# Maximal order of Hooley's Delta function?

There is a large literature on Hooley's $$\Delta(n)=\max_u\sum_{d|n,\ e^u\le d< e^{u+1}}1$$ giving its normal and average order. What is known of its maximal order?

Clearly $\Delta(n)\le d(n)$ and so the usual upper bounds apply, but these are not very close. By exhaustive calculation I have determined that $\Delta(n)\le138$ for $n\le5\cdot10^8$ with equality only for $465585120=2^5\cdot3^2\cdot5\cdot7\cdot11\cdot13\cdot 17\cdot19.$

# References

Paul Erdős, On abundant-like numbers, Canad. Math. Bull. 17 (1974), pp. 599-602.

C. Hooley, On a new technique and its applications to the theory of numbers, Proc. London Math. Soc. 3 38:1 (1979), pp. 115-151.

Helmut Maier and Gérald Tenenbaum, On the set of divisors of an integer, Invent. Math. 76 (1984), pp. 121-128.

Helmut Maier and Gérald Tenenbaum, On the normal concentration of divisors, J. London Math. Soc. 2 31:3 (1985), pp. 393-400.

Helmut Maier and Gérald Tenenbaum, On the normal concentration of divisors. II., Math. Proc. Cambridge Philos. Soc. 147:3 (2009), pp. 513-540.

Gérald Tenenbaum, Sur la concentration moyenne des diviseurs, Commentarii Mathematici Helvetici 60:1 (1985), pp. 411-428.

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How close is the maximum u to log(n)/2? Is there anything in the literature on the location of u? Or can it be proved that u is closer to log(delta(n))? Gerhard "Perhaps Jacobsthal's Function Is Related" Paseman, 2013.06.24 –  Gerhard Paseman Jun 24 '13 at 18:32
@Gerhard Paseman: In the case of the example I gave, $u\approx9.4796$ and $\log n\approx19.9588054,$ so $u$ is very close to $(\log n)/2+1/2.$ –  Charles Jun 24 '13 at 19:20
Since the interval $[1,n]$ may be partitioned into about $\log n$ intervals $[e^{k},e^{k+1}]$ we see that $\Delta(n) \ge d(n)/\log n$. Therefore the maximal order of $\Delta(n)$ is essentially just the maximal order of $d(n)$ -- about $2^{\log n/\log \log n}$. –  Lucia Aug 21 '13 at 22:24