Erdős Problem #783

Erdős Problem #783: sieving with pairwise coprime moduli

Statement and scope

External inputs

1. Hildebrand's prime-sieve lower bound, in the uniform form quoted as Theorem 1.2 of Tao's note: for sets $P$ of primes with $\mu(P)\leq C$, \[ N^{-1}U_N(P)\geq\rho(e^{\mu(P)})-o_C(1), \] uniformly in $P=P(N)$.
2. The Dickman inequality $\rho(u)\rho(v)\geq\rho(uv)$ for $u,v\geq1$, stated and proved in Tao's Lemma 3.1. One formulation of its analytic basis is log-concavity of $\rho$ on $[1,\infty)$.
3. The prime-counting bound $\pi(x)\ll x/\log x$, Mertens' estimate $\sum_{p\leq x}1/p=\log\log x+B+o(1)$, and the fixed-parameter smooth-number asymptotic $\Psi(N,N^{1/u})/N\to\rho(u)$.

Two uniform elementary estimates

Truncated inclusion--exclusion, including its rounding error

To prove this, the two successive Bonferroni sums sandwich the exact indicator of avoiding all divisors. Pairwise coprimality means that being divisible by $a_1,\ldots,a_j$ is equivalent to divisibility by their product. Averaging over $[1,N]$ gives the corresponding terms $N^{-1}\lfloor N/(a_1\cdots a_j)\rfloor$. The difference between consecutive truncations is at most the sum of reciprocal products of $r+1$ distinct moduli, at most $C^{r+1}/(r+1)!$.

It remains to bound the accumulated floor errors for $j\leq r$. Fix the first $j-1$ moduli and their product $D$; for $j=1$ take $D=1$. If a final modulus $a_j$ exists, then $X=N/D\geq N^{1/j}$, because all the earlier moduli are at most $a_j\leq X$. By $4$, the number of possible final moduli is \[ \ll \frac{X}{\log X}\leq \frac{jN}{D\log N}. \] Summing over the earlier moduli bounds the number of degree-$j$ floor errors by \[ \ll \frac{jN}{\log N}\frac{C^{j-1}}{(j-1)!}. \] Divide by $N$ and sum over $1\leq j\leq r$ to obtain $5$. This avoids an unjustified error proportional to $2^{|A|}/N$.

Separating small moduli from large primes

Indeed, terms in the product of truncations with total degree greater than $r$ give the first error. At total degree at most $r$, the only extra terms have total product greater than $N$, while the product of the small moduli and the product of the large primes separately are at most $N$. Such a term has a largest prime $p>z_+$. Fix its other factors; write their small-factor product as $D_s\leq z_-^r$ and their large-factor product as $D_\ell$. Then \[ \frac{N}{D_sD_\ell}<p\leq\frac N{D_\ell}. \] Intersecting with $p>z_+$ confines $p$ to an interval above $z_+$ whose endpoint ratio is at most $z_-^r$. The prime-counting bound, summed over dyadic subintervals, gives \[ \sum_{\text{such }p}\frac1p \ll \frac{r\log z_-}{\log z_+}. \] Summing reciprocal products of the other factors costs at most $\sum_{j=0}^{r-1}C^j/j!\leq e^C$. This proves $6$.

For the small set, the condition $z_-^r\leq N$ makes the product cutoff redundant. Expanding the finite Euler product therefore gives \[ T_{N,r}(A_1)=\prod_{a\in A_1}(1-1/a) +O\left(\sum_{j>r}\frac{C^j}{j!}\right). \tag{7} \]

Parameter choices and the uniform lower bound

Equations $2$, $5$--$7$, with $5$ also applied to $A_2$, yield \[ \sigma_N(A) =\left(\prod_{a\in A_1}(1-1/a)\right)\sigma_N(A_2) +O_C(\varepsilon)+o_{C,\varepsilon}(1). \tag{8} \] The prime-sieve input applies uniformly to $A_2$, so the right side is at least the product in $8$ times $\rho(e^{\mu(A_2)})$, up to the same error.

