2r n+r+λ 1 n+r t 1+r n+t+ 1 ( 1) − 2 − − 2 − 4 λ − n+r 2r r n r M2n(2t)= λ 3 λ 3 − , 2t n t+ 2 4 n+t+ 2 4 r=0 n r t− n − X − and λ 1 n n r 2r+1 n+r+λ n+r+1 t+r n+t+ 2 + 4 λ λ ( 1) − 2 n+r+1 2r+1 r n r M (2t +1) = − λ λ − . 2n+1 t + 1 n t+ + 1 n+t+ + 1 r=0 2 4 2 4 X r t+1 n Theorem 2.7. Let ε = (1 ( 1)n)/2. Then the Mellin transforms are of the form − − λ 1 s+ǫ λ Γ 2 + 4 Γ 2 λ Mn (s)= s+n+λ 1 pn(s), 2(n!)Γ 2 + 4 λ and the polynomial factors pn(s) satisfy the diﬀerence equations in s (s + 2)(2λ + 4n 2s 5)pλ (s + 4) 8λn + 2λ + 8n2 4s2 12s 11 pλ (s + 2) − − 2n − − − − 2n 9 (s + 1)(2λ + 4n + 2s + 1)pλ (s) = 0, − 2n and

(s + 3)(2λ + 4n 2s 3)pλ (s + 4) 8λn + 6λ + 8n2 + 8n 4s2 12s 9 pλ (s + 2) − − 2n+1 − − − − 2n+1 s(2λ + 4n + 2s + 3)pλ (s) = 0. − 2n+1 .

λ Theorem 2.8 (Binomial ‘critical polynomial’ theorem). (a) The polynomials pn(s), can be written (up to multiplication by a constant) in terms of binomial coeﬃcients and powers of 2, as a variant of a Gould S:4/1 combinatorial function, such that

n n r 2r 1 n+r+λ 1 n+r 1 (s 2)+r n+ 1 (s+λ) 3 2 2 − 4 λ ( 1) − 2 − n+r− 2r −r n r p2n(s)= n!(2n)! − n − , r=0 r X 1 3 n n r 2r n+r+λ n+r+1 1 (s 1)+r n+ (s+1+λ) ( 1) 2 2 − 2 − 4 λ − − n+r+1 2r+1 r n r p2n+1(s)= n!(2n + 1)! n − , r Xr=0 or as a Gould S:3/2 combinatorial function variant, such that

1 3 n n r 2r 1 n+r+λ 1 n+r 1 (s 2)+r n + λ 1 n + (s + λ) ( 1) − 2 − − 2 − pλ (s)= n!(2n)! − 2 − 4 − r 2r r , 2n n n n+r 1 (s+λ) 3 +r r=0 2 − 4 X r r 1 1 1 n n r 2r n+r+λ n+r+1 (s 1)+r n + λ n + (s + λ) ( 1) − 2 2 − pλ (s)= n!(2n+1)! 2 − 4 − r 2r+1 r , 2n+1 n+r+1 1 (s+λ) 1 +r n + 1 n 2 − 4 Xr=0 r r thus establishing the binomial ‘critical polynomial’ relationships (7) and (8). λ (b) Let qn(s) denote the rational function in s derived from the S:3/2 form of the ‘critical λ polynomial’ pn(s) such that

n+λ 1 n n r 2r 1 n+r+λ 1 n+r 1 (s 2)+r 2n − ( 1) − 2 − − 2 − qλ (s)= n − r 2r r , 2n 2n+2λ 1 n+r 1 (s+λ) 3 +r λ 2n 1− r=0 2 − 4 − X r r n+λ n n r 2rn+r+λ n+r+1 1 (s 1)+r (2n + 1) ( 1) − 2 2 − qλ (s)= n+1 − r 2r+1 r . 2n+1 2n+2λ n+r+1 1 (s+λ) 1 +r λ 2n r=0 2 − 4 X r r Then we have λ 2n+1 λ λ 2 p2n(s) 2 p2n(s) q2n(s)= 1 3 = n , 2n+2λ 1 n+ 2 (s+λ) 4 (2λ)2n (2s + 2λ + 4j 3) λ(n 1)!(2n)! 2n 1− n − j=1 − − − Q 10 λ 2n+1 λ λ p2n+1(s) 2 p2n+1(s) q2n+1(s)= 1 1 = n , 2n+2λ n+ 2 (s+λ) 4 (2λ)2n+1 (2s + 2λ + 4j 1) λ(n)!(2n)! 2n n − j=1 − λ Q where both numerator and denominator polynomials of qn(s) are of degree n/2 . λ ⌊ ⌋ (c) For λ > 1/2, λ = 0, and s > 0, the rational function qn(s) has no singularities, − 6 ℜ λ R and has the same ‘critical zeros’ as the polynomial pn(s), so that for t , the roots of λ λ ∈ pn(1/2+ it) and qn(1/2+ it) are identical. R λ λ When s >1, qn(s) takes values on (0, 1), with lims qn(s) = 1 (from below), and ∈ →∞ with ε = 0 for n even and ε = 1 for n odd, obeys the functional equation

