Talk:Landau's function

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I disagree with the preceding recursive formula. IMNSHO, it gives the wrong value for g(29) (it gives 2310, not the correct 2520). (comment by User:128.174.253.191 moved from body of article)

The recursion formula is ~~straightforward to prove~~; it's just the properties of LCM applied to the definition of the Landau function. If that formula gives g(29)=2310, then g(29)=2310. Maybe Sloane's has the error. What evidence do you have that g(29) is really 2520? An actual partition of 29 whose LCM really is 2520? GTBacchus 08:32, 29 August 2005 (UTC)[reply]

Feel free to mock me. 2^3 + 3^2 + 5 + 7 = 29, 2^3 * 3^2 * 5 * 7 = 2520. User:128.174.253.191

Mock you? Not at all. Since you found a partition whose LCM is 2520, that means that g(29) really is at least 2520, and the recursion formula, as you note, fails to give that result. I've just been playing with it, and 29 is the first natural number for which it fails. If you write out the first 29 maximal partitions, I think it's clear why it fails. We have the partition 29=5+7+8+9, whose LCM is 2520. But, there is no k for which 2520=lcm(k,g(29-k)). For example, if we look at lcm(5,g(24)), it turns out that 24=7+8+9 wasn't a maximal partition for 24, but it's the one that gives the best lcm with 5. I dunno how much of that made sense, but the recursion formula is certainly wrong. I'll just remove it now. GTBacchus 18:51, 29 August 2005 (UTC)[reply]

I think asserting it is straightforward to prove and then using "really" twice is rather condescending, however having a pissing match about this is a waste of our time. Amling, formerly User:128.174.253.191.

Goodness, I didn't mean that at all, nor do I want a "pissing match". I apologize. I have no desire to condescend to you. What an unfortunate miscommunication. :-( GTBacchus 06:21, 30 August 2005 (UTC)[reply]

On an unrelated note, I don't agree with what you've added about g(n) growing like exp(sqrt(n log(n))). It is unclear what you mean by grows like, but log(f(n)) ~ sqrt(n log(n)) does not imply f(n) ~ exp(sqrt(n log(n))), e.g. for f(n) = n exp(sqrt(n log(n))). I do not know whether g(n) ~ exp(sqrt(n log(n))). Amling

Good point. Oops. Unsubstantiated claim removed. Good night. GTBacchus 06:21, 30 August 2005 (UTC)[reply]

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I disagree. f(n) = n exp(h(n)) does NOT mean log(f(n)) ~ h(n). You're missing a log(n) in there, which can't be removed since it is neither constant nor -> 1...

Informally speaking, ln(g(n)) ~= sqrt(n ln(n)) +/- epsilon (epsilon ~= 0), so g(n) ~= exp(sqrt(n ln(n))) * exp(epsilon), and since exp(epsilon) ~= 1, g(n) ~= exp(sqrt(n ln(n))). Is something very wrong about this pseudo-proof? User:128.174.253.191 19:24 28 January 2005 (UTC)

Re. equivalence with the Riemann Hypothesis, this article claims "for all n"; the Riemann Hypothesis article claims "for sufficiently large n". The citation in that article is in French, so I am not sure of what it says, although my friend (who is actually a mathematician) believes that it only proves "for sufficiently large n". (I should not actually say "only"; as the claim is equivalence, changing from "sufficiently large" to "all" strengthens one direction and weakens the other.) 66.31.30.18 (talk) 23:32, 27 June 2011 (UTC)[reply]