January 2011 - mathfun@mailman.xmission.com

[math-fun] Ombudsman needed for OEIS
by N. J. A. Sloane 25 Jan '11

25 Jan '11

Dear Math Fun, I have been getting messages from people complaining that their submissions to the OEIS (both new sequences and comments) have been unfairly rejected. In other cases I see this happening, but people do not like to complain to me. Perhaps the OEIS needs an ombudsman, whose task will be to resolve disputes like this. Someone - not one of the editors-in-chief - who will listen sympathetically to complaints. It needs to be someone with a knowledge of mathematics and programming. It will be very interesting. Would someone like to volunteer to try this for a few weeks? Neil

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[math-fun] Calendars and continued fractions
by Adam P. Goucher 25 Jan '11

25 Jan '11

Dear all: In 45 BC, Julius Caesar reformed the Roman calendar to closely adhere to the tropical year. The Julian calendar has a leap year every four years, and thus has an average year length of: 1461/4 = 365.25 days. This is not a particularly accurate approximation to the actual tropical year, which is 365.2421897 days. Our current calendar, the Gregorian calendar, has the following rules: * If 400 | N, then N is a leap year, otherwise; * If 100 | N, then N is not a leap year, otherwise; * If 4 | N, then N is a leap year, otherwise; * N is not a leap year. It is not too difficult to ascertain that this leads to a mean year of: 146097/400 = 365.2425 days. This is much better than the Julian calendar, but far from perfect. What we actually desire is a *best rational approximation* to the length of the mean tropical year. Calculating the continued fraction expansion of the mean tropical year yields the following: 365; 4, 7, 1, 3, 27, ... The large value 27 means that it is wise to truncate immediately before it, and we obtain the following approximation: 365; 4, 7, 1, 3 = 46751/128 = 365.2421875. This is impressively close to the mean tropical year, and corresponds to the following rule system: * If 128 | N, then N is not a leap year, otherwise; * If 4 | N, then N is a leap year, otherwise; * N is not a leap year. This system is simpler than the Gregorian calendar, has a shorter period, and is 140 times more accurate. Additionally, a period of 128 years is a power of two, and therefore it is trivial to determine the number of days in the year based on the year's binary expansion. The next discrepancy between my calendar and the Gregorian calendar occurs in the year 2048, which is a leap year in the Gregorian calendar but not in mine. We have now identified the fact that best rational approximations form good calendars. What if a sadistic deity wanted to create a solar system to make it intentionally difficult for the inhabitants to design an accurate calendar? Obviously, we would want a number that has poor rational approximations. The basis is again continued fractions, but a value of Phi = 1.618033... would be optimal for this purpose. It has a continued fraction of 1; 1, 1, 1, 1, ..., with the best rational approximations equal to ratios of consecutive Fibonacci numbers. Let's presume we had a mean tropical year of 364+Phi days, so a year can have 365 or 366 days, in the ratio 1 : Phi. When should the 'leap years' occur? This is rather interesting, and corresponds to the Golden String: L R L L R L R L L R L L R L R L L R L R ... (L = leap year, R = regular year) Sincerely, Adam P. Goucher

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[math-fun] Plouffe's 2 surprising approximations and 1 exact formula for pi.
by Bill Gosper 25 Jan '11