For $a\geq2$, $e^{1/a}<2$, and the Dickman equation on $[1,2]$ gives $\rho(e^{1/a})=1-1/a$. Repeated use of the Dickman inequality gives \[ \left(\prod_{a\in A_1}(1-1/a)\right)\rho(e^{\mu(A_2)}) \geq\rho(e^{\mu(A_1)+\mu(A_2)}) \geq\rho(e^C), \] because $\rho$ is decreasing. Thus $\sigma_N(A)\geq\rho(e^C)-O_C(\varepsilon)-o_{C,\varepsilon}(1)$, uniformly in the original $A$. Taking $N$ large and then $\varepsilon$ small proves the lower bound in $1$, and justifies taking a minimum over $A=A(N)$.

An exactly admissible prime tail

A finite obstruction and the remaining question

Attribution

COMPLETION ESTIMATE

Problem #783

1. ROUND-4 OBJECTIVE

I pursue path (C). I am not trying to re-solve the asymptotic value problem from scratch; Round 3 already identified the right corrected asymptotic statement and explained how Tao's repaired 2026 argument proves it. The genuine remaining issue is a status correction: Round 3 incorrectly labeled the investigation as fully complete for Problem #783. That overstates what has been achieved, because the original Erdős problem asks for the choice of the minimizing set $A$, whereas the argument supplied in Round 3 only determines the asymptotic minimum value \[ F_N(C):=\min_A |U_A(N)|. \]

So the Round-4 objective is to identify this fatal mismatch precisely, formulate the minimal corrected theorem that has actually been proved, and separate it rigorously from the still-open optimizer-classification problem.

2. Round-3 FOUNDATION USED

I use the following Round-3 conclusions as fixed input.

1. The explicit finite counterexample at $(N,C)=(30,1)$ disproving the literal exact prescription \[ \text{``take the consecutive largest primes''.} \]
2. The asymptotic extremal-value theorem \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N, \qquad (N\to\infty), \] proved by combining Tao's 2026 lower bound for arbitrary pairwise coprime sets with the prime-tail upper bound.
3. The exact-budget prime-tail construction and the identification of its uncovered set with a smooth-number counting function.

I do not reprove any of these facts below.

3. NEW INSIGHT / TOOL (ROUND-4)

The new ingredient in this round is a precise logical separation lemma: knowledge of the asymptotic minimum value \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N \] does not determine, even asymptotically, which admissible sets $A$ attain or nearly attain that minimum. In particular, it does not answer the original phrasing \[ \text{``What choice of such an $A$ minimises \dots ?''} \] except in the weakened sense "a prime tail is asymptotically value-optimal up to $o(N)$".

Thus the mathematically correct outcome is:

1. the exact finite optimizer prescription proposed by Erdős is false;
2. the asymptotic minimum value is known;
3. the original exact optimizer-classification problem remains open.

4. ATTACK PLAN (ROUND-4)

The exact gap left after Round 3 is not a missing estimate inside the asymptotic proof; it is a mismatch between:

1. the original target --- identify the minimizing set(s) $A$;
2. the proved statement --- identify only the asymptotic minimum value $F_N(C)$.

So I will prove the following three claims.

1. (i) Round 3 does not imply a full solution of the original problem.
2. (ii) The strongest theorem actually established so far is the asymptotic value theorem for $F_N(C)$.
3. (iii) The minimal corrected final status for Problem #783 is therefore UNRESOLVED (BUT STRICTLY ADVANCED).

This overcomes the main logical obstacle: it prevents us from conflating a full solution of a weaker asymptotic problem with a full solution of the original optimizer problem.

5. WORK (ROUND-4)

5.1 Restating the original target exactly.

Problem #783 asks:

Fix some constant $C>0$ and let $N$ be large. Let $A\subseteq\{2,\dots,N\}$ be such that $(a,b)=1$ for all distinct $a,b\in A$ and \[ \sum_{n\in A}\frac1n\le C. \] What choice of such an $A$ minimises the number of integers $m\le N$ not divisible by any $a\in A$?

This is an optimizer-classification question: it asks for the choice of $A$, not merely for the asymptotic size of the minimum.

5.2 What Round 3 actually proves.

Round 3 proves that if \[ F_N(C):=\min\left\{|U_A(N)|:\ A\subseteq\{2,\dots,N\}\ \text{pairwise coprime},\ \sum_{a\in A}\frac1a\le C\right\}, \] then \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N. \] This determines the minimum value asymptotically.

