daz> this approximation suggests that9^(4^(7*6)) and 3^(2^85) must be fairly good approximations to one another. Indeed! sin(11) is a bit of a cheat, since it takes the closeness of 11 to 7pi/2 and squares it. But sin(2) would be perfectly reasonable, so maybe instead of banning "asymptotic" functions, the problem should be recast as finding the appropriate price (in complexity units). Ed's example of cos(1+cos(1+cos(...))) shows you can square the epsilon at a constant cost, but if we charged according to 1/derivative, this could be fixed. He also hasn't provided for including function definitions, non-named infinite series/products/limits/sups, and functional inverses. MaxCF doesn't seem to be defined. If it's supposed to be the maximum CF partial quotient, some care is needed: it's likely infinite for everything except quadratic surds. If it's the sup of the average p.q., this is usually bounded and can be guessed from the first few CF terms. Rich rcs@cs.arizona.edu
"Once, in the taxi from London, Hardy noticed its number, 1729. He must have thought about it a little because he entered the room where Ramanujan lay in bed and, with scarcely a hello, blurted out his disappointment with it. It was, he declared, 'rather a dull number,' adding that he hoped that wasn't a bad omen. 'No, Hardy,' said Ramanujan, `it is a very interesting number. It is the smallest number expressible as the sum of two [positive] cubes in two different ways." I've long thought that the fourth root of 9.1 was a dull number, despite a strange continued fraction. Of all the potentially dull numbers in the world, I had to chose that one. Richard Sabey wrote to me.
You say [I paraphrase, for lack of a root-symbol] "I considered 9.1^(1/4) weird, but nothing particularly special". You are too modest, this being an alternative presentation of the largest 2 consecutive 19-smooth numbers, 11859210 and 11859211, which you published in Mathpuzzle on the 23rd of December.
I didn't see the connection at all, so I asked Richard to explain. (http://www.research.att.com/projects/OEIS?Anum=A002072 , by the way, is largest consecutive n-smooth numbers)) Richard wrote: As you reported, the largest 2 consecutive 19-smooth numbers are: 11859210 ~~ 11859211 => 7*13*19^4 ~~ 2*3^4*5*11^4 => 91*19^4 ~~ 10*33^4 => 9.1 ~~ 33^4/19^4 => 9.1^(1/4) ~~ 33/19 Now I get it. So 9.1^(1/4) turns out to be interesting after all. Ed Pegg Jr
I can prove A002072(1) = 1 and A002072(2) = 8 (but I have to invoke the Catalan Conjecture to get the latter). Subsequent elements seem a lot more difficult to prove, and I suspect that they are conjectural. Is this the case? ----- Original Message ----- From: "ed pegg" <ed@mathpuzzle.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Thursday, March 10, 2005 12:18 AM Subject: [math-fun] My Ramanujan moment
I didn't see the connection at all, so I asked Richard to explain. (http://www.research.att.com/projects/OEIS?Anum=A002072 , by the way, is largest consecutive n-smooth numbers))
-2.50290787509... is the start of Feigenbaum alpha. 2.50290786097... is the start of phi+(7/8)^G G is the Catalan constant, and phi is the golden ratio. Sent to me by Derek Ross. Ed Pegg Jr
participants (3)
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David Wilson -
ed pegg -
Schroeppel, Richard