GPT 6 Astra Ultra attempt on Math-808

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%.