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Definition: n is said to be a highly composite number if and only if $d(n)>d(m)$ for all $m<n$, where $d(n)$ denotes number of divisors of n.

Questions:
1) Are there any theorems about highest prime that divides n, assuming that n is highly composite?. It seems that the highest prime is of the order of ln(n).
2) Are there any theorems about the highest power of 2 which divides n, assuming that n is highly composite?

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Have you checked out the references in OEIS? There are quite a few. –  Ross Millikan Dec 16 '10 at 16:34
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2 Answers

up vote 8 down vote accepted

This paper by Alaoglu and Erdős proves that if $p$ is the largest prime divisor of a highly composite number (HCN) $n$ then the exponent $k_q$ on another prime $q$ dividing $n$ satisfies $$\log\left(1 + \frac{1}{k_q}\right) > \frac{\log q \log 2}{\log p} + O\left(\frac{(\log \log p)^3}{(\log p)^3}\right),$$ $$\log\left(1 + \frac{1}{k_q+1}\right) < \frac{\log q \log 2}{\log p} + O\left(\frac{(\log \log p)^3}{(\log p)^3}\right).$$ (See Theorem 12.)

For sufficiently large $q$ they claim that $k_q$ is either $\Lambda$, $\Lambda + 1$, or $\Lambda - 1$, where $$\Lambda = \left\lfloor \left(2^{\log q/\log p} - 1\right)^{-1} \right\rfloor.$$ (See Theorem 13.)

When $q = 2$ this latter formula appears to be close, although it is not exactly correct, as there are more than three possible values for the power of 2 that divides a highly composite number with largest prime factor $p$. See, for example, this list of the first 1200 HCN's, which includes their prime factorizations.

This, together with Derek Jennings' answer on the order of $p$, should give you the estimate of the highest power of 2 that divides $n$.

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Thanks, it should be enough to solve my problem. –  Tomek Tarczynski Dec 16 '10 at 20:29
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In this paper On Highly Composite Numbers P. Erdős uses Bertrand's postulate and the Prime Number Theorem to show that if $p$ is the highest prime that divides $n,$ a highly composite number, then

$$ c_1 \log n < p < c_2 \log n $$

for some constants $c_1$ and $c_2 .$

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Thanks, it helps a lot to know that there is formal proof of what I expected. –  Tomek Tarczynski Dec 16 '10 at 20:30
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