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Statement

Setting. $Q(X)$ is the number of highly composite numbers less than $X$ (inférieurs à $X$); highly composite and superior highly composite numbers are defined as on the Théorème 1 page.

Théorème 3 (p. 125). Let $N=N_\epsilon$ and $N'$ be two consecutive superior highly composite numbers. There is a constant $c$ for which $Q(N')-Q(N)=O((\log N)^c)$.

The print writes the bound as $O(\log N)^c$. The proof (p. 127) gives the bound $O(x^c)$, $x=2^{1/\epsilon}$, for every

$$c>\frac{\log\tfrac12(k'+1)}{2\log2},\qquad k'=\Bigl\lfloor\frac1{e^{\gamma\log2}-1}\Bigr\rfloor,$$

with $\gamma$ the constant of Théorème 1, and uses $x\sim\log N$ (Ramanujan, reference [8], § 39).

Proof pointer

Pp. 125–127. A highly composite $A$ between $N$ and $N'$ has benefit below $Cx^{-\gamma}$ (Théorème 1), so by Proposition 6 its exponents agree with those of $N$ except at primes near the thresholds $x_k$. For primes $\lambda\le x_{k''}$, where consecutive thresholds lie within $2$ of each other, Proposition 5 leaves at most three choices of exponent; for $x_{k''}<\lambda<x_{k'}$ each zone holds at most one prime, with two choices; for $2\le k\le k'$ Proposition 4 leaves at most $2(x_k\log x)^{1/2}$ choices of the largest prime with exponent $k$; and the largest prime factor has at most two choices. Multiplying these counts (display (21), p. 126) and using $\sum_{k=2}^{k'}\tfrac12\log x_k=\frac{\log\frac12(k'+1)}{2\log2}\log x$ gives the theorem.

Dependencies

Théorème 1; the paper's Propositions 4, 5 and 6 (pp. 120, 123); Ramanujan's estimate $x\sim\log N$ (reference [8], § 39).

Read depth

Claims checked: the statement and the value of $c$ were read on the page images of the print. The counting argument was read for its structure and is not reconstructed or independently reviewed here.

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 paper sums this bound over the superior highly composite numbers below $X$ to obtain the upper bound of Théorème 4, which answers the problem's question no.