1 1 s+λ+ǫ 3 s+λ+ǫ 3 − λ n/2 n/2 + − 2 4 n/2 + 2 4 λ q (s) = ( 1)⌊ ⌋ ⌊ ⌋ − ⌊ ⌋ − q (1 s). (18) n − n/2 n/2 n − ⌊ ⌋ ⌊ ⌋ As s along the positive real axis it follows that qλ(s) 1, from below, as does 1/ζ(s). →∞ n → Corollary 2.9. The polynomial factors arising from the Mellin transform of the Chebyshev polynomials have the simpler form as a variant of a Gould S:2/1 combinatorial function, such that

n 1 s 1 n r 2r 1 n+r 2 (s 2)+r n + 2 4 ( 1) − 2 − 2r −r p2n(s)= n!(2n)! − − s , n 1 +r r=0 2 −4 X r n n r 2r n+r+1 1 (s 1)+r s 1 ( 1) 2 2 − n + 2 + 4 − 2r+1 r p2n+1(s)= n!(2n + 1)! − s . + 1 +r n 2 4 Xr=0 r For s = t Z an integer, and with = 1 2n , the nth Catalan number, the polynomi- ∈ Cn n+1 n als 4 n 1p2n(t) and np2n+1(t) yield integers with only odd prime factors. Moreover the C − C polynomials 22n+1 22n+1 p (t), and Tn+1 p (t), (19) (2n)! 2n (2n + 2)! 2n+1 with the largest odd factor of n + 1, yield odd integers with fewer prime factors. Tn+1 Example 2.10. As an example we give respectively tables of the expressions in (19), for 0 n 5, with t an integer in the range 4 t 4. The functional relation can be ≤ ≤ − ≤ ≤ clearly observed in both cases. n t 4 3 2 1 012 3 4 \ − − − 0 1 1 1 111111 1 27 21 15 9 3 3 9 15 21 − − − − − 2 421 261 141 61 21 21 61 141 261 3 7119 3969 1995 861 231 231 861 1995 3969 − − − − − 4 154665 80361 36729 13401 3465 3465 13401 36729 80361 5 4029795 1946637 828135 293073 65835 65835 293073 828135 1946637 − − − − − 11 n t 4 3 2 1 012 3 4 \ − − − 0 1 1 1 111111 1 9 7 5 3 11 3 5 7 − − − − − 2 279 183 111 63 39 39 63 111 183 3 1341 819 465 231 69 69 231 465 819 − − − − − 4 128637 72765 37581 17325 8157 8157 17325 37581 72765 5 1809459 959805 465975 197505 52731 52731 197505 465975 959805 − − − − − Theorem 2.11 (Perfect reﬂection property theorem). We say that f(s) has the ‘perfect reﬂection property’ to mean f(s) = f(s), f(s)= χf(1 s), with χ = 1, f(s) = 0, only − ± when s = 1/2. ℜ Then the polynomials

n+s Γ 2 1 n n (n + s) pn(s; β)= 3F2 1 β, − , ; 2(1 β), 1 ; 1 , Γ s+ε − 2 − 2 − − 2 2 have the perfect reﬂection property with χ(n) = ( 1) n/2 , wherein ε = 0 for n even − ⌊ ⌋ and = 1 for n odd, β < 1, of degree n/2 , satisfy the functional equation p (s; β) = ⌊ ⌋ n ( 1) n/2 p (1 s; β). These polynomials have zeros only on the critical line, and all zeros − ⌊ ⌋ n − = 1/2 occur in complex conjugate pairs. 6 Corollary 2.12. (a) The properties of Theorem 2.11 are satisﬁed by the polynomials

n+s 2(n + s) Γ 2 (n + 1) n (n + s) pn(s;0) = 1 2F1 , 1; ; 1 (n + 1)(n + 2) Γ s+ε − − 2 − 2 − − 2 2 (n+s) n+s Γ n+3 s 2(n + s) Γ 2 − 2 Γ 2− = s+ε 1 1 s s . (n + 1)(n + 2) Γ  − Γ − Γ 1  2 2 − 2 (b) More generally, for β a negative integer,  the properties of Theorem 2.11 are satisﬁed by the polynomials p (s; m), and these polynomials may be written in terms of elementary n − factors and the Gamma function.

3 Proof of the Main Results

λ 1 Proof of Theorem 2.1. Setting a = 1, α = s, β = 2 + 4 , and c = 1 in (4), and taking ǫ = 0 and then ǫ = 1 respectively gives us (12) and (13). The displays in (14) then follow by setting n = 0 in (12) and (13). λ To see the hypergeometric form (15) of Mn (s), we use the series representation ([26], p. 278 (6)) [n/2] n 2k 2 k λ (2λ)nx − (x 1) Cn(x)= k − , 4 k!(λ + 1/2)k(n 2k)! Xk=0 − 12 and recall a form of the Beta integral,

1 a 1 2 b 1 1 a x − (1 x ) − dx = B , b , a> 0, b> 0. 0 − 2 2 ℜ ℜ Z Then

[n/2] k 1 λ (2λ)n( 1) s+n 2k 1 2 k+λ/2 3/4 Mn (s)= k − x − − (1 x ) − dx 4 k!(λ + 1/2)k(n 2k)! 0 − Xk=0 − Z