25 Jan '11

By LatticeReduce, it appears that these are all of the form -1/240 + <algebraic>*Gamma[1/4]^8/Pi^6, e.g. for k=6, Sum[n^3/(E^(2*Pi*n/k)-1),{n,Infinity}== -1/240 + ((231 + 120 Sqrt[2] 3^(1/4) + 140 Sqrt[3] + 60 Sqrt[2] 3^(3/4)) Gamma[1/4]^8)/(5120 Pi^6) and Sum[n^3/(E^(2*Pi*n/7)-1),{n,Infinity}== 1/4 3/4 1 8 (301 + 210 Sqrt[2] 7 + 120 Sqrt[7] + 90 Sqrt[2] 7 ) Gamma[-] 1 4 -(---) + ------------------------------------------------------------------ 240 6 5120 Pi In[135]:= N[%, 33] Out[135]= 10.0000000000000001901617678886627 For k=13, the algebraic is (1/402653184000)(188822323200 + 91226112000 Sqrt[13] + 3145728000 (2128581 + 50505 Sqrt[3] + 610709 Sqrt[13] + 22176 Sqrt[39])^(1/3) + (66260343848333767461568512000000000 + 19010687557237458566381568000000000 Sqrt[13] - 31128880624384868352000000000 (50505 Sqrt[3] + 22176 Sqrt[39]))^(1/3)), which I suspect simplifies. k=163 might be, to quote Prof Watson, "Too cumbrous to be of any importance". --rwg Simon says:-> Hello everybody, these days I am working on exponential sums and I have found something of interest, like infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 13 = 119.00000000000000000000000000000009593745851025547335588584913... the precision is 31 digits. Another one is infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 7 = 10.0000000000000001901617678886626755843593058554453334802548978434061099438\ The precision here is 15 digits. For an argument of 2*Pi*n/163, the precision is 435 digits! I don't know why I took 163... (of course). in general, these sums will be near an integer if the argument is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ? These are the simplest I could find, if the exponent is of the form 4m-1 then the sum is often an integer with the same conditions. This, I believe extends a little bit the known formula of Ramanujan/Berndt/etc. A good question is : does someone has a simple explanation of this ? I don't. I am preparing a paper on these results. Because, I do have another one which is EXACT, namely for pi: In general these sums with a fractional exponent are very close to integers or pi, I have a couple of series for 1/pi, 1/pi^2, etc. here is the EXACT formula with fractional arguments mixed with integer exponents. /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | 10 | ) -----------------| - 40 | ) -------------------| | / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)| | ----- | | ----- | \ n = 1 / \ n = 1 / /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 10 | ) -------------------| - 10 | ) -----------------| | / n (exp(4 Pi n) - 1)| | / / Pi n \| | ----- | | ----- n |exp(----) - 1|| \ n = 1 / \ n = 1 \ 5 // /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 40 | ) -------------------| - 10 | ) -------------------| | / / 2 Pi n \| | / / 4 Pi n \| | ----- n |exp(------) - 1|| | ----- n |exp(------) - 1|| \ n = 1 \ 5 // \ n = 1 \ 5 // > evalf(%); 3.14159265358979323846264338327950288419716939937510582097494459230781640628\ 62089986280348253421170679821480865132823066470938442... well, I verified up to 1000 digits and it holds. This is trivial, ? I do not see how. If someone has a piece of information on why this exist, I would be glad to ear from it, references, known results, etc. Me, I never saw these kind of formulas before, have a good day, Simon Plouffe

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[math-fun] Slightly harder 15 puzzle
by Neil Bickford 23 Jan '11

23 Jan '11

If you happen to have a 15 puzzle lying around with a removable piece, you can make a harder puzzle by bandaging the pieces as follows: + - - - - + | A A E b | | F c E E | | F D g | | D h i | + - - - - + Puzzle: Move AA from the upper-left corner to lower-right. If a move is counted as moving 1 piece any direction (including L-shaped paths)any number of squares, 131 moves are minimum. Second puzzle: Is this the hardest 4x4 Sliding Block Puzzle there is? --Neil Bickford

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[math-fun] 2 surprising approximations and 1 exact formula for pi.
by Simon Plouffe 22 Jan '11

22 Jan '11

Hello everybody, these days I am working on exponential sums and I have found something of interest, like infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 13 = 119.00000000000000000000000000000009593745851025547335588584913... the precision is 31 digits. Another one is infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 7 = 10.0000000000000001901617678886626755843593058554453334802548978434061099438\ The precision here is 15 digits. For an argument of 2*Pi*n/163, the precision is 435 digits! I don't know why I took 163... (of course). in general, these sums will be near an integer if the argument is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ? These are the simplest I could find, if the exponent is of the form 4m-1 then the sum is often an integer with the same conditions. This, I believe extends a little bit the known formula of Ramanujan/Berndt/etc. A good question is : does someone has a simple explanation of this ? I don't. I am preparing a paper on these results. Because, I do have another one which is EXACT, namely for pi: In general these sums with a fractional exponent are very close to integers or pi, I have a couple of series for 1/pi, 1/pi^2, etc. here is the EXACT formula with fractional arguments mixed with integer exponents. /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | 10 | ) -----------------| - 40 | ) -------------------| | / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)| | ----- | | ----- | \ n = 1 / \ n = 1 / /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 10 | ) -------------------| - 10 | ) -----------------| | / n (exp(4 Pi n) - 1)| | / / Pi n \| | ----- | | ----- n |exp(----) - 1|| \ n = 1 / \ n = 1 \ 5 // /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 40 | ) -------------------| - 10 | ) -------------------| | / / 2 Pi n \| | / / 4 Pi n \| | ----- n |exp(------) - 1|| | ----- n |exp(------) - 1|| \ n = 1 \ 5 // \ n = 1 \ 5 // > evalf(%); 3.14159265358979323846264338327950288419716939937510582097494459230781640628\ 62089986280348253421170679821480865132823066470938442... well, I verified up to 1000 digits and it holds. This is trivial, ? I do not see how. If someone has a piece of information on why this exist, I would be glad to ear from it, references, known results, etc. Me, I never saw these kind of formulas before, have a good day, Simon Plouffe

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[math-fun] series for pi and approximations
by Simon Plouffe 22 Jan '11

