June 2005 - mathfun@mailman.xmission.com

[math-fun] Re: Packing circles into a finite rectangle
by James Propp 09 Feb '06

09 Feb '06

>> If you have circles of radii 1/1, 1/2, 1/3, ..., 1/n for all n >= 1, >> then their total area is finite, although the sum of their radii is >> infinite. >> >> Can one fit these circles into a finite-sized rectangle? > > yes, they will fit inside a 1 x (11/6) rectangle for example. put each > circle inside a square, and then pack the squares like this: > http://www.stetson.edu/~efriedma/harmonic/ It is apparently still unknown whether one can pack squares of sides-lengths 1/1, 1/2, 1/3, etc. into a 1-by-(pi squared over six) rectangle. See Clive Tooth's web-page www.pisquaredoversix.force9.co.uk/Tiling.htm for an example of how one might get started. I asked Clive for background on this tiling, and he kindly forwarded to me copies of some relevant sci.math postings from last year (one by Clive himself and one by David Cantrell): ************************************************************************** From: "The Last Danish Pastry" <TheLastDanishPas...(a)yahoo.com> Newsgroups: sci.math Subject: Packing squares into a 1 x Zeta[2] rectangle Date: Sun, 14 Mar 2004 13:41:34 -0000 Since the sum of the reciprocals of the squares of the positive integers is pi^2/6, the question arises as to whether squares with sides 1, 1/2, 1/3, etc can be packed into a rectangle of size 1 by pi^2/6. A picture of such a packing appears at http://www.pisquaredoversix.force9.co.uk/Tiling.htm I know of no proof that such a packing is possible or impossible. A program which I wrote has recently packed the first one million such squares into such a rectangle. I wrote the program in Mathematica and it uses exact arithmetic throughout. The algorithm is simple: The program maintains a list of rectangles. Each member of this list is itself a list containing exactly two members: the short side of a rectangle followed by the long side of the rectangle. The program does not store the position or orientation of any rectangle. The list is stored in ascending short side order. Initially the list contains one entry: {1, pi^2/6}. The program inserts the squares in decreasing size order, starting with the 1x1 square. At each stage, to insert the square with side 1/n, the program finds the rectangle with the shortest short side which will accommodate the square. [In the case that there are two or more such rectangles with equal short sides, the program will pick the youngest one.] Suppose that this rectangle is a x b, where 1/n<=a<b. The program deletes this rectangle from the list and inserts two new rectangles (b-1/n) x a and (a-1/n) x 1/n into the list. - If any side of one of these new rectangles is zero, the rectangle is not inserted. - Each new rectangle is stored in {short side, long side} form. Thus, the list of rectangles grows by one per inserted square, except in the rare cases in which a rectangle of exactly the right width is found. A trivial optimisation, which took me a long time to notice, is that if the number of squares to be inserted is decided at the beginning, as N, say, then any rectangles narrower than 1/N can be discarded as soon as they appear. This has a dramatic effect on the memory requirements and running time of the program. In the case of N=10^6 the maximum length of the rectangle list is only 7,493 which occurs just after inserting the square with side 1/55,205. Thereafter the length of the list slowly decreases. After inserting the last square, with side 10^-6, there are just 7 rectangles in the list. Their dimensions are, roughly: 0.000001000264 x 0.000371508565 [1,000,000] 0.000001000289 x 0.000006561855 [ 999,736] 0.000001000332 x 0.000011950546 [ 999,710] 0.000001000424 x 0.000099911008 [ 999,668] 0.000001001705 x 0.000013366392 [ 999,575] 0.000001001771 x 0.000555285230 [ 998,297] 0.000621399972 x 0.000621402936 [ 998,232] The number in square brackets is the square insertion step on which the rectangle was created. Note that each rectangle, except the last, has a short side just exceeding 10^-6. The last one is huge in comparison to the others. Also, it is almost square, having an aspect ratio of about 1.00000477 . The exact size of that last rectangle may be expressed as: (pi^2/6 - ShortNumerator/ShortDenominator) by (LongNumerator/LongDenominator) Where ShortNumerator = 88148577597184641314023698687804198322622003067950 90277185502203375757868045576236778623499141581835 91095283100445795053089253938670874859361703421694 93589311823604254188777249571387340885945805010588 59158565482204748239750452173829861295626045336259 37611823762149119593932203343599960000593198902456 62151237628335467123582313067030461952701063806541 09667253352708582423851542412389977650553557102776 48867766805593783542952432276691372915179943637567 30921614687225494242879103415386639772227105415075 60643644416702161056703911777435451470868967261015 03285333687384626370723064421360623232628962182637 