[n/2] k (2λ)n( 1) 1 s + n λ 1 = k − B k, k + + 4 k!(λ + 1/2)k(n 2k)! 2 2 − 2 4 Xk=0 − [n/2] k s+n λ 1 (2λ)n ( 1) Γ 2 k Γ k + 2 + 4 = s+n+λ 1 −k − 4 k! (λ + 1/2)k(n 2k)! 2Γ 2 + 4 k=0 X − λ 1 s+n Γ 2 + 4 Γ 2 2λ + n 1 λ 1 1 n n 1 (n + s) = − 3F2 + , − , ; + λ, 1 ; 1 , 2Γ s+n+λ + 1 n 2 4 2 − 2 2 − 2 2 4 as required, where the above has used the duplication formula for the Gamma function so that 1 1 4k n 1 n = = − . (n 2k)! Γ(n 2k + 1) n! − 2 k 2 k − −

Proof of Theorem 2.2. (a) follows from ([2], p. 303 or [19], p. 1030)

(n + 2)Cλ (x) = 2(λ + n + 1)xCλ (x) (2λ + n)Cλ(x), n+2 n+1 − n λ λ C0 (x) = 1, and C1 (x) = 2λx. (b) follows from ([2], p. 302 or [19], p. 1029)

2 λ ∞ λ n (1 2xt + t )− = C (x)t . − n n=0 X (c) The Mellin transforms may be computed explicitly by several means. In particular, from (λ) n 1 n 1 Cλ(x)= n (2x)n F , − ; 1 n λ; , n n! 2 1 − 2 2 − − x2 2 transformation of the 2F1 function to argument x and use of [19] (p. 850), we ﬁnd that

λ (λ)n n 1 n λ 1 M (s)= 2 − √π( i) Γ + Γ(1 λ n) n n! − 2 4 − −

13 s+1 1 n n+1 s+1 3 3 λ+s 2iΓ F − , λ + , ; , + ; 1 − 2 3 2 2 2 2 2 4 2 1 n n λ+s 3 × (Γ λ Γ Γ + 2 − − 2 − 2 2 4 Γ s F n , λ + n , s ;1 , 1 + λ+s ; 1 2 3 2 − 2 2 2 2 4 2 + 1 n n λ+s 1 . Γ − Γ 1 λ Γ + ) 2 − − 2 2 4 The second line of the right member provides the transform for n odd and the third line for n even. Then transformation of the 3F2 functions and arguments similar to the proof of Theorem 2.4 (b) may be used to show that the degree of the polynomial factors of the transforms is again n/2 , and that they satisfy the functional equation pλ(s) = ⌊ ⌋ n ( 1) n/2 pλ(1 s). − ⌊ ⌋ n − An alternative proof of the functional equation can be deduced directly from the first remark, concerning the fact that Bm = 0 in the recurrence relation, following on from the proof of part (e) of this theorem. λ (d) To obtain the difference equation for Mn (s), we apply the ordinary differential equation satisfied by Gegenbauer polynomials (e.g., [19], p. 1031)

2 (x 1)y′′(x) + (2λ + 1)xy′(x) n(2λ + n)y(x) = 0. − − If f(x) Cλ(x)/(1 x2)3/4 λ/2, we then substitute Cλ(x) = (1 x2)3/4 λ/2f(x) into this ≡ n − − n − − diﬀerential equation. We then ﬁnd that

1 2 1/4 λ/2 2 2 2 (1 x )− − (6 4(λ + 2λn + n ) + ( 9 + 4(λ + n) )x )f(x) 4 − − − 2 2 +4(x 1)( 4xf ′(x) + (1 x )f ′′(x)) = 0. − − − s 1 It follows that the quantity in square brackets is zero. We multiply it by x − and integrate from x = 0 to 1, integrating the f ′ term once by parts, and the f ′′ term twice by parts. We determine that the Mellin transforms satisfy the following diﬀerence equation:

[6 4(λ + 2λn + n2) 16s + 8s(s + 1)]M λ(s) − − n +[ 9 + 4(n + λ)2 + 16(s + 2) 4(s + 2)(s + 3)]M λ(s + 2) − − n 4(s 1)(s 2)M λ(s 2) = 0, − − − n − and hence the result. (e) To show that the resulting zeros of pλ(s) occur only on s = 1/2 we apply a n ℜ connection with continuous Hahn polynomials, which are given by (e.g., [2] p. 331, [3]) (a + c) (a + d) h (x; a, b, c, d)= im m m F ( m,m + a + b + c + d 1, a + ix; a + c, a + d; 1) . m m! 3 2 − − (20) By using the second transformation of a terminating 3F2(1) series given in the Appendix, we have im h (x; a, b, c, d)= (a + b + c + d + m 1) (1 b m ix) m m! − m − − − m

14 F (1 b c m, 1 b d m, m; 2 a b c d 2m, 1 b m ix; 1). × 3 2 − − − − − − − − − − − − − − − Then comparing with the 3F2(1) function of Theorem 2.1

