October 2003 - mathfun@mailman.xmission.com

[math-fun] finite Q-series
by Richard Schroeppel 09 Oct '03

09 Oct '03

rwg> The integerhood of a 3F2 summand I mentioned follows from the conjecture that n /===\ k - j + 1 k + 2 j + 2 k + 2 j + 3 | | (1 - q ) (1 - q ) (1 - q ) | | ------------------------------------------------------ | | j j + 1 j + 2 j = 1 (1 - q ) (1 - q ) (1 - q ) is a polynomial in q for all n,k >0. This seems to be hard!? I'll bet there's a theorem that for an expression like this, you only need to check a dozen or so consecutive n and k --rwg Chorasmian maraschino harmonicas inspissate antisepsis. Should the exponent on the first q in the numerator be k+j+1? As it stands, it looks like k=1 n=3 (and hence the j=3 term) will create a q^-1 that won't disappear. --- Wrt the pyramid puzzle, I recall seeing an article about it Science News, maybe 5-10 years ago. It may have been the Math SAT? Rich rcs(a)cs.arizona.edu

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[math-fun] Rational Group
by Jon Perry 09 Oct '03

09 Oct '03

{Q,+} (the rationals with the addition operator) is a group. How come the infinite sum of rationals is not then necessarily a member of this group, e.g. zeta(2)? Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] nonobviously integer sequence
by R. William Gosper 09 Oct '03

09 Oct '03

Let a and c be integers. "Shew that" F := 1, F := a c - 1, F := 0 1 n 2 (2 n + a - 3) ((2 c - 1) (2 n + a - 4) (2 n + a - 2) + (a - 2) ) F n - 1 (----------------------------------------------------------------------- 2 - (n - 1) (n + a - 3) (2 n + a - 2) F )/(n (n + a - 2) (2 n + a - 4)) n - 2 (note cubic denominator) is an integer sequence. Distantly related: it appears that [ n + 4 n + 5 ] [ - n, -----, ----- | ] F [ 2 2 | -4] = 1,6,57,701,10147,164317,2888282,54047434, 3 2 [ | ] [ 3, 2 ] is an integer sequence. The smallest prime factors go {},2,3,701,73,37,2,2,7,2,3,3,5,2,23,37,65306671610636210891,2, 31,3,3,19,2,2,3,2,3,7,3,3,2,2,7,2,67,5,... . Likewise, [ n + 5 n + 6 ] [ - n, -----, ----- | ] 4 F [ 2 2 | -4] = 4,25,228,2620,35164,527663,8613004, 3 2 [ | ] [ 4, 2 ] [ n + 6 n + 7 ] [ - n, -----, ----- | ] 15 F [ 2 2 | -4] = 15,99,891,9825,125085,1772775,27303603, 3 2 [ | ] [ 5, 2 ] [ n + 7 n + 8 ] [ - n, -----, ----- | ] 5 F [ 2 2 | -4] = 5,29,230,2260,25921,334105,4717653, 3 2 [ | ] [ 5, 3 ] More generally, [ n + k + 1 n + k + 2 ] [ n + k + 2 n + k + 3 ] [-n,---------,---------| ] [-n,---------,---------| ] F [ 2 2 |-4] & F [ 2 2 |-4] 3 2[ | ] 3 2[ | ] [ k, 2 ] [ k, 3 ] appear to have bounded denominators. --rwg Does OEIS list sequences not proven integral?

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[math-fun] manners
by Richard Schroeppel 09 Oct '03

09 Oct '03

Let's keep the discussion civil, please. Rich rcs(a)cs.arizona.edu

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RE: [math-fun] Rational Group
by Jon Perry 09 Oct '03

09 Oct '03

'But it doesn't give you a way to add *infinitely many* elements together. The usual way to define infinite sums is as a limit of finite sums. But a group structure is not enough to be able to define limits; you need an additional, topological, structure, to do that. And even once you have such a topological structure, that doesn't guarantee that *every* series or *every* sequence converges. Some do, some don't. If your topological structure comes from a metric structure (The standard topological structure on Q does), and the metric space is complete (Q isn't), then any Cauchy sequence will converge. Since the Reals are complete, this guarantees that the series for zeta(2) converges to some Real number (but not necessarily to a number in Q, since Q is not complete).' So how do we know that the incomplete nature of Q produces Reals, or indeed how can we determine the 'completetion' of an incomplete group (R is the completion of Q). Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] Hodge conjecture
by Jon Perry 09 Oct '03

09 Oct '03

While we are on the subject of the $1million questions up for grabs, has anyone seen the lecture at Clay for the Hodge conjecture? The lecturer waffles on about spaces and equivalence and doughnuts for ages, and then right at the end he is asked a question, and says that the Hodge conjecture boils down to something about skew matrices or something else. Does anyone know this - it takes ages to get to on my linear RealPlayer, and I have found no references to it elsewhere. Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] Insomniacs Problem Revisited:
by otto2hoh 08 Oct '03

08 Oct '03

John Conway posted the original insomniacs problem below, which assumes a 12 hour clock with no leading zeros. The following trickier problem also assumes a 12 digit clock. """ Here in washington state I am watching palindromic numbers go by in local time on my on my computers 12-hour clock without leading zeros. I count 281 seconds between two palindromes. How long will it be before I count 281 seconds between two number again? Regards Otto Original Problem: > There's a digital clock at my bedside that shows > hours minutes and seconds, and which I usually look > at the moment I wake up. A few days ago, I happened > to wake up at a palindromic time, and out of interest, > waited until the next palindromic time, and noticed the > interval between them. > > Last night exactly the same thing happened, but the > interval was exactly four times as long as in the previous > case. What did the clock read when I woke up last night? > > John Conway

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[math-fun] Poincare's Conjecture
by Jon Perry 08 Oct '03

08 Oct '03

I think I have proved this, and I am looking for a respectable journal to publish it. Does anyone on this list have any suggestions? I live in the UK and would prefer a UK based journal although this is not a requirement. Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] Re: Cedric Smith - Herbert Wright - 13-coin problem
by Richard Guy 06 Oct '03

06 Oct '03

Sorry to hear of your computer problems. I'm copying this quick reply to `math-fun' and to Rich Schroeppel, who will make sure that you're reinstated on the math-fun list. More later. R.

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[math-fun] 3x+1 problem
by Richard Guy 03 Oct '03

03 Oct '03

I seem to have deleted a recent message which gave the bound to which the 3x+1 problem has been checked (something like 10^17) and an address of a Jeff Lagarias web page which gave info on the problem. Could someone kindlily resend to me? Thanks! R.

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