Erdős Problem #416

Erdős Problem 416: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

Let $\mathcal V=\{\varphi(m):m\ge1\}$ and $V(x)=|\mathcal V\cap[1,x]|$. Must $V(2x)/V(x)\to2$? Source: https://www.erdosproblems.com/416. We give an exact decomposition into doubling chains and identify the regularity missing from the known order-of-magnitude theorem. The limit remains unresolved here.

The earlier GPT-5.2 draft's basic prime-value bound is retained. No asymptotic ratio is inferred from coarse estimates, and no novelty is claimed for the deductions below.

Doubling chains and an exact identity

Every doubled totient is a totient. If $t=\varphi(m)$ with $m$ even, then $\varphi(2m)=2t$. If $m$ is odd, multiplicativity gives $\varphi(4m)=2t$. Define the primitive totients \[ \mathcal B=\{t\in\mathcal V:t/2\notin\mathcal V\}, \qquad B(x)=|\mathcal B\cap[1,x]|, \] where nonintegers are not in $\mathcal V$. Each totient can be halved repeatedly until it reaches a unique member of $\mathcal B$. Conversely every repeated doubling of a member of $\mathcal B$ remains a totient. Thus there is a disjoint decomposition \[ \mathcal V=\bigsqcup_{j\ge0}2^j\mathcal B, \qquad V(x)=\sum_{j\ge0} B(x/2^j). \tag{1} \] The sum is finite for each $x$. Subtracting the version with $x$ from the version with $2x$ gives \[ V(2x)-V(x)=B(2x),\qquad \frac{V(2x)}{V(x)}=1+\frac{B(2x)}{V(x)}. \tag{2} \] Equivalently, for each odd part that occurs among totients, there is a minimum exponent of two and then every larger exponent occurs. The question is exactly whether primitive chains are initiated at the rate $B(2x)\sim V(x)$.

Primes give distinct totients $p-1$, while $\varphi(m)$ is even for $m>2$, so the elementary bounds are \[ \pi(x+1)\le V(x)\le1+\lfloor x/2\rfloor. \tag{3} \] They do not control the quotient in $2$.

The precise known estimate and its limitation

Ford's The distribution of totients, Theorem 1, gives $V(x)\asymp Z(x)$, where \[ Z(x)=\frac{x}{\log x} \exp\!\left(C(\log_3 x-\log_4 x)^2+D\log_3 x -(D+\tfrac12-2C)\log_4 x\right). \tag{4} \] Here $\log_j$ denotes the $j$fold iterated natural logarithm, and the constants are $C=0.817814\ldots$ and $D=2.176968\ldots$, defined precisely in equations $1.3$--$1.6$ of the paper. Primary source: https://arxiv.org/abs/1104.3264v2. The notation $\asymp$ means bounds above and below by fixed positive constant multiples, not a ratio tending to one. Ford's theorem is imported, not reproved here.

Direct differentiation of $4$ gives \[ \frac{xZ'(x)}{Z(x)}=1+o(1),\qquad \frac{Z(2x)}{Z(x)}\to2, \qquad \frac{Z(x)}x\to0. \tag{5} \] For example $x(\log_j x)'=(\log x\cdots\log_{j-1}x)^{-1}$ for $j\ge2$; every derivative term other than $x(\log x)'=1$ vanishes, even after the displayed iterated-log factors. Integrating the first relation over $[x,2x]$ proves the middle one. The last follows because the exponent in $4$ is $o(\log_2 x)$.

But $4$ by itself cannot transfer $5$ to $V$. To demonstrate the logical gap, consider the comparison function \[ H(x)=Z(x)\left(1+\tfrac14\sin(\log x)\right) \] for sufficiently large $x$. It is positive and comparable to $Z$. It is eventually strictly increasing, since, with $L=xZ'/Z\to1$, \[ \frac{xH'(x)}{Z(x)} =L\left(1+\tfrac14\sin\log x\right) +\tfrac14\cos\log x \ge1-\tfrac{\sqrt2}{4}+o(1)>0. \] Nevertheless \[ \frac{H(2x)}{H(x)} =\frac{Z(2x)}{Z(x)} \frac{1+\tfrac14\sin(\log x+\log2)} {1+\tfrac14\sin\log x} \] has different subsequential limits: take phases $\log x\equiv0$ and $\pi$ modulo $2\pi$. The limits are $2(1+\tfrac14\sin\log2)$ and $2(1-\tfrac14\sin\log2)$, which differ.

