GPT 6 Astra Ultra attempt on Math-5616

FORMAL RESTATEMENT

Erdős #381; structural counting and Nicolas's disproof, 2026-09-24. A positive integer $n$ is highly composite when $\tau(m)<\tau(n)$ for every $m<n$. Let $Q(x)$ count these integers up to $x$. Is $Q(x)\gg_k(\log x)^k$ true for every fixed $k\geq1$?

QUICK LITERATURE AND CONTEXT CHECK

Nicolas answered negatively in Répartition des nombres hautement composés de Ramanujan $1971$. His Theorem 4 supplies a fixed power of $\log x$ as an upper bound. The proof below explicitly uses that theorem, proves the elementary exponent structure and a separate partition bound, and explains the quantifier contradiction. The weaker elementary bound is not presented as sufficient for the disproof.

OBJECTIVE AND ATTACK PLAN

First control the form of a highly composite integer without analytic inputs. Then state the precise extra estimate needed to distinguish a fixed logarithmic power from the much larger partition bound.

WORK

Prime factors and exponent order. If $n>1$ is highly composite, its prime divisors form an initial segment $p_1=2,\ldots,p_s$ of the primes. Otherwise replacing an occurring prime $q$ by a missing smaller prime $p$, with the entire exponent of $q$, would produce a smaller integer with the same divisor count. Write $n=\prod_{j=1}^s p_j^{a_j}$. Its positive exponents satisfy \[ a_1\geq a_2\geq\cdots\geq a_s\geq1. \tag{1} \] Indeed, if $p<q$ have exponents $a<b$, swapping these exponents multiplies the integer by $(p/q)^{b-a}<1$ and preserves $\tau$. Both contradictions use the strict record condition in the definition.

For $n\leq x$, we also have $\sum a_j\leq L=\lfloor\log_2x\rfloor$. Thus these exponent lists inject into integer partitions of numbers at most $L$. If $p(j)$ is the partition function, for any $h>0$, \[ p(j)e^{-hj}\leq\prod_{r\geq1}(1-e^{-hr})^{-1}. \] Expanding the logarithm and using $e^{h\ell}-1\geq h\ell$ gives \[ \log\prod_{r\geq1}(1-e^{-hr})^{-1} =\sum_{\ell\geq1}\frac1{\ell(e^{h\ell}-1)} \leq\frac{\pi^2}{6h}. \] Choosing $h=L^{-1/2}$ for $L\geq1$ yields the entirely elementary bound \[ Q(x)\leq\sum_{j=0}^{L}p(j) \leq(L+1)\exp\!\big((1+\pi^2/6)\sqrt L\big). \tag{2} \] This is subpolynomial in $x$, but it still grows faster than every fixed power of $\log x$ and therefore cannot answer the question.

The decisive external estimate. Nicolas's Theorem 4 states that there is a fixed finite exponent $A$ and constant $C$ such that \[ Q(x)\leq C(\log x)^A \tag{3} \] for all sufficiently large $x$. The paper writes $A=1+c$ with the constant from its Theorem 3. Its proof controls how many ordinary highly composite numbers lie between successive superior highly composite numbers. The estimates underlying $3$ are imported, not inferred from $1$ or $2$.

Choose an integer $k>A$. If the proposed assertion held for this $k$, some $c_k>0$ would satisfy $Q(x)\geq c_k(\log x)^k$ for all sufficiently large $x$. Combining with $3$ would give \[ 0<c_k\leq C(\log x)^{A-k}\longrightarrow0, \] an impossibility. Equivalently, $3$ shows $Q(x)/(\log x)^k\to0$ for every $k>A$. This directly negates the universal quantifier over $k$.

RESULT AND LIMITATIONS

Attempt status: solved relative to Nicolas's Theorem 4. The answer is no. The exponent normal form, partition estimate, and quantifier argument are proved in full; the logarithmic-power bound is the named external dependency. No explicit best exponent or asymptotic formula for $Q(x)$ is claimed.

SOURCES

https://www.erdosproblems.com/381; Nicolas, Canadian Journal of Mathematics 23 $1971$, 116--130, Theorem 4 on printed page 127: https://doi.org/10.4153/CJM-1971-012-6.

Completion rate estimate: 100% relative to the stated external theorem.