λ 1 s+n λ Γ 2 + 4 Γ 2 2λ + n 1 λ 1 1 n n 1 (n + s) M (s)= − 3F2 + , − , ; + λ, 1 ; 1 , n 2Γ s+n+λ + 1 n 2 4 2 − 2 2 − 2 2 4 we see when n= 2m is even, that setting

i 1 1 λ 1 x = s + , a = c = m, b = d = , 2 − 2 2 − 2 − 4 λ our polynomial factors pn(s) are proportional to continuous Hahn polynomials such that

λ 2 2m 1 m m + λ 1 i 1 1 λ 1 1 λ 1 p (s) = (m!) 2 − ( i) − h − s ; m, , m, . n − m m 2 − 2 2 − 2 − 4 2 − 2 − 4 (21) Similarly when n = 2m + 1 is odd, setting

i 1 λ 3 x = s + , a = c = m, b = d = , 2 − 2 − 2 − 4 λ our polynomial factors pn(s) are proportional to continuous Hahn polynomials such that m + λ i 1 λ 3 λ 3 pλ(s) = (m!)222m( i)mλ h − s ; m, , m, . (22) n − m m 2 − 2 − 2 − 4 − 2 − 4 For ﬁxed values of a, b, c, d, the continuous Hahn polynomials are an orthogonal system of polynomials which satisfy the recurrence relation (e.g see 6.10.11 of [2])

Amhˆm+1(x) = ((a + ix)+ Am + Cm)hˆm(x) Cmhˆm 1(x), − − where

ˆ ˆ m! hm(x) := hm(x; a, b, c, d)= Dmhm(x)= m hm(x; a, b, c, d) i (a + c)m(a + d)m

m so that Dm = m!/(i (a + c)m(a + d)m), hˆm(x) is the 3F2 hypergeometric function given in (20), and

(m+a+b+c+d 1)(m+a+c)(m+a+d) m(m+b+c 1)(m+b+d 1) Am = (2m+a+b+c+−d 1)(2m+a+b+c+d) , Cm = (2m+a+b+c+d −2)(2m+a+b−+c+d 1) . − − − − In the case that a =c ¯ and b = d¯, the recurrence relation simpliﬁes to

Amhˆm+1(x)= ixhˆm(x) Cmhˆm 1(x), − −

15 and substituting for hˆm and rearranging we have

ixDm CmDm 1 hm+1(x)= hm(x) − hm 1(x)= Gm x hm(x) Hm hm 1(x), (23) Dm+1Am − Dm+1Am − − − say, where iDm CmDm 1 Gm = , and Hm = − . Dm+1Am Dm+1Am As the a, b, c, d values must be constant for the conditions of the orthogonality theorems to be met, we set m = u a constant in the variable a, so that in (23) we take a = c = 1 λ 1 u, b = d = to obtain hm+1(x)= Gm x hm(x) Hm hm 1(x), with 2 − 2 − 4 − − ( 2λ + 4m 4u + 1)( 2λ + 4m 4u + 3) G = − − − − , (24) m 2(m + 1)( 2λ + 2m 4u + 1) − − and (2m 1)(2λ 4m + 4u + 1)( 2λ + 4m 4u + 3)( λ + m 2u) H = − − − − − − . (25) m 16(m + 1)( 2λ + 2m 4u + 1) − − Similarly setting a = c = λ u, b = d = 3 in (23) gives us h (x) = G x h (x) − 2 − 4 m+1 m m − Hm hm 1(x), with Gm the same as in (24), but where Hm is now given by − (2m + 1)(2λ 4m + 4u + 1)( 2λ + 4m 4u + 3)( λ + m 2u 1) H = − − − − − − . (26) m 16(m + 1)( 2λ + 2m 4u + 1) − − For the above coeﬃcient Gm, and the two choices for the coeﬃcient Hm, both of the resulting recurrence relations hm+1(x)= Gm x hm(x) Hm hm 1(x) are of the form − −

Pm+1(x) = (Amx + Bm)Pm(x) CmPm 1(x), − − with Bm = 0, which is the form of the recurrence relation satisﬁed by a system of orthogonal polynomials that are not monic (e.g. see (4.2) p19 of [9]). More rigorously, if P˜m(x) is the monic (scaled) polynomial corresponding to Pm(x), so that Pm(x)= kmP˜m, where k 1 = 1, then in generality one ﬁnds that (e.g. see (4.3) p19 − of [9])

km+1 km+1 km+1 Am = , ,Bm = cm+1 , Cm = µn = µnAmAm 1, km − km km 1 − − so that cm = 0, µm = Cm/(AmAm 1), and where the corresponding monic recurrence − relation can be written as

P˜m(x) = (x cm)P˜m 1(x) µmP˜m 2(x), P˜ 1(x) = 0, P˜0(x) = 1, m = 1, 2, 3,... − − − − − (27)

16 For our two recurrence relations we find that µm = Hm/(GmGm 1), so that in the case − n = 2m is even we have m(2m 1)(2λ 2m + 4u + 1)( λ + m 2u) µ = − − − − , (28) m 4( 2λ + 4m 4u 3)( 2λ + 4m 4u + 1) − − − − − and when m = 2m +1 is odd m(2m + 1)(2λ 2m + 4u + 1)( λ + m 2u 1) µ = − − − − . (29) m 4( 2λ + 4m 4u 3)( 2λ + 4m 4u + 1) − − − − − For fixed values of u and λ, and m = 0, 1, 2, 3,..., with µ = 0 and well defined, we obtain m 6 a pair of families of Hahn polynomials given by