22 Jan '11

Hello, sorry mr Schroepel, I posted a pdf which does not pass the filters, here is the same thing on a web page http://www.plouffe.fr/simon/Identities%20inspired%20from%20the%20Ramanujan%… bonne journÃ©e, simon plouffe

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[math-fun] article on new partitions results by ken ono and collaborators that are to be revealed tomorrow
by Thane Plambeck 21 Jan '11

21 Jan '11

not too much detail here, but sounds interesting http://esciencecommons.blogspot.com/2011/01/new-theories-reveal-nature-of-n… -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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[math-fun] NYTimes: Statistical significance testing questioned
by Henry Baker 17 Jan '11

17 Jan '11

FYI -- Coincidentally, I've been trying to find someone to co-author a paper with me on the correlation of prostate cancer & BMW driving -- where the incidence of prostate cancer is directly related to the first digit of the BMW model: 1,3,5,6,7. Of course, this digit is also correlated with the price of the BMW, and therefore the wealth of the driver, which is also correlated with the age of the driver (usually men). --- http://www.nytimes.com/2011/01/11/science/11esp.html January 10, 2011 You Might Already Know This ... By BENEDICT CAREY They should have seen it coming. In recent weeks, editors at a respected psychology journal have been taking heat from fellow scientists for deciding to accept a research report that claims to show the existence of extrasensory perception. The report, to be published this year in The Journal of Personality and Social Psychology, is not likely to change many minds. And the scientific critiques of the research methods and data analysis of its author, Daryl J. Bem (and the peer reviewers who urged that his paper be accepted), are not winning over many hearts. Yet the episode has inflamed one of the longest-running debates in science. For decades, some statisticians have argued that the standard technique used to analyze data in much of social science and medicine overstates many study findings Â often by a lot. As a result, these experts say, the literature is littered with positive findings that do not pan out: ÂeffectiveÂ therapies that are no better than a placebo; slight biases that do not affect behavior; brain-imaging correlations that are meaningless. By incorporating statistical techniques that are now widely used in other sciences Â genetics, economic modeling, even wildlife monitoring Â social scientists can correct for such problems, saving themselves (and, ahem, science reporters) time, effort and embarrassment. ÂI was delighted that this ESP paper was accepted in a mainstream science journal, because it brought this whole subject up again,Â said James Berger, a statistician at Duke University. ÂI was on a mini-crusade about this 20 years ago and realized that I could devote my entire life to it and never make a dent in the problem.Â The statistical approach that has dominated the social sciences for almost a century is called significance testing. The idea is straightforward. A finding from any well-designed study Â say, a correlation between a personality trait and the risk of depression Â is considered ÂsignificantÂ if its probability of occurring by chance is less than 5 percent. This arbitrary cutoff makes sense when the effect being studied is a large one Â for example, when measuring the so-called Stroop effect. This effect predicts that naming the color of a word is faster and more accurate when the word and color match (ÂredÂ in red letters) than when they do not (ÂredÂ in blue letters), and is very strong in almost everyone. ÂBut if the true effect of what you are measuring is small,Â said Andrew Gelman, a professor of statistics and political science at Columbia University, Âthen by necessity anything you discover is going to be an overestimateÂ of that effect. Consider the following experiment. Suppose there was reason to believe that a coin was slightly weighted toward heads. In a test, the coin comes up heads 527 times out of 1,000. Is this significant evidence that the coin is weighted? Classical analysis says yes. With a fair coin, the chances of getting 527 or more heads in 1,000 flips is less than 1 in 20, or 5 percent, the conventional cutoff. To put it another way: the experiment finds evidence of a weighted coin Âwith 95 percent confidence.Â Yet many statisticians do not buy it. One in 20 is the probability of getting any number of heads above 526 in 1,000 throws. That is, it is the sum of the probability of flipping 527, the probability of flipping 528, 529 and so on. But the experiment did not find all of the numbers in that range; it found just one Â 527. It is thus more accurate, these experts say, to calculate the probability of getting that one number Â 527 Â if the coin is weighted, and compare it with the probability of getting the same number if the coin is fair. Statisticians can show that this ratio cannot be higher than about 4 to 1, according to Paul Speckman, a statistician, who, with Jeff Rouder, a psychologist, provided the example. Both are at the University of Missouri and said that the simple experiment represented a rough demonstration of how classical analysis differs from an alternative approach, which emphasizes the importance of comparing the odds of a study finding to something that is known. The point here, said Dr. Rouder, is that 4-to-1 odds Âjust arenÂt that convincing; itÂs not strong evidence.Â And yet classical significance testing Âhas been saying for at least 80 years that this is strong evidence,Â Dr. Speckman said in an e-mail. The critics have been crying foul for half that time. In the 1960s, a team of statisticians led by Leonard Savage at the University of Michigan showed that the classical approach could overstate the significance of the finding by a factor of 10 or more. By that time, a growing number of statisticians were developing methods based on the ideas of the 18th-century English mathematician Thomas Bayes. Bayes devised a way to update the probability for a hypothesis as new evidence comes in. So in evaluating the strength of a given finding, Bayesian (pronounced BAYZ-ee-un) analysis incorporates known probabilities, if available, from outside the study. It might be called the ÂYeah, rightÂ effect. If a study finds that kumquats reduce the risk of heart disease by 90 percent, that a treatment cures alcohol addiction in a week, that sensitive parents are twice as likely to give birth to a girl as to a boy, the Bayesian response matches that of the native skeptic: Yeah, right. The study findings are weighed against what is observable out in the world. In at least one area of medicine Â diagnostic screening tests Â researchers already use known probabilities to evaluate new findings. For instance, a new lie-detection test may be 90 percent accurate, correctly flagging 9 out of 10 liars. But if it is given to a population of 100 people already known to include 10 liars, the test is a lot less impressive. It correctly identifies 9 of the 10 liars and misses one; but it incorrectly identifies 9 of the other 90 as lying. Dividing the so-called true positives (9) by the total number of people the test flagged (18) gives an accuracy rate of 50 percent. The Âfalse positivesÂ and Âfalse negativesÂ depend on the known rates in the population. In the same way, experts argue, statistical analysis must find ways to expose and counterbalance all the many factors that can lead to falsely positive results Â among them human nature, in its ambitious hope to discover something, and the effects of industry money, which biases researchers to report positive findings for products. And, of course, the unwritten rule that failed studies Â the ones that find no effects Â are far less likely to be published than positive ones. What are the odds, for instance, that the journal would have published Dr. BemÂs study if it had come to the ho-hum conclusion that ESP still does not exist?