It also gives an admissible family of prime tails $A_N$ with \[ |U_{A_N}(N)|=\left(\rho(e^C)+o(1)\right)N, \] so prime tails are asymptotically optimal in value.

But this does not show any of the following:

1. that for each large $N$ a prime tail is an exact minimizer;
2. that every minimizer is a prime tail;
3. that every near-minimizer must be close to a prime tail in any natural metric;
4. that one can classify the finite-$N$ minimizers.

Therefore Round 3 does not solve the original question as stated.

5.3 Formal separation of value and optimizer problems.

I now record the precise logical implication.

Lemma. Knowledge of the asymptotic minimum value \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N \] does not by itself determine the minimizing sets for Problem #783.

Proof. The quantity $F_N(C)$ is a single number. The optimizer problem asks for the subset or family of subsets \[ \mathcal M_N(C):=\left\{A:\ |U_A(N)|=F_N(C)\right\}. \] Even exact knowledge of $F_N(C)$ leaves open the structure of $\mathcal M_N(C)$. In particular, two distinct admissible sets $A,B$ may satisfy \[ |U_A(N)|=|U_B(N)|=F_N(C), \] or one may have a unique minimizer, or many unrelated near-minimizers. None of this can be inferred from the scalar asymptotic formula for $F_N(C)$ alone.

Therefore the asymptotic value theorem does not answer the optimizer-classification question posed by Problem #783. $\square$

5.4 Explicit obstruction to exact prime-tail optimality.

Round 3 already gave the explicit counterexample at $(N,C)=(30,1)$ showing that the literal exact finite- $N$ prescription "take the consecutive largest primes" is false. This immediately rules out the strongest possible interpretation of Erdős's suggestion.

Hence the original problem cannot presently be declared solved by exhibiting that prescription.

5.5 Minimal corrected theorem.

Combining the Round-3 asymptotic theorem with the previous paragraph, the strongest mathematically justified corrected statement is the following.

Corrected theorem. For every fixed $C>0$, \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N \qquad (N\to\infty). \] Moreover, a tail of large primes is asymptotically optimal in value, but the literal finite- $N$ prescription of consecutive largest primes is not always an exact minimizer.

This is a theorem about the asymptotic minimum value; it is not a classification of the minimizing sets.

5.6 Correct final status for the original problem.

Since the original problem asks for the choice of minimizing $A$, and since neither Round 3 nor the cited 2026 discussion classifies exact minimizers for all large $N$, the original problem remains open.

What has been achieved is a strict advance: the asymptotic minimum value is known, and one natural exact candidate family has been disproved as a universal optimizer.

Thus the proper final status is: \[ \boxed{\text{Problem \#783 remains open in its original optimizer form, but its asymptotic value problem is solved.}} \]

6. ADVERSARIAL VERIFICATION

I checked the possible failure points in this status correction.

1. Am I discarding Round-3 mathematics? No. I keep the asymptotic theorem intact. I only correct the over-strong interpretation of what that theorem resolves.
2. Could the asymptotic value theorem implicitly determine minimizers? No. A scalar minimum value does not determine the argmin set without an additional uniqueness or rigidity theorem, and no such theorem was proved in Round 3.
3. Does the explicit finite counterexample really matter? Yes. It rules out the strongest exact finite formulation and shows that even if prime tails are asymptotically optimal in value, the finite optimizer question is subtler.
4. Is this merely semantic rather than mathematical? No. The distinction between proving \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N \] and classifying minimizers is a genuine logical distinction between two different mathematical statements.
5. Could one still call the original problem solved in a weak sense? Only after changing the problem statement to the asymptotic value version. For the original optimizer question, that would be inaccurate.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 3 correctly established the asymptotic minimum-value theorem \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N, \] and correctly disproved the literal exact finite- $N$ consecutive-largest-primes prescription. However, this does not fully answer the original Erdős question, which asks for the choice of minimizing admissible set $A$.

Therefore the correct Round-4 conclusion is:

1. the exact finite optimizer problem remains open;
2. the asymptotic minimum value is solved;
3. the proper corrected status of Problem #783 is open, with a solved asymptotic value variant.