1 λ 1 1 λ 1 h x; u, , u, , (30) m 2 − 2 − 4 2 − 2 − 4 and λ 3 λ 3 h x; u, , u, . (31) m − 2 − 4 − 2 − 4 In both (30) and (31), we find that h 1(s)=0 and h0(s) = 1, and applying Favard’s The- − orem (see for example [9] Theorem 4.4, p21) concerning polynomial sequences satisfying the three-term recurrence relation given in (27), for c R and µ = 0 it follows that m ∈ m 6 (30) and (31) form two families of orthogonal polynomial systems. It is a well known fact (e.g. [9], Section 5, Chapter 1, or [27, 28]) that systems of orthogonal polynomials have only real zeros which interlace. Hence the polynomial families corresponding to (30) and (31) have only real zeros which interlace on the real i 1 line. Setting x = −2 (s 2 ) in (30) and (31), the resulting polynomials will therefore − 1 π have their zeros dilated by the factor of 2 on the real line; rotated by 2 clockwise onto the 1 imaginary axis by the factor of i, and then translated by 2 from the imaginary axis to 1 − i − 1 the critical line s = . This means that setting x = − (s ) in (30) and (31), results ℜ 2 2 − 2 in families of polynomials which have zeros only on the critical line s = 1 . ℜ 2 Restricting u N, to be a positive integer, analysis of (30) and (31) shows that for ∈ λ = 1 (the Legendre polynomial case proven by the authors in [14]) and λ = 3 , µ is well 6 2 6 2 m defined for u m, and when λ = 1, µ is well defined for u< 2m. If λ is not an integer ≤ m or half-integer then µ is well defined for all u N. m ∈ Therefore, for each integer value u m, and fixed λ > 1 , λ = 1 , λ = 3 , we get a ≤ − 2 6 2 6 2 family of Hahn polynomials in (30), and also in (31). We can associate with each Hahn polynomial a lattice point (m, u) in R2 in the positive quadrant of the plane, so that an orthogonal polynomial family associated with a specific value of u generated by (30), corresponds to lattice points on the line m = u, parallel to the horizontal axis. For each particular family (so parallel line), there will exist one value of m, namely m = u, where i 1 replacing the polynomial variable with − (s ), the Hahn polynomial corresponds to 2 − 2 that given for pλ(s) given in (21). Hence when n = 2m is even, the family of polynomials n

17 λ pn(s) can be visualised as points on a line diagonally cutting through the parallel lines corresponding to the diﬀerent families of Hahn polynomials with the polynomial variable i 1 change x = − (s ). Conversely, for each family of Hahn polynomials determined by 2 − 2 the integer u, there will exist exactly one polynomial in the family which is proportional to λ a pn(s) polynomial. A similarly argument holds in the odd case n = 2m + 1, relating the λ pn(s) polynomials to families of Hahn polynomials generated by (31) with the polynomial i 1 variable change x = − (s ). 2 − 2 The above argument implies that all the zeros of our polynomial families pλ(s) lie on n the critical line s = 1 , as required. ℜ 2

Remark 1. By Theorem 4.3 of [9], as cm = 0 in the recurrence (27), it follows that h ( x) = ( 1)mh (x), thus giving another proof of the functional relation pλ(s) = m − − m n ( 1) n/2 pλ(1 s). − ⌊ ⌋ n − λ Remark 2. From the above geometric interpretation of the pn polynomial line diagonally λ cutting through the parallel Hahn polynomial lines, it follows that scaled pn(s) will not satisfy the three term recurrence relation obeyed by the Hahn polynomials. Remark 3. If a family of polynomials with only critical zeros whose distribution of zeros is proportional to that of the Riemann zeta function is identiﬁed, then if possible it might be of interest to apply the above arguments and see for which values the recurrence coeﬃcients satisfy Favard’s Theorem.

Proof of Theorem 2.4. (a) According to Lemma 2.3, the Mellin transforms are of the form

s+ε 3 pn(s)Γ 2 Mn(s)=Γ , (32) 4 Γ s + 2n+3 2 4 where ε = 0 for n even and = 1 for n odd. The recursions (16) and (17) then follow by λ inserting the form (32) into the recursion for Mn (s) given in part (a) of Theorem 2.2, and repeatedly applying the functional equation of the Gamma function, Γ(z +1) = zΓ(z). (b) We give a proof which makes use of properties of the Gamma function. From Lemma 2.3(a), up to factors not involving s, the polynomials pn may be taken as

3 n+s Γ 4 Γ 2 3 1 n n 3 (n + s) pn(s) = (n + 1) 3F2 , − , ; , 1 ; 1 . (33) 2 Γ s+ε 4 2 − 2 2 − 2 2

