Counting distinct totients and least preimages

Source and verification

This article explains family 024 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Counting distinct totient values

Euler’s totient $\phi (n)$ counts integers from one to $n$ coprime to $n$. Let

$$\begin{array}{c}\mathcal{V}=\{\phi (n):n\ge 1\},\\ V(x)=\#(\mathcal{V}\cap [1,x]).\end{array}$$

Each value is counted once, regardless of how many preimages it has. For example, $\phi (5)=\phi (8)=\phi (10)=\phi (12)=4$, contributing only one to $V(x)$.

Since $\phi (p)=p-1$, the prime number theorem supplies $V(x)\ge \pi (x+1)\sim x/\log x$. The full count is larger by a slowly varying factor, while still having density zero.

The manuscript claims an explicit asymptotic equivalent and regular variation:

$$\begin{array}{c}\frac{V(cx)}{V(x)}⟶c\\ \text{for every fixed }c>0.\end{array}$$

The new coefficient retains discrete arithmetic data and is not asserted to be a single constant. (OpenAI, 2026)

🏷️ The counting scale

In this post ${\log }_j$ means the $j$-fold iterated natural logarithm. Define

$$a_j={\int }_j^{j+1}\log tdt,j\ge 1.$$

These numbers are positive. There is a unique $\rho \in (0,1)$ satisfying ${\sum }_{j\ge 1}{a}_j{\rho }^j=1$. Put

$$\begin{array}{c}\lambda =\log (1/\rho ),\\ \gamma ={\left(\sum _{j\ge 1}j{a}_j{\rho }^j\right)}^{-1},\\ g_0=1,g_j=\sum _{d=1}^j{a}_d{g}_{j-d}.\end{array}$$

The recurrence is a renewal relation. If $p_d=a_d{\rho }^d$, then ${\sum }_d{p}_d=1$, and ${\rho }^j{g}_j$ satisfies the renewal recurrence with step distribution $p_d$.

For sufficiently large $x$, set

$$\begin{array}{c}B={\log }_{2}x,\\ t=\frac{\log B-{\log }_{2}B}{\lambda },\\ m=\lfloor t\rfloor ,\theta =t-m,\\ G_m=\frac{B^m}{m!{\prod }_{i=1}^m{g}_i}.\end{array}$$

The source’s main theorem is

$$V(x)\sim \frac{x}{\log x}{G}_{m}A(1;\theta ),$$

where $A(1;s)$ is bounded above and bounded away from zero for $0\le s<1$. The phase $\theta$ comes from the integer location of a cutoff in the prime-factor decomposition.

This is more precise than a formula with an unspecified bounded multiplicative factor.

🏷️ Why the coefficient counts unions

A typical preimage is decomposed into a long ordered prefix of large prime factors and a shorter arithmetic tail. The double logarithms of the prefix primes occupy a simplex, so much of their count is approximated by volume. Tail primes near the cutoff remain discrete.

Different tails can give the same totient. Summing their permitted prefix volumes would count some output values more than once. Instead, for each tail value $d$, the source takes the volume of the union of the regions permitted by all its witnesses.

The construction can be described probabilistically. For a truncation $H$ and phase $s$, let ${\mathcal{W}}_{H,s}(d)$ be the finite collection of permitted tail factorizations with totient $d$. Each witness $\eta$ determines thresholds $t_h(\eta )\ge 0$ for $h\ge H$. For independent mean-one exponential variables $E_h$, its event is

$${\mathcal{E}}_{\eta }=\{E_h\ge t_h(\eta )\text{ for all }h\ge H\}.$$

The thresholds have a summable tail. For a finite nonempty set $T$ of witnesses,

$$\begin{array}{c}\mathbb{P}\left(\bigcap _{\eta \in T}{\mathcal{E}}_{\eta }\right)\\ =\exp \left(-\sum _{h\ge H}\underset{\eta \in T}{\max }{t}_h(\eta )\right).\end{array}$$

Finite inclusion–exclusion therefore computes each union probability exactly. The approximant $A_H(1;s)$ is a normalized sum of these probabilities weighted by $1/d$. Its normalization is ${\rho }^{H(H-1)/2}(\gamma /{\alpha }_s{)}^H$, where ${\alpha }_s=\lambda e^{\lambda s}$.