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Re: [math-fun] FW: A conjectural series for 1/pi of a new type
by Bill Gosper 15 Jan '11

15 Jan '11

DanA>I've never studied infinite matrix products, but would have expected that for one to converge, its general term would need to approach the identity. That's the less common case. It happened with those Sierpinski gasket Fourier series of a few yrs ago. DanA>No? (Or perhaps some entries "diverge to 0" ?) --Dan Yes. You can see that happened in the UL 2x2 of the RHS, below. The determinant approaches -63/256 and determines the convergence rate. Easier example: 1/n 1 0 e Prod ( ) = ( ) . n>0 0 1 0 1 --rwg I've found a 2F1[9/16] limiting case of the "Stone" that gives 12/pi and is tantalizingly close to Sun's matrix, below. Bashing it to coincide would prove the conjecture. << . . . prod(matrix([-(k+1/2)^3/(8*(k+1)^3),-(k+1/2)^2/(16*(k+1)^2),30*(k+7/30)],[-63*k*(k+1/2)^2/(16*(k+1)^3),0,0],[0,0,1]),k,0,inf) = matrix([0,0,24/%pi],[0,0,0],[0,0,1]) [ 1 3 1 2 ] [ (k + -) (k + -) ] [ 2 2 7 ] [ - ---------- - ----------- 30 (k + --) ] inf [ 3 2 30 ] [ 24 ] /===\ [ 8 (k + 1) 16 (k + 1) ] [ 0 0 --- ] | | [ ] [ %pi ] | | [ 1 2 ] = [ ] | | [ 63 k (k + -) ] [ 0 0 0 ] k = 0 [ 2 ] [ ] [ - ------------- 0 0 ] [ 0 0 1 ] [ 3 ] [ 16 (k + 1) ] [ ] [ 0 0 1 ] . . .>>

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Re: [math-fun] FW: A conjectural series for 1/pi of a new type
by Dan Asimov 14 Jan '11

14 Jan '11

I've never studied infinite matrix products, but would have expected that for one to converge, its general term would need to approach the identity. No? (Or perhaps some entries "diverge to 0" ?) --Dan << . . . prod(matrix([-(k+1/2)^3/(8*(k+1)^3),-(k+1/2)^2/(16*(k+1)^2),30*(k+7/30)],[-63*k*(k+1/2)^2/(16*(k+1)^3),0,0],[0,0,1]),k,0,inf) = matrix([0,0,24/%pi],[0,0,0],[0,0,1]) [ 1 3 1 2 ] [ (k + -) (k + -) ] [ 2 2 7 ] [ - ---------- - ----------- 30 (k + --) ] inf [ 3 2 30 ] [ 24 ] /===\ [ 8 (k + 1) 16 (k + 1) ] [ 0 0 --- ] | | [ ] [ %pi ] | | [ 1 2 ] = [ ] | | [ 63 k (k + -) ] [ 0 0 0 ] k = 0 [ 2 ] [ ] [ - ------------- 0 0 ] [ 0 0 1 ] [ 3 ] [ 16 (k + 1) ] [ ] [ 0 0 1 ] . . . >> Those who sleep faster get more rest.

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