Zeta-Regularization of Arithmetic Sequences

Home , Dirichlet series

EPJ Web of Conferences 244, 01008 (2020) https://doi.org/10.1051/epjconf/202024401008 Complexity and Disorder Meetings 2018–2020

Zeta-regularization of arithmetic sequences

Jean-Paul Allouche1,∗ 1CNRS, IMJ-PRG, Sorbonne Université 4 Place Jussieu F-75252 Paris Cedex 05 France

Abstract. Is it possible to give a reasonable value to the infinite product 1 × 2 × 3 × · · · × n × · · · ? In other words, can we define some sort of convergence of the finite product 1 × 2 × 3 × · · · × n when n goes to infinity? One way is to relate this product to the Riemann√ zeta function and to its analytic continuation. This approach leads to: 1 × 2 × 3 × · · · × n × · · · = 2π. More generally the “zeta-regularization” of an infinite product consists of introducing a related Dirichlet series and its analytic continuation at 0 (if it exists). We will survey some properties of this generalized product and allude to applications. Then we will give two families of possibly new examples: one unifies and generalizes known results for the zeta-regularization of the products of Fibonacci, balanced and Lucas-balanced numbers; the other studies the zeta-regularized products of values of classical arithmetic functions. Finally we ask for a possible zeta-regularity notion of complexity.

1 Introduction It is worth noting that zeta-regularization of infinite products is used in physics, for example to define the deter- Mathematicians and physicists like to bypass the “rules” minant of certain infinite-dimensional operators. We cite that objects in their fields have to satisfy: they often want only one reference, namely a 1977 paper by Hawking [10], to explore whether there is something interesting beyond which, in particular, gives a way of calculating the Casimir these rules. For example, they “invented” negative num- effect between two parallel reflecting planes. bers to give a meaning√ √ to 3 − 5. They legitimated expressions like 2/3, 2, or −1. They even introduced expres- 2 Definition and first properties P 2 sions like n≥1 1/n . Another example is the theory of dis- tributions: does there exist a function which is zero every- 2.1 The regularized product where except at 0 where its value is ∞, and whose integral Suppose that Λ = (λi)1≤i≤n is a finite sequence of positive on the real line is equal to 1? Sure! it evens has a name real numbers. Define (the Dirac function of course). In the same spirit, how is it Xn 1 possible to give a reasonable meaning to non-convergents ζ (s) = · × × × · · · × × · · · Λ (λ )s infinite products, e.g., 1 2 3 n ? This is i=1 i a typical example of a product for which there exists a√ so- called “zeta-regularization", which assigns the value 2π We clearly have to this product. n n 0 X Y 0 In the sequel, we will first recall a “natural” defini- ζΛ(0) = − log λi i.e., log λi = −ζΛ(0) tion of zeta-regularizations of infinite products, and give i=1 i=1 some properties and “classical” examples. For this part which can be written the reader is referred to the references given at the end, n Y 0 and to the references given in [1]. Then we will give −ζ (0) λi = e Λ . two families of examples, one providing a unified treat- i=1 ment of Fibonacci-like sequences; the other showing zeta- regularized products that we did not find in the literature, Now suppose that Λ = (λi)i≥1 is an infinite non-decreasing where arithmetic functions like d(n), the number of divi- sequence of positive real numbers. Let ζΛ(s) the associated sors of the integer n, enter the picture. Finally we will Dirichlet series, i.e., mention our (still unsuccessful) attempts to concoct a com- X 1 · ζΛ(s) = s plexity measure from the zeta-regularized product of se- (λi) quences of integers. i≥1 By analogy with what precedes, we give the following def- ∗e-mail: [email protected] initions (see, [6] or [25, p. 96–97]):

© The Authors, published by EDP Sciences. This is an open access article distributed under the terms of the Creative Commons Attribution License 4.0 (http://creativecommons.org/licenses/by/4.0/). EPJ Web of Conferences 244, 01008 (2020) https://doi.org/10.1051/epjconf/202024401008 Complexity and Disorder Meetings 2018–2020