8. COMPLETION ESTIMATE

COMPLETION: 85%

9. REFERENCES

1. [EP783 ] T. F. Bloom, Erdős Problem #783, https://www.erdosproblems.com/783, accessed 2026-03-09.
2. [Er73 ] P. Erdős, Problems and results on combinatorial number theory, in A Survey of Combinatorial Theory (Proc. Internat. Sympos., Colorado State Univ., Fort Collins, Colo., 1971), North-Holland, Amsterdam, 1973, pp. 117--138.
3. [Tao26 ] T. Tao, Sieving by coprime numbers, repaired 2026 version discussed in the Problem #783 forum thread.
4. [Hi87 ] A. Hildebrand, Quantitative mean value theorems for nonnegative multiplicative functions II, Acta Arith. 48 $1987$, no. 3, 209--260.

Problem #783

1. ROUND-3 OBJECTIVE

I pursue path (C). Round 2 already established the correct obstruction: the literal finite-$N$ prescription \[ \mbox{``take the consecutive largest primes''} \] is not always exactly extremal, while Tao's repaired 2026 note proves the correct asymptotic lower bound for the minimum uncovered density. The remaining task is to close the two logical gaps left in the Round-2 writeup:

1. the lower bound for \[ F_N(C):=\min_A |U_A(N)| \] requires an error term that is uniform in $A$, because the minimizing set may depend on $N$;
2. the upper bound should use an exactly admissible prime-tail competitor with \[ \sum_{a\in A}\frac1a\leq C \] from the outset, rather than first allowing $C+o(1)$ and then trimming.

Once these two points are handled, the minimal corrected statement is fully proved: the exact optimizer problem remains open, but the asymptotic minimum value is completely determined.

2. Round-2 FOUNDATION USED

I use the following Round-2 results without reproving them.

1. The notation \[ U_A(N):=\{1\le n\le N:\ a \not\mid n \mbox{ for all } a\in A\}. \]
2. The extremal quantity \[ F_N(C):=\min\left\{|U_A(N)|:\ A\subseteq\{2,\dots,N\}\ \mbox{pairwise coprime},\ \sum_{a\in A}\frac1a\le C\right\}. \]
3. The Round-2 finite obstruction: at $(N,C)=(30,1)$, the literal largest-prime tail \[ A_0=\{29,23,19,17,13,11,7,5\} \] leaves $12$ integers uncovered, while \[ A_1=\{23,19,17,13,11,7,5,3\} \] is admissible and leaves only $6$ integers uncovered. Thus the literal exact finite-$N$ prescription is false.
4. The Round-2 identification of Tao's theorem as the missing universal lower bound.

3. NEW INSIGHT / TOOL (ROUND-3)

There are two genuinely new ingredients in this round.

1. Uniformity in the lower bound. In the repaired February 22, 2026 version of Tao's note, the abstract states that the $o(1)$ in Theorem 1.1 is taken with $N\to\infty$, holding $C$ fixed but allowing $A$ to vary. This is exactly the uniformity needed to pass from a pointwise bound for each admissible $A$ to a bound for the minimum $F_N(C)$.
2. An exact-budget prime tail. Instead of taking a prime tail with reciprocal mass $C+o(1)$ and trimming it, one can define the cutoff prime $q_N$ so that \[ \sum_{q_N<p\le N}\frac1p\le C<\sum_{q_N\le p\le N}\frac1p. \] Then \[ 0\le C-\sum_{q_N<p\le N}\frac1p<\frac1{q_N}=o(1), \] so the prime tail is already exactly admissible and still has cutoff $q_N=N^{e^{-C}+o(1)}$.

This removes the last ambiguity in the Round-2 proof and upgrades the correction to a complete one.

4. ATTACK PLAN (ROUND-3)

The exact gaps left after Round 2 are:

1. justify rigorously that Tao's lower bound may be minimized over all admissible $A=A(N)$;
2. give an upper-bound construction with the budget condition $\le C$ satisfied exactly;
3. separate clearly the solved asymptotic minimum-value problem from the still-open optimizer-classification problem.

