February 2003 - mathfun@mailman.xmission.com

[math-fun] A051532 help please?
by David Wilson 12 Feb '03

12 Feb '03

I know I started a flurry with my Ulam message, but can anyone give me an extension of A051532 up to a(n) <= 1001?

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Re: [math-fun] Ulam(1,2)
by Jud McCranie 12 Feb '03

12 Feb '03

At 12:43 PM 2/11/2003, Edwin Clark wrote: > Do you still have the sequence up to n = 5 million? If so I would >be interested in knowing whether or not the frequency of i such >that u(i) = u(i-1) + 2 continues to hold up. Gaps between successive >such i are quite small in my limited data (up to n = 2000). > >Also u(i) = u(i-1) + 4 only occurs in my data for i = 18 and 28. >And u(i) = u(i-1) + 13 only occurs once (at i = 53) in my list. Here are the small gaps for the 370,468 Ulam numbers <= 5,000,000 gap freq percent 1 4 0.0% 2 136930 37.0% 3 50710 13.7% 4 2 0.0% 5 4057 1.1% 6 0 0.0% 7 806 0.2% 8 4524 1.2% 9 3 0.0% 10 373 0.1% 11 0 0.0% 12 18417 5.0% 13 1 0.0% 14 0 0.0% 15 12557 3.4% 16 0 0.0% 17 31669 8.5% 18 0 0.0% 19 750 0.2% 20 28652 7.7% 21 0 0.0% 22 14623 3.9% 23 0 0.0% 24 748 0.2% 25 20120 5.4% Gap of 1 occurs only once, 4 only twice, 13 only once, and 9 only 3 times. Gaps of 6, 11, 13, 14, 18, 21, 23 have not occurred.

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Re: [math-fun] Ulam(1,2)
by Dan Hoey 11 Feb '03

11 Feb '03

Thanks to corroboration from Jud McCranie, Klaus Brockhaus, and my own new code, and partial corroboration from Edwin Clark, I'm satisfied I had the right answers (though occasionally the wrong question, due to confusion between f(n) mod 21.6 and f(n)/21.6 mod 1 . Oh, and occasionally dividing 21.6 by 2 and getting 10.3 instead of 10.8, in case you were wondering. It's amazing any of my code works the way I'm dropping bits. I'm still working on the faster version of the code. The idea is to code the "lookahead numbers" from u(n)+1 to 2u(n) in order as k when k is not representable as u(i)+u(j) and [m,n] when all numbers m <= k <= n are representable as u(i)+u(j) in just one way. The numbers representable in two or more ways are not mentioned. This is based on the observation that about 1/3 of the lookahead numbers are in the first case, and 2/3 in the third case. Almost all of the first case are singletons, while the run length in the second case averages about 2. The second case accounts for a vanishing fraction of the numbers, but they appear in larger and larger runs. For instance, at u(53584)=724384, the 724384 lookahead numbers are representable in 0 ways, 1 way, more than 1 way 246539 4059 473786 times, in 245835 266 249533 runs. Dan

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Re: [math-fun] Ulam(1,2)
by Dan Hoey 11 Feb '03

11 Feb '03

Jud McCranie <judmccr(a)bellsouth.net> > I'm getting in on this in the middle, so I don't know what you're > discussing, but maybe I could see what I get with my program. Thanks for the offer. There are two things I want verified. First, is it true that: u(1000)=12336 u(1091)=13533 u(1210)=14894 u(1730)=21870 u(1748)=22151 u(1947)=24851 u(1999)=25511 ? This is with zero-based indexing (u(0)=1, u(1)=2, u(2)=3, ...), so maybe you need to add 1 to the indices. Second, letting f(n) = (u(n)/21.6) - floor(u(n)/21.6, for what n in 1000...1999 is .5 <= f(n) < .55? Thanks. Dan Hoey

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Re: [math-fun] Ulam(1,2)
by Dan Hoey 11 Feb '03

11 Feb '03

Sorry, the second query should be: Second, letting f(n) = (u(n)/21.6) - floor(u(n)/21.6, for what n in 1000...1999 is 10.3 <= f(n) < 11.33? Thanks. Dan Hoey

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Re: [math-fun] Ulam(1,2)
by Dan Hoey 11 Feb '03

