Define f(k)=|{(m,n)∈N2:[m,n]=k}|f(k) = |\{(m, n) \in \mathbb{N}^2 : [m,n] = k\}| and F(s)=∞∑i=1f(i)isF(s) = \sum_{i=1}^\infty \frac{f(i)}{i^s} where s=σ+it∈C.s= \sigma + it \in \mathbb{C}.

Write FF in Reimann zeta function form and determine its half-plane of absolute convergence.

I would like to study this problem. First, I notice that F(s)=∞∑i=1f(i)is=∞∑m=1∞∑n=11[m,n]s=∑m>1∞∑n=1[m,n]−s+∞∑i=11is=∑m>1,n=1,2,3…[m,n]−s+ζ(s).F(s) = \sum_{i=1}^\infty \frac{f(i)}{i^s} =\sum_{m = 1}^\infty \sum_{n=1}^\infty \frac{1}{[m, n]^{s}} = \sum_{m>1}\sum_{n=1}^\infty [m, n]^{-s} + \sum_{i=1}^\infty \frac{1}{i^s} = \sum_{m>1, n= 1,2,3…} [m, n]^{-s} + \zeta(s). For m=2m = 2,

∞∑n=11[m,n]s=12s+12s+16s+…=12s+∞∑k=11(2k)s+∞∑k=31(2k)s=2∞∑k=11(2k)s−14.\sum_{n=1}^\infty \frac{1}{[m, n]^s} = \frac{1}{2^s} + \frac{1}{2^s} + \frac{1}{6^s} + … = \frac{1}{2^s} + \sum_{k=1}^\infty \frac{1}{(2k)^s} + \sum_{k=3}^\infty \frac{1}{(2k)^s} = 2 \sum_{k=1}^\infty \frac{1}{(2k)^s} – \frac{1}{4}.

Not sure what is the Riemann zeta form of that. I also think that this is not a good idea to acheive the formula (if it has pattern, it might suggest the formula.)

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1 Answer

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We first claim the following:

Lemma. For any n∈Z+n\in\mathbf Z_+ it holds that

f(n)=d(n2),f(n)=d(n^2),

where d(⋅)d(\cdot) is the divisor function.

Proof. We prove it by induction on the number of factors of kk. If kk is a power of a prime then the result is plainly true, so suppose that it holds when kk has NN distinct prime factors, and we will show that

f(k⋅pγ)=f(k)f(pγ)f(k\cdot p^\gamma)=f(k)f(p^\gamma)

for any prime pp that does not divide kk. Indeed, a pair (a,b)(a,b) satisfies that [a,b]=kpγ[a,b]=kp^\gamma iff either pγ|ap^\gamma|a, pj|bp^j|b (j=0,…,γj=0,\dots,\gamma) and [a/pγ,b/pj]=k[a/p^\gamma,b/p^j]=k or pj|ap^j|a (j=0,…,γ−1j=0,\dots,\gamma-1), pγ|bp^\gamma|b and [a/pj,b/pγ]=k[a/p^j,b/p^\gamma]=k. Since there are f(k)⋅(γ+1)f(k)\cdot (\gamma+1) possibilities in the former case, and f(k)⋅γf(k)\cdot \gamma in the latter, we have that the number of possible pairs is

f(k)[γ+1+γ]=f(k)(2γ+1)=d(k2)d(p2γ)=d(k2p2γ)=f(kpγ).f(k)[\gamma+1+\gamma]=f(k)(2\gamma+1)=d(k^2)d(p^{2\gamma})=d(k^2p^{2\gamma})=f(kp^\gamma).

This proves the lemma. ◼\blacksquare

From the lemma, we see then that

∑n≥1f(n)ns=∑n≥1d(n2)ns.\sum_{n\geq1}\frac{f(n)}{n^s}=\sum_{n\geq1}\frac{d(n^2)}{n^s}.

Since d(n2)d(n^2) is a multiplicative function, we know that

∑n≥1d(n2)ns=∏p{1+d(p2)ps+d(p4)p2s+…}\sum_{n\geq1}\frac{d(n^2)}{n^s}=

\prod_{p}\left\{1+\frac{d(p^2)}{p^s}+\frac{d(p^4)}{p^{2s}}+\dots\right\}

where the product is extended to all the primes. This is,

∑n≥1d(n2)ns=∏p{1+3ps+5p2s+…}=∏p∞∑l=02l+1psl=∏pps(ps+1)(ps−1)2=∏p11−ps∏p(1+p−s)∏p11−p−s=ζ(s)ζ(s)ζ(2s)ζ(s)=ζ(s)3ζ(2s).\begin{align}\sum_{n\geq1}\frac{d(n^2)}{n^s}

&=\prod_{p}\left\{1+\frac{3}{p^s}+\frac{5}{p^{2s}}+\dots\right\}\\

&=\prod_p\sum_{l=0}^\infty\frac{2l+1}{p^{sl}}\\

&=\prod_p

\frac{p^s \left(p^s+1\right)}{\left(p^s-1\right)^2}\\

&=\prod_p\frac1{1-p^s}\prod_p(1+p^{-s})\prod_p\frac1{1-p^{-s}}\\

&=\zeta(s)\frac{\zeta(s)}{\zeta(2s)}\zeta(s)\\

&=\frac{\zeta(s)^3}{\zeta(2s)}.

\end{align}

Hence

∑n≥1f(n)ns=ζ(s)3ζ(2s),whenever ℜ(s)>1.\sum_{n\geq1}\frac{f(n)}{n^s}=\frac{\zeta(s)^3}{\zeta(2s)},\qquad\text{whenever }\Re(s)>1.

I would start with : if gcd(k,k′)=1gcd(k,k’) = 1 and lcm(m,n)=k,lcm(m′,n′)=k′lcm(m,n) = k, lcm(m’,n’) = k’ then lcm(mm′,nn′)=kk′lcm(mm’,nn’) = kk’ and conversely, if lcm(a,b)=kk′lcm(a,b) = kk’ then a=mm′,b=nn′,lcm(m,n)=k,lcm(m′,n′)=k′a = mm’,b = nn’, lcm(m,n)=k,lcm(m’,n’)=k’, thus f(k)f(k) is multiplicative.

– user1952009

Oct 20 at 19:06

@iqcd I try to convince myself about your Lemma, but I am not sure if it is correct. If I do not calculate anything wrong, f(6)=|{(1,6),(2,6),(3,6),(6,6),(6,1),(6,2),(6,3)}|=7f(6) = |\{(1,6), (2,6), (3,6), (6,6), (6,1), (6,2), (6,3)\}| = 7 but d(36)=d(22)d(32)=9?d(36) = d(2^2)d(3^2) = 9 ?

– Both Htob

Oct 20 at 19:16

Oh, okay. It is just silly that I forget (2,3)(2, 3) and (3,2)(3, 2). Okay, so it seems that your Lemma is perfectly fine. Could you tell me the intuition ? I mean how can you recognize that this two functions are the same. I really have no idea that the number of ordered pairs for [m,n][m, n] is exactly the number of positve divisor of a square.

– Both Htob

Oct 20 at 19:21

@user1952009 Definitively a quicker and much better solution. I don’t know why I didn’t start trying that out first

– iqcd

Oct 20 at 19:22

@BothHtob I thought “If this function is multiplicative, then it is easy”. After proving it, I noticed that f(pγ)=2γ+1f(p^\gamma)=2\gamma+1 for each prime, and recalled that d(pγ)=γ+1d(p^\gamma)=\gamma+1, so the identification became evident.

– iqcd

Oct 20 at 19:24