The plan is therefore:

1. use Tao's uniform theorem to obtain \[ F_N(C)\ge (\rho(e^C)+o(1))N; \]
2. construct an exact-budget prime tail $A_N$ and evaluate its uncovered set by the Dickman--de Bruijn asymptotic, giving \[ F_N(C)\le (\rho(e^C)+o(1))N; \]
3. conclude that the corrected asymptotic statement is proved, while the original exact choice-of-$A$ problem remains open.

5. WORK (ROUND-3)

5.1 The literal finite-$N$ optimizer prescription is false.

By the Round-2 computation at $(N,C)=(30,1)$, the consecutive-largest-primes choice $A_0$ is admissible but leaves $12$ integers uncovered, whereas the admissible set $A_1$ leaves only $6$. Thus the literal finite statement \[ \mbox{``the extremal set is the consecutive tail of largest primes''} \] is false.

So the exact optimizer-classification problem cannot be solved by that prescription.

5.2 The corrected asymptotic statement.

For fixed $C>0$, define \[ F_N(C):=\min\left\{|U_A(N)|:\ A\subseteq\{2,\dots,N\}\ \mbox{pairwise coprime},\ \sum_{a\in A}\frac1a\le C\right\}. \]

The minimal corrected statement is:

Theorem. For every fixed $C>0$, \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N \qquad (N\to\infty), \] where $\rho$ is the Dickman function.

I now prove this theorem.

5.3 Uniform lower bound.

Let $C>0$ be fixed. Tao's repaired Theorem 1.1 states that if $A$ is a finite pairwise coprime set not containing $1$ and $\mu(A):=\sum_{a\in A}1/a\le C$, then \[ \sigma_N(A):=\frac{|U_A(N)|}{N}\ge \rho(e^{\mu(A)})+o(1), \] where the $o(1)$ is with respect to $N\to\infty$, holding $C$ fixed and allowing $A$ to vary.

Also, for $u>1$ the Dickman equation \[ u\rho'(u)+\rho(u-1)=0 \] implies \[ \rho'(u)=-\frac{\rho(u-1)}{u}\le 0, \] so $\rho$ is decreasing on $[1,\infty)$. Since $\mu(A)\le C$, we obtain \[ \rho(e^{\mu(A)})\ge \rho(e^C). \] Hence, uniformly for every admissible $A$, \[ \sigma_N(A)\ge \rho(e^C)+o(1). \]

Equivalently: for every $\varepsilon>0$, there exists $N_0(C,\varepsilon)$ such that for all $N\ge N_0(C,\varepsilon)$ and all admissible $A$, \[ |U_A(N)|\ge (\rho(e^C)-\varepsilon)N. \] Taking the minimum over admissible $A$ gives \[ F_N(C)\ge \left(\rho(e^C)+o(1)\right)N. \]

5.4 Exact-budget prime-tail upper bound.

Let $q_N$ be the minimal prime for which \[ \sum_{q_N\le p\le N}\frac1p>C, \] and define \[ A_N:=\{p\ \mbox{prime}: q_N<p\le N\}. \] Then $A_N\subseteq\{2,\dots,N\}$, its elements are pairwise coprime, and by minimality of $q_N$, \[ \mu(A_N)=\sum_{q_N<p\le N}\frac1p\le C<\mu(A_N)+\frac1{q_N}. \] Thus $A_N$ is exactly admissible.

Since $\sum_{p\le N}1/p\to\infty$, we necessarily have $q_N\to\infty$, and therefore \[ 0\le C-\mu(A_N)<\frac1{q_N}=o(1). \] So \[ \mu(A_N)=C+o(1). \]

By Mertens' theorem for the prime harmonic series, \[ \sum_{q_N<p\le N}\frac1p = \log\log N-\log\log q_N+o(1), \] hence \[ \log\log N-\log\log q_N=C+o(1). \] Exponentiating, \[ \log q_N=e^{-C+o(1)}\log N, \] or equivalently \[ q_N=N^{e^{-C}+o(1)}. \] Define \[ u_N:=\frac{\log N}{\log q_N}. \] Then \[ u_N=e^C+o(1). \]

Now an integer $n\le N$ is not divisible by any prime in $A_N$ if and only if every prime factor of $n$ is at most $q_N$. Therefore \[ |U_{A_N}(N)|=\Psi(N,q_N), \] where \[ \Psi(x,y):=\#\{n\le x:\ P^+(n)\le y\} \] is the smooth-number counting function.