11 Feb '03

"David Wilson" <davidwwilson(a)attbi.com> wrote: > There are three differences between my computations and yours. > (1) I use Perl (2) I index from zero and (3) I wrote the algorithm > myself. If you index from zero, i.e. u(0)=1, u(1)=2, then that is the same thing I did. My program is written in Common Lisp. I wrote it with enough complexity that I could easily have an error. > Any of these may account for discrepancies between our > results, and since you and Brockhaus agree, I will let you be right, > it isn't worth it to me to find my error. Well, it's worth it to me if you can find it, because I want to know if I made an error. I wonder if you might have the correct sequence but some problem with the histogram, say with the boundaries of the bins or something. For that reason, I would really like to know what you calculate as the contents of bin [.5,.55). But this wish is not critical--don't bother if it's really a big problem. For my part, I'm going to rewrite my code (a) simply and slowly, just to test up to u(1999), and (b) with a slightly different method of optimization, to see if I can carry it out further. > All I was interested in was whether the observed bunching > phenomenon was real, or whether I was totally off the wall. > It appears that both you and Brockhaus have confirmed the > bunching with period about 21.6. Well, I wouldn't call what I confirmed "bunching", but a periodic predominance. Maybe that's just a difference of vocabulary, and maybe there's something else here we haven't really tested for. > Do you think this is a constant? In carrying it out to u(70000) it seemed to be a constant, but the value is more like 21.60157 as I said. Definitely between 21.6015 and 21.6016. > If so, what would cause this regular bunching? Here's a phenomenon that might explain it. It is similar to a phenomenon seen in "Take or Break" games such as Grundy's Game. Suppose that the real numbers mod K are divided into a sparse space S and a common coset C such that for all s1,s2 in S, c1,c2 in C, we have s1+s2 and c1+c2 in S and s1+c2 in C. Suppose further that the sequence values are in C with finitely many exceptions X, a subset of S. Then we might see a situation where - almost all values in S are be prohibited because they appear as c1+c2 for more than one pair c1,c2 in the sequence, and - for any sequence value c1 in C and x1 in X, the value c1+x1 is in C and will be represented at least once as a sum of sequence values. So as long as there is no c2+x2=c1+x1, the value c1+x1 will be in the sequence. There seems to be something more going on here, because the density within C is not uniform. But possibly that's just because the buckets are too big to show the exact structure of C. Dan Hoey Hoey(a)AIC.NRL.Navy.Mil

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Re: [math-fun] Ulam(1,2)
by Jud McCranie 11 Feb '03

11 Feb '03

At 12:43 PM 2/11/2003, Edwin Clark wrote: >Do you still have the sequence up to n = 5 million? If so I would >be interested in knowing whether or not the frequency of i such >that u(i) = u(i-1) + 2 continues to hold up. Gaps between successive >such i are quite small in my limited data (up to n = 2000). No I don't still have the data but I do have the program that did the test, and now I have a faster computer (although it gets slower and slower as n increases). I think the frequency you mention holds up, because I seen to remember that the density holds pretty constant. I'll dust off my old program and let you know.

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Re: [math-fun] Ulam(1,2)
by Jud McCranie 11 Feb '03

11 Feb '03

At 09:43 AM 2/11/2003, you wrote: > However based on looking at the first 2000 terms I boldly >conjecture that there are only 4 instances where u(n)-u(n-1) = 1. I tested that up to n=5,000,000 in June 2001, and commented about it in A002858.

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Re: [math-fun] Ulam(1,2)
by asimovd＠aol.com 11 Feb '03

11 Feb '03

David Wilson wrote: << I took a look at Ulam(1,2), the Ulam sequence starting with (1, 2) and including every subsequent number which is a unique sum of distinct earlier terms. This is Sloane's A002858. >> Actually A002858 is described in EIS as "Ulam numbers: next is uniquely the sum of 2 earlier terms" It's listed in EIS as beginning 1,2,3,4,6,8,11,13,..., (least such) but one could get a whole nother sequence via 1,2,3,5,8,13,21,34,..., (largest such) or 1,2,3,5,4,9,8,17,..., (largest alternating with least) for instance. ------------------------------------------------------------------------------ --------------- To well-define A002858 perhaps one should define it as "[Sequence beginning with 1,2 whose] next term is *the least* unique sum of 2 previous terms" QUESTION: If "2 previous terms" were replaced with "any number of previous terms" in this definition, would it be a different sequence? --Dan A.

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[math-fun] Math Corners in pop sci magazines?
by Helger Lipmaa 11 Feb '03

11 Feb '03

Hi, there are people here who should know an almost complete answer to that. I am interested in what popular science magazines publish (regular) math corners, and how well those are respected? What are most active, most fun, which ones accept papers by community and which ones only by the regular editors? Helger

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