33785897530162588640506103091960735379749803817168 27337105343328553249474140253583305664128829433056 13496001065570446619534699711595669897241621797339 93388113729020710706312912651983440135427698863738 63455691221381048718834049233446561869585869886892 11066449759482338440176987851999633382542943944743 83052271691048660471857179765488027759900589960305 37378583137068870187987573969424873832345344118539 33368635075465276924658419905001027235726511505469 90353609271920928926254763071708700239940565565014 69168650106503612867362486635886531701170797485510 99375741498187773039269448405524049571692907463933 52292512770766034352574830725344567781634199947344 39128864912738319438009565992988317045941602472518 84296172734535520954249457165155850357846464984168 55299222999724423825677947429849212042460805178788 23214473791347512926915739195451188103089711423995 69891250713418835019305494967439261535807066257765 86799633983386818438909781178910531233498247023179 22489232680161587870238609995903489290522620579734 27913072409961919211050524749783993630732833119534 51221464170732847284504244076797981093088600479209 053071348915420741889 ShortDenominator = 53608160645434703140761211994317971700783074368336 23325975705041657962199994733358382994458318582660 77598233503372093023408990096331259105144227503324 99931830157705570560424971647310855286927378535483 51519419895212616165032657829920140448566211296935 08925318960134998789890442610040864611429495599467 55671385275409484022080990210675967224987014208598 03335236373663242927780013944648864149878730486125 39041613975393755631755935261748017063478349180906 68435262002209072815068783227098957825899390123056 87116154057047656878878907753207321889684985791605 16008019019744195226005265347943729214648741668979 90732236259531674523664568538104488968273268732774 74645971065348455713992347183400863276778739619943 89625588842652615699820699583835957017913827580111 39619239553717364061586421301662144411944880262864 13395969932087668263688142360819354641427777701996 11575083489100468980371536759296335301136691198138 39559100156995546068054947710363960181181289364408 47219039882502818687162480694856587042340603392188 93192097190331493671006154660882637398429894947140 66724834302759980279336075917538263518035481261929 09959571761983813160895335069812335369906345916646 57307263315174319702690110108769119880417800559541 00846379467033278426493445187132240090150811316237 11870955392306082427560405688265993364931620972062 42335013731582665984021398103261291575096750373611 78793366106368150944101270721191063173365978409196 32419016511795819355629216137822167151147383280072 17692953976640188298210375720290857531988737370969 01207985676688854219832216404101906724620701738590 65777947161752685898547108814129100143613045915161 23374032620084188029394058765229087587568694264890 44233796349785411756140522085279648064146173069741 478636801148014080000 LongNumerator = 86047691660315035868968439680618750650385496662337 69341618423224868557176707904056350747739644607152 19341301092244142985694427567208621603825424715808 50112490209730250303812592410205852634552809617406 54471220118615832220663814127702644186714156802514 89272173167125109946591599082250813475432136844437 68434150604827750301142594110541508119955806775639 48463821703879298880841441517861092382325647927555 17115984413353561811152746794319549286695056542231 75370181749913246697604501180515750468588947850814 96706860842334628047016110265861987131951269186913 55908484624794372227266755207694315503681976948427 59732973255837108405693520922709400140190847636749 60091554508265275883766083172672994896381234279724 35485615485280999872894822261731570552709443192476 07685741096374165452217373319152585412622236248087 75856424918727148070254116048744362672852782317829 99207549345144244124571110563211160343448412194752 79110755652853486307738886342254777707077572587992 29405091182393418319601369763056120887765700618802 27139753549245993550447831136020976532760818138519 98216552674127086312240961717818956038535226187300 43026597993187644795910836481706823332909020130603 09729716420449811214331443400182398534278484763571 27904525989212662382850630094162630962359551947273 97311922973034069310043168935737831265241199089047 46559558404697050977201086017144902884602248144108 71005578441670661613107026657746377160408235952122 70087369832576028781964223879066392055590320740847 64872057694761838719127407266757016871496087930551 58594303661605591550655979493954945142285417155556 97061447617516531128930865552526142619566789459050 16802730874016347300046435875463493515247956092289 80788952041557673741641666771939979618817508420910 11050430185421809507580791626321726210797393934779 33362836600904869583181 LongDenominator = 13847326202197605511604348541676281864027795818043 31494335574789145450920249049910013265993162156505 60831474662540178414423726822350848233301461067293 21368995360010230519749558542277057071473577106811 23548536664652068929533780687547853336157369555086 