Deﬁnition 1 Suppose that ζΛ(s) converges for

Deﬁnition 2 Suppose that ζΛ(s) converges for 1. (Compare with the famous ≥ (i) If Λ = {λ1, λ2, ··· , λk, λk+1 ...} with positive λi’s, n 0 X Ya∞ Yk Ya∞ “Ramanujan sum” n = −1/12, see, e.g., [27].) then λi = λ j λi. n≥1 i=1 j=1 i=k+1 Ya n = 2π, where S is the set of squarefree integers. (ii) If If A ∪ B is a partition of the positive integers, then n∈S ∞ Ya Ya Ya Ya (Note that we also have n2 = 2π.) λi = λi λi. n≥1 i=1 i∈A i∈B Ya 0 0 nn = e−ζ (−1) = Ae−1/12, where A = e−ζ (−1)+1/12 is Ya d Ya d (iii) If d is a positive integer, then λi = λi . n≥1 known as the Glaisher-Kinkelin constant. Ya∞ Ya∞ ζΛ(0) (iv) For a > 0, (aλi) = a λi. Remark 6 i=1 i=1 ∗ From, e.g., [4, p. 6] one has (compare with Remark 5):

Ya Ya Ya Q` 4 4 ≥ (n − 1) Y n − 1 Remark 5 In general (λiµi) , λi µi . n 2 = · Q` (n4 + 1) n4 + 1 n≥2 n≥2 Q` 3 Classical examples ∗ To prove n∈sqf n = 2π, let µ be the Möbius function. P |µ(n)| ζ(s) From, e.g., [9], Q(s):= n≥1 ns = ζ(2s) for ~~1. 0 0 − log(2π)/2 We give some classical examples. Then Q (0) = −ζ (0)/ζ(0) = − −1/2 = − log(2π).~~

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Q` n −ζ0(−1) ∗ The example n≥1 n = e might need some The function H is holomorphic, so that we have the mero- explanation: it should be considered as the zeta- moprhic continuation of ζ , with poles located at s such X regularized product of the sequence of integers s s+k−1 that λ = 1. Furthermore H(0) = 0 (since k = − 1 2 2 3 3 3 ... n n ... n ... s(s + 1) ... (s + k 1)/k!). So that |{z} |{z} |{z} | {z } 1 factor 2 factors 3 factors n factors H(s) X X 1 H0(0) = lim = γk(bλ−2)nk More generally, if (an) is a sequence of integers, then s→0 s k Q` n≥1 k≥1 nan should be considered as the zeta-regularized n≥1 Y −2 n product of the following sequence of integers = − log (1 − γ(bλ ) ). n≥1 Λ = 1 ... 1 2 ... 2 3 ... 3 ... n ... n ... |{z} |{z} | {z } | {z } a factors a factors a factors a factors On the other hand, a quick computation shows that the 1 2 3 n 1 coeﬃcient of s in the Laurent series expansion of s s with associated Dirichlet series P an = ζ (s). α (λ −1) n≥1 ns Λ is log α 1 1 + log α + log λ. 4 Fibonacci and Fibonacci-like numbers 2 log λ 2 12 The zeta-regularized product of the Fibonacci numbers Corollary 8 With the notation of Theorem 7 we have (i.e., the sequence A000045 = 0, 1, 1, 2, 3 ... in the OEIS Ya 2 Y [20]) and the zeta-regularized product of the balancing and −1/2 −1/12 − log α/2 log λ −2 n Xn = α λ e (1 − γ(bλ ) ). Lucas-balancing numbers (i.e., the sequences A001109 n≥1 n≥1 and A001541 in [20]) are computed in [8, 13] and in [7]. Of course the proofs involve the zeta functions associated In particular we have:

with these sequences of integers. We give here a unified * (see [13]) Let Fn be the Fibonacci numbers, defined proof based on “explicit” expressions for the meromorphic by F0 = 0,F√ 1 = 1, and for all n ≥ 0,Fn+2 = Fn+1 + Fn. continuations of these sequences. Our approach is similar Let ϕ = 1+ 5 . Then to the ones given, e.g., in [19, 23] or, in a different con- 2 text, in [2] (also see, e.g., [1, 3, 26], and [5] which gives Ya  1/4 −(log2 5)/(8 log ϕ)  Y 5 e  2 −n interesting historical remarks). Fn =   (1 − (−ϕ ) ).  ϕ1/12  n≥1 n≥1 Theorem 7 Let a and b be two rational integers with a2 − 4b > 0 such that its two (real) roots λ and λ0 sat- ∗ (see [7]) Let B and C be the balancing and Lucas- isfy λ > 1 and |λ0| = |b|/λ < 1. Let (X ) be a se- n n n n≥0 balancing numbers, defined by B = 0,B = 1 ,and for quence of integers that satisfies the recurrence equation 0 1 all n ≥ 0,B = 6B −B , respectively C = 0,C = 3 X = aX − bX , for all n ≥ 0. Let X = αλn + βλ0n. n+2 n+1 n 0 1 √ n+2 n+1 n n and for all n ≥ 0,C = 6C − C . Let µ = 3 + 8. We let γ = −β/α and we suppose that α > 0 and that n+2 n+1 n 2 P s Then |γb|/λ < 1. Define ζX(s) = n≥1 1/Xn. Then ζX has a meromorphic continuation given by Ya  5/4 −(log2 32)/(8 log µ)  Y 2 e  −2n ! Bn =   (1 − µ ) 1 1 X s + k − 1 γkβk  µ1/12  + · n≥1 n≥1 αs(λs − 1) αs k λ2k+s − bk k≥1 and Furthermore the coefficient of s in the Laurent series ex- √ Ya  −(log2 2)/(2 log µ)  Y pansion of this continuation in the neighborhood of s = 0  2e  −2n Cn =   (1 + µ ). is equal to  µ1/12  n≥1 n≥1 log α 1 1 Y + log α + log λ − log 1 − γ(bλ−2)n . 2 log λ 2 12 −2n n≥1 Remark 9 Note that there is a misprint in [13]: (−ϕ) should be replaced with (−ϕ2)−n. n 0n n −2 n Proof. We have Xn = αλ + βλ = αλ (1 − γ(bλ ) ), hence X 1 1 X 1 5 Some more arithmetic examples ζ (s) = = (1 − γ(bλ−2)n)−s X Xs αs λns n≥1 n n≥1 The first result of this section addresses the product of all ! 1 X 1 X s + k − 1 k −2 nk primes. The definitions above for zeta-regularized prod- = γ (bλ ) P s αs λns k ucts cannot be used: the Dirichlet series p prime 1/p can- n≥1 k≥0 ! not be extended meromorphically to a neighborhood of 0 1 X 1 1 X 1 X s + k − 1 = + γk(bλ−2)nk since the imaginary axis is its natural boundary [16]. But, αs λns αs λns k n≥1 n≥1 k≥1 extending once more the zeta-regularization of a product, ! 1 1 X 1 X s + k − 1 the notion of “super-regularization” was defined in [18] = + γk(bλ−2)nk (also see [24]) yielding: αs λs − αs λns k ( 1) ≥ ≥ n 1 k 1 Ya 1 p = 4π2. = + H(s). αs(λs − 1) p prime

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A second arithmetic example is the product of all odious X d(n2) ζ3(s) = , for 1 numbers. Recall that an integer is said to be odious if its ns ζ(2s) n≥1 binary expansion contains an odd number of 1’s and evil X θ(n) ζ2(s) otherwise. One has [1]: = , for 1 ns ζ(2s) n≥1 Let Λ = 1 2 4 7 8 11 13 14 16 ... be the sequence X r (n) 2 = ζ(s)β(s), for <(s) > 1. of odious numbers (sequence A000069 in the OEIS [20]). ns Then n≥1 Ya p n = π1/4 2ϕe−γ n∈Λ It is then easy to ﬁnish the proofs using the well-known values ζ(0) = −1/2, ζ0(0) = − log(2π)/2, ζ(−1) = −1/12, where γ and ϕ are respectively the Euler-Mascheroni and ζ0(−1) = − log A+1/12, as well as the possibly less-known the Flajolet-Martin constants. √ values β(0) = 1/2 and β0(0) = log(Γ2(1/4)/2π 2) which can be obtained from values of the Hurwitz zeta function The following examples might not be in the literature. ζ(s, u) = P 1/(n + u)s, namely ζ(0, u) = −u + 1/2, They involve arithmetic functions (as was previously the n≥0 √ ζ0(0, u) = log(Γ(u)/ 2π), since β(s) = (ζ(s, 1/4) − case with the Dirichlet series P |µ(n)| ; see both parts of Re- ns ζ(s, 3/4))/4s. mark 6 above) and largely use the catalog of Dirichlet series [9] and the references therein, from which it is easy to compute their derivatives at 0. We give only a few exam- 6 Towards a notion of complexity of ples; several other examples can be deduced from [9] and sequences of integers? its references.