From the form of the numerator parameters, the degree of pn is evident. By a ‘Beta transformation’ [19] (p. 850) we have the following integral representation: 3 1 n n 3 (n + s) F , − , ; , 1 ; 1 3 2 4 2 − 2 2 − 2 1 √π 1/4 1/4 1 n n (n + s) = (1 x)− x− F − , ; 1 ; x dx. 2Γ2(3/4) − 2 1 2 − 2 − 2 Z0 18 We use (33), together with an x 1 x transformation of the F function [19] (p. 1043). → − 2 1 Owing to the poles of the Γ function, and that n is a nonnegative integer, the 2F1(x) function then transforms to a single F (1 x) function, and there results 2 1 − (n+s) n+s Γ 1 1+n s (n + 1) Γ 2 √π − 2 Γ 2− pn(s)= s+ε s 1 s 4 Γ Γ(3/4) Γ 1 Γ − 2 − 2 2 1 1/4 1/4 1 n n s n + 1 (1 x)− x− F − , ; − ; 1 x dx × − 2 1 2 − 2 2 − Z0 1+n s (n + 1) π √π 1 Γ 2− = s+ε n+s s 1 s 4 Γ sin π Γ(3/4) Γ 1 Γ − 2 2 − 2 2 1 1/4 1/4 1 n n s n+ 1 (1 x)− x− F − , ; − ; x dx. × − 2 1 2 − 2 2 Z0 The following observations lead to veriﬁcation of the functional equation. When n is even, ε = 0, s s π Γ Γ 1 = , 2 − 2 sin π(s/2) 1 s leaving the denominator factor Γ −2 . When n is odd, ε = 1, s + 1 1 s π Γ Γ − = , 2 2 cos π(s/2) leaving the denominator factor Γ 1 s . − 2 Hence the factor ( 1) n/2 emerges as sin(πs/2)/ sin[π(n + s)/2] = ( 1)n/2 when n is − ⌊ ⌋ − even and as cos(πs/2)/ sin[π(n + s)/2] = ( 1)(n 1)/2 when n is odd, and the functional − − equation of pn(s) follows. (c) The fact that the zeros all lie on the critical line s = 1/2 is a corollary of Theo- ℜ rem 2.2 (e).

λ Proof of Theorem 2.5. The binomial forms of Mn (s) are obtained by rewriting the hypergeometric forms given in Theorem 2.1 in terms of Pochhammer symbols, and then (in a variety of orders) applying the ﬁve identities a + s 1 a(a + 1) . . . (a + s 1) (a) − = − = s , s s! s! 1 1 n + m n − (n + m) n n − (n m k + 1) = m , = − − m , k k (n k + 1) k + m k (k + 1) − m m 1 1 1 n n − (n k + 1) c a + b − a + c − a + c = − m , = . k m k (k m + 1) b b c c b − − − 19 λ λ Corollary 2.6 then follows immediately by by replacing M0 (s) and M0 (s + 1) with their equivalent binomial coeﬃcients forms when s is respectively an even or an odd integer.

Remark 4. With a simple change of variable, the Mellin transforms of (5) may be obtained from [19] (p. 830). Alternative forms of these transforms and their generating functions may be realised by using various expressions from [26] (pp. 279–280).

Proof of Theorem 2.7. It follows from either part (c) of Theorem 2.2 or the hypergeometric form in Theorem 2.5, that the Mellin transforms are of the form

λ 1 s+ǫ λ Γ 2 + 4 Γ 2 λ Mn (s)= s+n+λ 1 pn(s). 2(n!)Γ 2 + 4

λ To demonstrate the diﬀerence equation for pn(s) s + ǫ s + n + λ 1 [6 4(λ + 2λn + n2) 16s + 8s(s + 1)] 1 + pλ(s) − − 2 − 2 4 n s + ǫ s + ǫ +[ 9 + 4(n + λ)2 4(s 1)(s + 2)] 1 pλ(s + 2) − − − 2 2 − n s + n + λ 1 s + n + λ 3 4(s 1)(s 2) + pλ(s 2) = 0, − − − 2 4 2 − 4 n − wherein ǫ = 0 for n even and = 1 for n odd, as follows from either part (d) or Theorem 2.5, the Mellin transforms are of the form

λ 1 s+ǫ λ Γ 2 + 4 Γ 2 λ Mn (s)= s+n+λ 1 pn(s). 2(n!)Γ 2 + 4

λ 1 Noting that the factor Γ 2 + 4 /(2n!) is independent of s, and repeatedly applying the functional equation Γ(z +1) = zΓ(z), the result follows.

Proof of Theorem 2.8. (a) The S:4/1 type combinatorial expressions for the polynomial λ λ factors pn(s) are obtained from the S:4/2 type expressions for Mn (s) in Theorem 2.5, by respectively multiplying through by the factors

n!(2n)! n + s+λ 3 n!(2n + 1)! n + s+1+λ 3 2 − 4 , or 2 − 4 , 2M λ(s) n 2M λ(s + 1) n 0 0 depending on whether n is odd or even. λ (b) The two expressions for qn(s) given can be veriﬁed by inserting the explicit expressions λ λ for pn given in part (a) into the latter expression for qn(s) given in part (b) and rearranging. The degree of both numerator and denominator polynomials of qλ(s) being n/2 , then n ⌊ ⌋