The source specifies the bounded prime ranges and thresholds in its “explicit main term” section. They depend on $H,s$, not on $V(x)$; the coefficient is not defined by taking a ratio with the unknown answer.

The claimed uniform convergence $A_H(1;s)\to A(1;s)$ is the substantial limiting assertion. Exponential variables are a device for encoding prefix volumes, not a claim that actual prime factors are exactly independent.

🏷️ Prefix collisions and regular variation

Two facts convert a representation count into a value count. Outside a negligible exceptional set, every preimage has the required structure. Collisions between different long prefixes then have negligible total multiplicity. Tail collisions remain and are counted by the exact union above.

The source adapts layered shifted-prime comparisons to control all ordered prefix collisions. Recovering the tuple data inside residual factors has a cost, which must be included before replacing a tuple count by a count of distinct values.

For fixed dilation $x\mapsto cx$, the proof counts the largest prime using common endpoint parameters. This supplies the limiting ratio directly. The source does not assume continuity of $A(1;s)$, so the ratio should not be justified merely by saying that the phase changes little.

🏷️ Least preimages

Define $\ell (v)=\min \{n:\phi (n)=v\}$. For fixed positive integer $k$, let

$$\begin{array}{c}{N}_k(x)\\ =\#\left\{v\in \mathcal{V}:\begin{array}{c}v\le x\\ kx<\ell (v)\le (k+1)x\end{array}\right\}.\end{array}$$

The source gives a weighted version of the main asymptotic. Its coefficient is positive exactly when some totient $d$ satisfies $\ell (d)>kd$. If no such $d$ exists, then $N_k(x)=0$ identically: for every $v\le x$, one has $\ell (v)\le kv\le kx$.

The positive case is established for $k=1,2$. The manuscript does not settle whether $\ell (d)/d$ is unbounded, and hence does not claim positivity for every $k$.

🏷️ Key steps and proof outlines

The renewal recurrence determines the continuous prefix scale

Let $A(z)={\sum }_{j\ge 1}{a}_j{z}^j$ and $G(z)={\sum }_{j\ge 0}{g}_j{z}^j$. Multiplying the recurrence for $g_j$ by $z^j$ and summing gives $G(z)=1/(1-A(z))$ wherever the series converge.

Proof outline. At the positive root $A(\rho )=1$, the denominator has derivative $-A^{\prime }(\rho )<0$, so its pole contributes

$$\frac{1}{\rho A^{\prime }(\rho )}\frac{1}{1-z/\rho }=\frac{\gamma }{1-z/\rho }.$$

All $a_j$ are positive, so equality in $|A(z)|\le A(|z|)$ on $|z|=\rho$ forces $z=\rho$ when $A(z)=1$. Subtracting the displayed pole leaves an analytic function on a slightly larger disk, which gives $g_j=\gamma {\rho }^{-j}$ plus a smaller exponential error. This explains the geometric spacing of the prefix scales and the product ${\prod }_{i\le m}{g}_i$ in $G_m$. The arithmetic argument must still determine which points of that continuous region correspond to distinct totients. (OpenAI, 2026)

Prefix collisions must be counted with all their hidden data

The source records a representation by a tuple $(p_0,\dots ,p_R,d)$, mapped to the totient value $d{\prod }_{j=0}^R(p_j-1)$. Different preimages of the tail value $d$ are not different tuples.