By the classical Dickman--de Bruijn asymptotic for smooth numbers, whenever \[ \frac{\log x}{\log y}\to u\in[1,\infty), \] one has \[ \Psi(x,y)=x\rho(u)+o(x). \] Applying this with $x=N$, $y=q_N$, and $u=u_N=e^C+o(1)$, and using continuity of $\rho$, we conclude that \[ |U_{A_N}(N)|=\Psi(N,q_N)=\left(\rho(e^C)+o(1)\right)N. \] Since $A_N$ is admissible, it follows that \[ F_N(C)\le \left(\rho(e^C)+o(1)\right)N. \]

5.5 Conclusion.

Combining the lower and upper bounds gives \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N. \]

This completely proves the corrected asymptotic minimum-value theorem.

5.6 What is solved and what is not.

The argument above proves the minimum value \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N. \] It does not determine the exact extremizing set $A$ for each finite $N$, nor does it prove uniqueness or even asymptotic uniqueness of minimizers. Thus the original optimizer-classification problem remains open, but the minimal corrected asymptotic statement is now fully settled.

6. ADVERSARIAL VERIFICATION

I checked the new work against the most likely failure modes.

1. Uniformity in $A$. This was the main logical gap in Round 2. It is now explicitly addressed: Tao's repaired note states that the $o(1)$ is taken with $N\to\infty$, holding $C$ fixed but allowing $A$ to vary. Therefore the lower bound may indeed be passed to the minimum $F_N(C)$.

2. Exact budget versus $C+o(1)$. The new prime-tail competitor $A_N$ satisfies $\mu(A_N)\le C$ by construction. No trimming or Lipschitz estimate is needed.

3. Does $q_N\to\infty$? Yes. If $q_N$ stayed bounded, then $\sum_{q_N<p\le N}1/p$ would diverge like $\log\log N$, contradicting $\mu(A_N)\le C$.

4. Applicability of the smooth-number asymptotic. We only use the fixed-$u$ regime $u_N\to e^C$, which is exactly the classical Dickman regime.

5. What exactly has been disproved? The false statement is the literal finite-$N$ optimizer prescription "take the consecutive largest primes." The proven corrected statement concerns only the asymptotic minimum value. No stronger optimizer-classification claim is asserted.

7. FINAL

FULL SOLUTION -- COUNTEREXAMPLE / DISPROOF.

The literal exact finite-$N$ prescription "take the consecutive largest primes" is false, by the explicit Round-2 counterexample at $(N,C)=(30,1)$.

The minimal corrected statement is the asymptotic minimum-value theorem \[ F_N(C)=\left(\rho(e^C)+o(1)\right)N, \] and this is now proved completely:

1. the lower bound follows from Tao's repaired Theorem 1.1, using the crucial uniformity in $A$;
2. the matching upper bound is attained by an exactly admissible prime tail $A_N=\{p: q_N<p\le N\}$, whose uncovered set is counted by $\Psi(N,q_N)$.

So path (C) is complete: the exact optimizer prescription is disproved, the exact optimizer problem remains open, and the asymptotic minimum-value problem has a full proof.

8. COMPLETION ESTIMATE

COMPLETION: 100%

9. REFERENCES

1. [EP783 ] T. F. Bloom, Erdős Problem #783, https://www.erdosproblems.com/783, accessed 2026-03-08.
2. [Er73 ] P. Erdős, Problems and results on combinatorial number theory, in A Survey of Combinatorial Theory (Proc. Internat. Sympos., Colorado State Univ., Fort Collins, Colo., 1971), North-Holland, Amsterdam, 1973, pp. 117--138.
3. [Tao26 ] T. Tao, Sieving by coprime numbers, repaired third version dated 2026-02-22.
4. [Hi86 ] A. Hildebrand, On the number of positive integers $\le x$ and free of prime factors $>y$, J. Number Theory 22 $1986$, 289--307.
5. [Hi87 ] A. Hildebrand, Quantitative mean value theorems for nonnegative multiplicative functions II, Acta Arith. 48 $1987$, no. 3, 209--260.