Theorem 10 Let d(n) be the number of divisors of the in- It is tempting to try to use the zeta-regularized product of teger n, σ(n) be the sum of the divisors of the integer n, and a non-decreasing sequence of integers λA, λ2,... to “mea- ϕ(n) be the Euler totient function. Let θ(n) be the number sure how complex” this sequence is. Does this product of ordered paris a, b of positive integers such that n = ab detect whether these numbers are “dense” or whether they and gcd(a, b) = 1 (hence θ(n) = 2ω(n) where ω(n) is the tend quickly to infinity? One possible difficulty is that, number of distinct prime factors of n). Let ζ be the Rie- in a usual (finite) product, there might be some “compen- mann zeta function. Let A be the Glaisher-Kinkelin con- sation”: the product of “dense” numbers could be larger than the product of “scarce” numbers, but on the other stant. Let finally r2(n) be the number of decomposition of the integer n into the sum of the squares of two inte- hand scarce integers could grow extremely quickly mak- P (−1)n ing the product possibly huge. What happens with a zeta- gers, and β(n) = n≥0 (2n+1)s be the beta-Catalan (or beta- Dirichlet) function. Then regularized product? An example given above is deceiv- ing: the zeta-regularized product of the squarefree integers Ya is equal to the zeta-regularized product of the squares. nd(n) = (2π)−1/2 n≥1 Another approach could be to try to evaluate the tail of the Ya ∞ nσ(n) = e1/24(2π)−1/24A−1/2 Ya zeta-regularized product: let Ck(Λ) = λi. One has n≥1 Ya ϕ(n) 1/6 −2 i=k+1 n = (2πe) A    −1 Ya∞ Yk  n≥1     Ya Ck(Λ) =  λi  λ j . This means that the zeta- d2(n) 1/4     n = (2π) i=1 j=1 n≥1 regularized product could discriminate between two se- Ya 2 nd(n ) = (2π)−1/4 quences whose first k terms give products with the same Yan≥1 growth. But we have not found interesting examples yet. nθ(n) = 1 On the other hand, let (an)n be the characteristic function n≥1 √ Ya r (n) 1/4 of a sequence of integers Λ, and let us denote as previously n 2 = Γ(1/4)/(π 2). P s P s ζΛ(s) = n∈Λ 1/n = n≥1 an/n . The analytic density of n≥1 Λ is the limit (if it exists) of lims→1(s − 1)ζΛ(s) when s tends to 1 . (Note that if ζ has a simple pole at s = 1, Proof. Using, e.g., [9], we have: + Λ then this limit is equal to Res ζλ(s).) The logarithmic den- s=1 P X d(n) sity of Λ is the limit (if it exists) of ( 1/n)/ log n = ζ2(s), for 1 n∈Λ, n≤x ns when n goes to infinity: a nice result is that one of these n≥1 X σ(n) two densities exists for a sequence Λ if and only if the = ζ(s)ζ(s − 1), for 2 other one exists, and they must be equal (see, e.g., [21, p. ns n≥1 96]). Is it reasonable to think that there might exist a sort of X ϕ(n) ζ(s − 1) = , for 2 density associated with this residue (compare with Defini- ns ζ(s) ζΛ(s) n≥1 tion 2, where the quantity Res 2 occurs), or is it totally s=0 s X d2(n) ζ4(s) = , for ~~1 hopeless to evaluate the scarcity/density of a sequence of ns ζ(2s) n≥1 integers from its zeta-regularized product?~~

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