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Statement

Setting. $Q(X)$ is the number of highly composite numbers less than $X$; a number $A$ is highly composite when every $M<A$ has fewer divisors than $A$ (p. 116).

Théorème 4 (p. 127). $Q(X)=O((\log X)^{1+c})$, with $c$ the constant of Théorème 3.

The print writes the bound as $O(\log X)^{1+c}$. The introduction (p. 117) announces the result as $Q(X)\le(\log X)^{c'}$ for a constant $c'$.

Proof pointer

P. 127. Summing Théorème 3 over the superior highly composite numbers $N\le X$ bounds $Q(X)$ by $O\bigl(\sum_{N\le X}(\log N)^c\bigr)$, which is at most $(\log X)^c$ times the number of such $N$; Ramanujan's count of superior highly composite numbers (reference [8], § 44) makes the latter $\sim\log X/\log\log X$. The paper adds the sharper asymptotic $\sum_{N\le X}(\log N)^c\sim(\log X)^{c+1}/((c+1)\log\log X)$, citing Landau.

Dependencies

Théorème 3, and through it Théorème 1 and Feldman's bound for linear forms in logarithms.

Read depth

Claims checked: the statement and its one-paragraph proof were read on the page image of the print. Not independently reviewed.

Source. Jean-Louis Nicolas, Répartition des nombres hautement composés de Ramanujan, Canadian J. Math. 23 (1971), no. 1, 116–130, doi:10.4153/cjm-1971-012-6; the edition read is named on the source card.

Bears on

• Problem 381: the problem asks whether $Q(x)\gg_k(\log x)^k$ for every $k\ge1$, with $Q(x)$ counting the highly composite numbers in $[1,x]$. The theorem's count (numbers less than $X$) differs from that by at most one, and its bound $Q(X)=O((\log X)^{1+c})$ rules out $Q(x)\gg_k(\log x)^k$ for every $k>1+c$, so the answer is no.