Proof outline. Given two different tuples with the same value, cancel their common prime prefix and locate the first unequal prime. The remaining equality has the form

$$D\prod _{j=0}^J(p_{i+j}-1)=D^{\prime }\prod _{j=0}^J(q_{i+j}-1).$$

The layered shifted-prime comparison bounds these equal products using the separation of the prime-factor scales and strict slack in their simplex constraints. Its saving must also pay for recovering the tuple coordinates hidden inside $D,D^{\prime }$; otherwise a bound on residual integers would not bound the original ordered tuple pairs. After restoring the common prefix, the source obtains $o(x{G}_m/\log x)$ ordered collisions in its iterated limit. To see why this suffices for values, if a finite map has fiber sizes $r_v$, its overcount is ${\sum }_v(r_v-1{)}_{+}$, bounded by the ordered collision count ${\sum }_v{r}_v(r_v-1)$. Exceptional tuples are controlled separately. Tail witnesses are still allowed to collide and are handled by unions in the next step.

A triangular change of variables computes witness intersections

The continuous prefix coordinates $u_i$ obey linear constraints involving the coefficients $a_j$. Introduce slack coordinates

$$t_i=u_i-\sum _{r>i}{a}_{r-i}{u}_r.$$

The transformation is triangular with diagonal one, hence preserves volume. The recurrence for $g_i$ turns the available size into ${\sum }_{i=0}^R{g}_i{t}_i=B$.

Proof outline. A particular tail witness imposes lower bounds on the $t_i$. Intersecting several witnesses replaces each lower bound by their maximum $C_i$. Translate $t_i$ by $C_i$. The remaining simplex has available size $B-{\sum }_i{g}_i{C}_i$ and volume

$$\frac{(B-{\sum }_i{g}_i{C}_i{)}_{+}^R}{R!{\prod }_{i=1}^R{g}_i}.$$

Inclusion–exclusion now computes the volume of the witness union exactly. The limiting expression is represented by independent exponential variables, whose joint survival probability is $\exp (-{\sum }_i{C}_i)$ after the source’s rescaling. Independence belongs to this volume representation; it is not assumed for arithmetic prime factors. Reciprocal prime sums are compared with the union volume by dividing into boxes, using the prime number theorem in interior boxes and bounding boxes meeting the boundary.

One common comparison proves uniform convergence in the phase

Let $C(x)=V(x)\log x/(x{G}_m)$. The source first proves, for each truncation $H$,

$$\underset{x\to \infty }{\mathrm{limsup}}|C(x)-A_H(1;\theta (x))|\le {\zeta }_H,{\zeta }_H\to 0.$$

Proof outline. Fix a phase $s\in [0,1)$. The continuous increasing function $(\log B-\log \log B)/\lambda$ assumes every sufficiently large value, so one can choose an unbounded sequence with phase exactly $s$. Compare truncations $H,K$ along that same sequence. The triangle inequality gives $|A_H(1;s)-A_K(1;s)|\le {\zeta }_H+{\zeta }_K$, uniformly in $s$. Thus the finite arithmetic coefficients form a uniformly Cauchy sequence. The pre-existing two-sided counting bounds transfer along these exact-phase sequences to positive upper and lower bounds for the limit. This proves convergence and turns an additive error into the asserted asymptotic equivalent, without any continuity assumption on the phase dependence.

Regular variation uses the same tail data at both cutoffs

For fixed tail-and-prefix data apart from the largest prime, put $D=d{\prod }_{i=1}^R(p_i-1)$. The possible largest primes satisfy $x^{0.9}\le p_0\le 1+t/D$, with $t=x$ or $t=x/c$. The source controls $D\le \exp ((\log x{)}^{0.8})$.

Proof outline. The prime number theorem uniformly counts this interval as $t/(D\log x)(1+o(1))$. Sum over the same data set for both values of $t$. The collision and exceptional-tuple errors are negligible, so the two counts share the same reciprocal mass and differ only by the factor $t$. Their ratio tends to $c$. This argument remains valid if the phase coefficient has discontinuities; it does not replace the coefficient at one phase by a nearby phase on the basis of an unproved regularity assertion.

🏷️ Connections

Large totient fibers treats many preimages of one value. That result is not an input to this distinct-value asymptotic. Both problems require careful control of products of prime predecessors.

🏷️ Manuscripts and verification scope

• An asymptotic formula for the number of totients — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects the arithmetic main term, regular variation, and the conditional positive or identically zero least-preimage alternatives. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. An asymptotic formula for the number of totients.