20 λ follows from the degree of the polynomials pn(s) given in Theorem 2.2, and the number λ of s-linear factors appearing in the denominator product of qn(s). λ (c) The zeros of the denominator polynomials (and so poles of qn), correspond to the zeros of the linear factors 2s+2λ+4j 3, or 2s+2λ+4j 1, with 1 j n/2 . For λ> 1/2, − − ≤ ≤ ⌊ ⌋ −λ λ = 0 and s> 0, each linear factor is non-zero, ensuring that the rational function qn(s) 6 ℜ λ has no singularities. Hence the ‘critical zeros’ of the polynomials pn(s), are the same as for qλ(s), and so for t R, the roots of pλ(1/2+ it) and qλ(1/2+ it) are identical. n ∈ n n To see that the rational functions qλ(s) are normalised with limit 1 as s , we n → ∞ consider the limit term by term as s in the S:3/2 sums of (7) and (8), giving →∞ n n r 2r 1 n+r+λ 1 n+r 1 (s 2)+r n n r 2r 1 n+r+λ 1 n+r ( 1) − 2 − − 2 − ( 1) − 2 − − lim − r 2r r = − r 2r , s n+r 1 (s+λ) 3 +r n+r →∞ r=0 2 − 4 r=0 r X r r X 1 n n r2r n+r+λ n+r+1 (s 1)+r n n r 2r n+r+λ n+r+1 ( 1) − 2 2 − ( 1) − 2 lim − r 2r+1 r = − r 2r+1 . s n+r+1 1 (s+λ) 1 +r n+r+1 →∞ 2 − 4 r Xr=0 r r Xr=0 Applying the combinatorial identities

n n r 2r 1 n+r+λ 1 n+r 1 ( 1) − 2 − − 1 2n + 2λ 1 n + λ 1 − − r 2r = − − , n+r 2 2n 1 n 1 r=0 r X − − n n r 2r n+r+λ n+r+1 1 ( 1) − 2 n + 1 2n + 2λ n + λ − − r 2r+1 = , n+r+1 2n + 1 2n n r=0 r X λ we then have the upper bounds for the combinatorial sums, so that lims qn(s) = 1 from λ →∞ below, as required. The functional equation follows from that for pn(s), by considering the third and fourth displays in part (b) of the theorem. λ To see Corollary 2.9, substituting λ = 1 in the S/3:2 forms for pn(s), simpliﬁes the sums to the Gould S/2:1 combinatorial functions stated. Term-by-term analysis of the n + 1 terms in each sum then reveals that for s an integer, each term is an even integer apart form the r = 0 term, given by

n!(2n)! n + s 1 n!(2n + 2)! n + s + 1 2 − 4 , and 2 4 , 2 n 2 n depending of whether n is respectively even or odd. The binomial coeﬃcients contribute 2n 2n+1 the power of two 2− , so that the power of 2 in the r = 0 term is determined by (2n)!/2 when n is even, and (2n + 2)!/22n+1, when n is odd. Noting that the n! terms cancel between numerator and denominator, we see that multiplying through by the reciprocal of these respective powers of 2 will produce odd integer values for the r = 0 term, whilst leaving the others terms r = 1, 2,...,n even. Hence the summation results in integers having only odd prime factors, being the sum of n even numbers and one odd number.

21 Analysis of the nth Catalan number 1 2n = , Cn n + 1 n shows that the power of 2 in the is determined by 22n+1/(2n+2)! (A048881 in the OEIS) Cn so that 4 n 1 and n have the respective reciprocal powers of 2 to p2n and p2n+1. It follows C − C that for s Z we have 4 n 1p2n and np2n+1 are odd integers. A slight modiﬁcation of ∈ C − C this argument also removes the odd factors arising in the (2n)! and (2n + 1)! polynomial factors such that generated by 22n+1 22n+1 p (s), and Tn+1 p (s), (2n)! 2n (2n + 2)! 2n+1 where is the largest odd factor of n + 1. Therefore the above two expressions yield Tn+1 odd integers with fewer prime factors than 4 n 1p2n and np2n+1, as required. C − C Proof of Theorem 2.11 and Corollary 2.12. The proof follows that of Theorem 2.4, noting the integral representation

1 β β 1 n n (n + s) (1 x)− x− F − , ; 1 ; x dx − 2 1 2 − 2 − 2 Z0

2β 1 √πΓ(1 β) 1 n n (n + s) = 2 − − F 1 β, − , ; 2(1 β), 1 ; 1 , Γ(3/2 β) 3 2 − 2 − 2 − − 2 − with β < 1. The F (x) function is again transformed to a F (1 x) function and the 2 1 2 1 − other steps are very similar to before. The location of the zeros follows from Theorem 2.2, setting λ = 3/2 2β. − To see Corollary 2.12 (a) The initial β = 0 reduction of Theorem 2.11 to 2F1 form follows from the series deﬁnition of the 3F2 function with a shift of summation index and the relations (1)j /(2)j = 1/(j + 1) and (κ)j 1 = (κ 1)j /(κ 1). The second − − − reduction is a consequence of Gauss summation. (b) Similarly, with m a positive integer, (m + 1)j/(2(m + 1))j may be reduced and partial fractions applied to this ratio. Then with shifts of summation index, the 3F2 function may be reduced to a series of 2F1(1) functions. These in turn may be written in terms of ratios of Gamma functions from Gauss summation.