This is not a counterexample involving actual totients. It shows that comparability and monotonicity, even for the exact scale in Ford's theorem, do not imply the desired limit. An integer-valued counting model has the same obstruction: $5$ gives $H'(x)\to0$, so for large integral $N$ the increments of $\lfloor H(N)\rfloor$ are zero or one, and rounding has no effect on these ratios.

Remaining gap

The chain identity $2$ is exact. The established order of magnitude $4$ still permits bounded oscillation in a multiplicative factor. One needs finer information on primitive-chain counts, or suitable control of $V(x)/Z(x)$ between $x$ and $2x$. No such control is proved here.

Completion rate estimate: 35%.

FORMAL RESTATEMENT

Definitions.

• $\mathbb{N}=\{1,2,3,\dots\}$.
• Euler's totient function $\varphi:\mathbb{N}\to\mathbb{N}$ is \[ \varphi(m)=\#\{1\le a\le m : \gcd(a,m)=1\}. \]
• For real $x\ge 0$, define \[ V(x)=\#\{n\in\mathbb{N}: n\le x \ \text{and}\ \exists m\in\mathbb{N}\ \text{with}\ \varphi(m)=n\}. \] (Equivalently, $V(x)$ is the number of distinct totients $\le x$.)

Questions.

• $Q1$ Does $\displaystyle \lim_{x\to\infty} \frac{V(2x)}{V(x)}=2$?
• $Q2$ Does $V(x)$ admit an asymptotic formula as $x\to\infty$?

Edge cases. For $0\le x<1$ one has $V(x)=0$. For $1\le x<2$ one has $V(x)=1$ since $\varphi(1)=\varphi(2)=1$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement (as given in the extracted file) records progressively sharper information on $V(x)$: Pillai proved $V(x)=o(x)$, Erdős proved $V(x)=x(\log x)^{-1+o(1)}$, Maier--Pomerance proved \[ V(x)=\frac{x}{\log x}\exp\big((C+o(1))(\log\log\log x)^2\big), \] and Ford proved a two-sided estimate of the shape \[ V(x)\asymp \frac{x}{\log x}\exp\Big( C_1(\log\log\log x-\log\log\log\log x)^2+ C_2\log\log\log x- C_3\log\log\log\log x\Big) \] with explicit constants $C_i>0$. The remaining issue is that these do not yield an asymptotic constant, nor do they settle whether $V(2x)/V(x)\to 2$.

ATTACK PLAN

Proof-track ideas (for $V(2x)/V(x)\to 2$).

• Use the best-known description of $V(x)$ as $\frac{x}{\log x}$ times a "slowly varying" factor, and attempt to show that factor is asymptotically stable under $x\mapsto 2x$.
• Translate the question into a statement about how many new totients appear between $x$ and $2x$ (i.e. $V(2x)-V(x)$) and seek matching upper/lower bounds.

Disproof-track ideas.

• Try to exhibit sequences $x_j$ for which $V(2x_j)/V(x_j)$ oscillates, by forcing unusually many/few totients near the endpoints.
• Look for "rigidity" obstructions: for instance, totients are mostly even and have strong multiplicative constraints; see if those could force $V(2x)/V(x)$ away from $2$.

Given the deep context already cited in the statement, I focus below on rigorous elementary bounds and sanity-check computations; I do not attempt to reproduce the cited deep estimates.

WORK

Fast reality check (exact computation for small $x$)

Using the inequality from Problem #417 below (Lemma 417.1), any totient value $\le x$ occurs as $\varphi(m)$ for some $m\le 2x^2$. Thus $V(x)$ can be computed exactly for moderate $x$ by enumerating $\varphi(m)$ for $m\le 2x^2$ and counting distinct values $\le x$.

Exact values obtained (via a totient sieve) are:

 x    V(x)   V(2x)   V(2x)/V(x)
 10     6      10      1.6666666667
 20    10      17      1.7000000000
 50    21      38      1.8095238095
 100   38      72      1.8947368421
 200   72     131      1.8194444444
 500  158     291      1.8417721519
 1000 291     543      1.8659793814

These ratios are far from conclusive but show $V(2x)/V(x)<2$ for these $x$.