86435599910456774645119837370840414802536199558214 07527909448749766750406409444125057733978666254122 19353807926168688488715863115960118447262678159252 66974533884810419996115934738184899912380787401311 77856867483197701324817440576351765023155329523060 68640715686158929369832278778842403496825357945180 51999476544062632945516973292520225337991834072383 82846228280041100692245330583478338241014202245803 30323143236417275386997321382053499979484896390576 45426647084624716067636294896758072966317096343334 28169063530634898773783098553798071542939121086910 08006721058240228461124891602875033925435245224172 90909661066024272174911721070106393261965722571136 18608587173321485889447514408382277645342459166650 66976600733716756193252940889415454560334136529577 70915746535414564177067690689656587269581203051464 54295311030944948219808305598301111195229732416082 42817235468158091436079088616794858652804473035517 82817817155337872004025893330437425894039012377193 76447539452009335637045807403996380650616246802471 05691575600545667349590304669247493320414626833379 52612693846867728391114410877249463229225980473996 60123182871902607011087292407939606651288648107916 20645836880002779839862309998364888537462475216562 83515669632032306317521726424095159019732485843596 26389117354057755490411586373128011034072899097872 57238596906896860163242623358201011919041838989031 36072177994374094882330699298066695784371692394667 11940033790850742395135563324255474227834715815498 61014376098413906103604397447388393432391565219123 065068758213935671227536000 -- Clive Tooth http://www.clivetooth.dk ************************************************************************** Subject: Re: Packing squares into a 1 x Zeta[2] rectangle From: David W. Cantrell <DWCantr...(a)sigmaxi.org> Date: 18 Mar 2004 16:31:52 GMT Newsgroups: sci.math "The Last Danish Pastry" <TheLastDanishPas...(a)yahoo.com> wrote: > Since the sum of the reciprocals of the squares of the positive integers > is pi^2/6, the question arises as to whether squares with sides 1, 1/2, > 1/3, etc can be packed into a rectangle of size 1 by pi^2/6. > > A picture of such a packing appears at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > > I know of no proof that such a packing is possible or impossible. [snip] It would be good if people interested in this thread were aware of some of the history of this problem. To that end: _Unsolved Problems in Geometry_ by Croft, Falconer and Guy (Springer-Verlag, 1991) "D5. Packing unequal rectangles and squares in a square." (pp. 112-113) mentions a question similar to the one considered in this thread. D5 caused me, in 1996, to consider several packing problems of this sort. I found simple algorithms, which I implemented in Mathematica, which seemed to pack rectangles of decreasing size "perfectly". The algorithms worked for classic problems, such as packing the rectangles 1/n by 1/(n+1) into the unit square, as well as for interesting new packing problems. I did not publish my results then, hoping first to prove that the algorithms were adequate for packing all of the rectangles. In November 1999, spurred by Clive Tooth, I posted to sci.math a brief description of what I had done. See both parts of "Packing Rectangles of Decreasing Size" at <http://mathforum.org/discuss/sci.math/a/t/241167>. (Readers may also be interested in looking at the packing algorithm which Clive had used back then.) Unbeknownst to me in 1999, an algorithm for such packings had appeared previously in the literature: An Algorithm for Packing Squares, Marc M. Paulhus, _J. Combin. Theory Ser. A_ 82 (1997) 147-157. His algorithm is somewhat different from any of the algorithms which I had used (but I wonder if perhaps it is the same as Clive's current algorithm). For packing squares of side lengths 1/2, 1/3, 1/4,... into a rectangle 1/2 by 2(pi^2/6 - 1), Paulhus gives two rules: "Rule 1. Always place the next square in the corner of the smallest width rectangle into which it will fit." "Rule 2. After placing a square into the corner of a rectangle, always cut the remaining area into two rectangular pieces by cutting from the free corner of the square tot the longer side of the original rectangle." For older references, see those cited by Paulhus. The most recent articles known to me concerning such packings -- but again, I may not be up-to-date with the literature -- are the following: Perfect Square Packings, Adam Chalcraft, _J. Combin. Theory Ser. A_ 92 (2000) 158-172. Compactness Theorems for Geometric Packings, Greg Martin, <http://arxiv.org/PS_cache/math/pdf/0005/0005054.pdf>. Perfect Packings of Squares Using the Stack-Pack Strategy, Johan Wastlund, <www.mai.liu.se/~jowas/dcg.ps <http://www.mai.liu.se/~jowas/dcg.ps>>. Thoughtful comments are welcome. Regards, David W. Cantrell ************************************************************************** (Clive tells me that there's more discussion of this problem in sci.math, though I haven't checked.) Jim Propp