4 Discussion

λ λ Given the Gould variant combinatorial expressions obtained for pn(s) and qn(s), our results invite several other research questions, such as: is there a combinatorial interpretation of λ λ pn(s) or qn(s), and more generally, of pn(s; β)? Relatedly, is there a reciprocity relation for pn(s) and pn(s; β)?

22 Two instances when the combinatorial sums produce “nice” combinatorial expressions are

n n+r+λ 1 n+r r 1 1 n r 2r 1 2 2λ 3 2λ 1 − λ ( 1) − 2 − r − 2r −r 1 n + 4− n + 4− q (1) = − λ = , 2n n+r 1 +r 2 n n r=0 2 − 4 X r r n n r 2r n+r+λ n+r+1 r+ 1 1 ( 1) 2 2 2λ 3 2λ+3 − λ − r 2r+1 r n + 4− n + 4 q (2) = − λ = (n + 1) . 2n+1 n+r+1 + 3 +r n n r=0 24 X r r In fact, polynomials with only real zeros commonly arise in combinatorics and elsewhere. Two examples are Bell polynomials [21] and Eulerian polynomials [16] (p. 292). This suggests that there may be other relations of our results to discrete mathematics and other areas. In this regard, we mention matching polynomials with only real zeros. The matching polynomial M(G, x) = ( 1)kp(G, k)xk counts the matchings in a graph G. k − Here, p(G, k) is the number of matchings of size k, i.e., the number of sets of k edges P of G, no two edges having a common vertex. M(G, x) satisﬁes recurrence relations and has only real zeros and M(G v,x) interlaces M(G, x) for any v in the vertex set of − G (e.g., [18]). As regards [6, 12], we note that the classical orthogonal polynomials are closely related to the matching polynomials. For example, the Chebyshev polynomials of the ﬁrst two kinds are the matching polynomials of paths and cycles respectively, and the Hermite polynomials and the Laguerre polynomials are the matching polynomials of complete graphs and complete bipartite graphs, respectively. There are several open topics surrounding the recursions (16) and (17). These include: is it possible to reduce this three-term recursion to two terms, can a pure recursion be obtained, and, can some form of it be used to demonstrate the occurrence of the zeros only on the critical line? The Gegenbauer polynomials have the integral representation

Γ λ + 1 π λ 1 (2λ)n 2 2 n 2λ 1 Cn (x)= (x + x 1 cos θ) sin − θ dθ. (34) √π n! Γ(λ) 0 − Z p λ Then binomial expansion of part of the integrand of Mn (s) is another way to obtain this Mellin transform explicitly. The representation (34) is also convenient for showing further special cases that reduce in terms of Chebyshev polynomials Un or Legendre or associated m Legendre polynomials Pn . We mention as examples 1 C2(x)= [(n + 1)xU (x) (n + 2)U (x)], (35) n 2(x2 1) n+1 − n − and 1 3/2 (n + 1) Pn+1(x) C (x)= [xPn+1(x) Pn(x)] = . n (x2 1) − −√1 x2 − −

23 The Gegenbauer polynomials are a special case of the two-parameter Jacobi polyno- α,β mials Pn (x) (e.g., [2]) as follows:

λ (2λ)n λ 1/2,λ 1/2 Cn (x)= 1 Pn − − (x). λ + 2 n The Jacobi polynomials are orthogonal on [ 1, 1] with respect to the weight function − (1 x)α(1 + x)β. Therefore, it is also of interest to consider Mellin transforms such as − 1 α,β s 1 α,β α/2 1/2 β/2 1/2 M (s)= x − P (x)(1 x) − (1 + x) − dx, n n − Z0 especially as the Jacobi polynomials can be written in the binomial form

n n s s n + α n + β x 1 − x + 1 P α,β(x)= − . n k n s 2 2 Xj=0 − In fact this line of enquiry may provide a far more general approach to investigate ‘critical polynomials’ arising from combinatorial sums.

Appendix

Below are collected various transformations of terminating 3F2(1) series [5], where again (a)n denotes the Pochhammer symbol.

(c a)n(d a)n 3F2( n, a, b; c, d;1) = − − 3F2( n, a, a+b c d n+1; a c n+1, a d n+1; 1) − (c)n(d)n − − − − − − − −

(a)n(c + d a b)n = − − 3F2( n, c a, d a; 1 a n, c + d a b; 1) (c)n(d)n − − − − − − −

(c + d a b)n = − − 3F2( n, d a, d b; d, c + d a b; 1) (c)n − − − − −

n (a)n(b)n = ( 1) 3F2( n, 1 c n, 1 d n; 1 a n, 1 b n; 1) − (c)n(d)n − − − − − − − − −

n (d a)n(d b)n = ( 1) − − 3F2( n, 1 d n, a+b c d n+1; a d n+1, b d n+1; 1) − (c)n(d)n − − − − − − − − − −

(c a)n = − 3F2( n, a, d b; d, a c n + 1; 1) (c)n − − − −

(c a)n(b)n = − 3F2( n, d b; 1 c n; 1 b n, a c n + 1; 1). (c)n(d)n − − − − − − − −

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