Lemma 416.1 (Parity of totients)

Lemma. If $m>2$ then $\varphi(m)$ is even. Consequently, if $n>1$ and $\varphi(m)=n$ for some $m$, then $n$ is even. Therefore, for all real $x\ge 0$, \[ V(x)\le 1+\big\lfloor \tfrac{x}{2}\big\rfloor. \]

Proof. Fix $m>2$. Let \[ R=\{a\in\{1,2,\dots,m\}: \gcd(a,m)=1\}. \] Then $\#R=\varphi(m)$ by definition. Define a map $T:R\to R$ by $T(a)=m-a$. If $\gcd(a,m)=1$ then $\gcd(m-a,m)=\gcd(a,m)=1$, so $T$ maps $R$ to itself. Also $T$ is an involution: $T(T(a))=a$. If $T(a)=a$ then $m-a=a$, so $m=2a$ and in particular $m$ is even and $a=m/2$. But then $\gcd(a,m)=\gcd(m/2,m)=m/2>1$ (since $m>2$), contradicting $a\in R$. Hence $T$ has no fixed points on $R$, so it partitions $R$ into disjoint $2$-element orbits. Therefore $\#R$ is even, i.e. $\varphi(m)$ is even.

Now suppose $\varphi(m)=n>1$. Then $m\ne 1,2$ because $\varphi(1)=\varphi(2)=1$, hence $m>2$ and the first part implies $n$ is even. Thus the set of totients $\le x$ is contained in $\{1\}\cup\{2,4,6,\dots,2\lfloor x/2\rfloor\}$. This set has size $1+\lfloor x/2\rfloor$, so $V(x)$ is at most this size. $\square$

Lemma 416.2 (Primes give many distinct totients)

Lemma. For all real $x\ge 1$, \[ V(x)\ge \pi(x+1), \] where $\pi(y)$ is the number of primes $\le y$.

Proof. Let $p$ be any prime with $p\le x+1$. Then $\varphi(p)=p-1$ (since the reduced residues modulo a prime are exactly $1,2,\dots,p-1$). Hence $p-1$ is a totient value and $p-1\le x$. If $p\ne q$ are distinct primes then $p-1\ne q-1$, so the totients $\{p-1: p\le x+1\text{ prime}\}$ are all distinct. Therefore there are at least $\pi(x+1)$ distinct totients $\le x$, i.e. $V(x)\ge \pi(x+1)$. $\square$

VERIFICATION

• Lemma 416.1 explicitly excludes the exceptional cases $m=1,2$ where $\varphi(m)=1$ is odd; the proof checks that the involution $a\mapsto m-a$ has no fixed point because $a=m/2$ would not be coprime to $m$.
• Lemma 416.2 uses only the identity $\varphi(p)=p-1$ for primes $p$, and injectivity of $p\mapsto p-1$.
• The computation of $V(x)$ used the bound $m\le 2x^2$ for totients $\le x$ proved as Lemma 417.1 below; this is independently verifiable.

FINAL

UNRESOLVED

1. (i) Strongest proved partial result here. Elementary bounds \[ \pi(x+1)\le V(x)\le 1+\lfloor x/2\rfloor \] for all $x\ge 1$ (Lemmas 416.1--416.2), together with exact computed values of $V(x)$ for $x\le 1000$ (table above).
2. (ii) First gap (crisp). Prove or disprove \[ \lim_{x\to\infty}\frac{V(2x)}{V(x)}=2. \] Equivalently, show that for every $\varepsilon>0$ there exists $X$ such that for all $x\ge X$, $\big|V(2x)-2V(x)\big|\le \varepsilon V(x)$; or exhibit a sequence $x_j\to\infty$ along which the ratio fails to approach $2$.
3. (iii) Top 3 next moves.
   • Try to upgrade the two-sided estimates for $V(x)$ stated in the problem (in particular the explicit exponent structure in Ford's estimate) to a true asymptotic, or at least show the multiplicative correction factor is "slowly varying" enough to imply $V(2x)/V(x)\to 2$.
   • Compute $V(x)$ for much larger $x$ (e.g. up to $10^7$ or $10^8$) using a totient sieve plus the bound $m\le 2x^2$, and numerically examine $V(2x)/V(x)$ to search for systematic drift or oscillation.
   • Study the increment $V(2x)-V(x)$ by estimating the number of totients landing in $(x,2x]$; any matching upper and lower bounds of order $V(x)$ would be decisive.
4. (iv) What a minimal counterexample would likely look like. If $V(2x)/V(x)$ fails to converge to $2$, one expects a sequence $x_j$ where the "local density" of totients near $x_j$ is atypically high or low, producing persistent oscillation in $V(2x_j)-2V(x_j)$. A minimal counterexample would likely come from constructing intervals with anomalously many (or few) new totient values, rather than from parity-type constraints (already accounted for in Lemma 416.1).