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[math-fun] Found in the IHP cellar! 2/2
by Christian Boyer 21 Nov '05

21 Nov '05

---- FAUQUEMBERGUE computations And I have the chance to have discovered, again in the cellar of the Institut Henri Poincaré, the original handwritten computations on Mersenne numbers M101, M103, and M109 as sent by Fauquembergue to André Gérardin. http://www.utm.edu/research/primes/mersenne/LukeMirror/lit/lit_008s.htm These computations done in 1913-1914 were... false, as found by Robinson 40 years later: http://www.utm.edu/research/primes/mersenne/LukeMirror/lit/lit_024s.htm Within the huge number of handwritten 0s and 1s, the game would be to locate the first line where the error occurs! I can send .JPG photos to interested people. Christian.

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[math-fun] regular polygon solver
by R. William Gosper 30 Jun '05

30 Jun '05

Macsyma has a package of autoloading geometric formulae, e.g. (c2) area_regular_polygon(n,s) c:\macsyma\Macsyma2\share\geofuncs.fas being loaded. %pi 2 cot(---) n s n (d2) ------------- 4 but the regular polygon functions have two drawbacks: there are too many possible interconversions (e.g., inradius and side to area), and you can't remember which take which arguments (e.g., maybe that 2nd arg to area_regular_polygon was really the circumradius). Short of popping open a doc window, you can determine the argument semantics by loading the uncompiled package and peeking at the lhs of the definition: (c4) (load("c:\\macsyma\\Macsyma2\\share\\geofuncs.mac"), first(fundef(area_regular_polygon))) (d4) area_regular_polygon(numsides, edge) But that's gross. I wanted something like Dick Petti's contribution (c5) first(fundef(solve_triangle)) (d5) solve_triangle(arg1, arg2, arg3, problem_type) where the args are either angles or sides, and problem_type is ASA, SAS, etc. I finally wrote (c1) solve_regular_polygon() Batching the file c:\rwg\climax\share\geofuncs.mac %pi 2 %pi cot(---) n s csc(---) s n n (d1) [n = n, area = -------------, circumradius = ----------, 4 2 %pi %pi cot(---) s tan(---) s n 2 n inradius = ----------, perimeter = n s, sagitta = ----------, 2 2 side = s] As you see, with no args, it just recites the various dimensions in terms of the number of sides and side length. (c2) first(fundef(solve_regular_polygon)) (d2) solve_regular_polygon([n or n,side or eqns]) So you can supply one or both of these (c3) solve_regular_polygon(6) 2 3 sqrt(3) s (d3) [name = hexagon, n = 6, area = ------------, circumradius = s, 2 sqrt(3) s (2 - sqrt(3)) s inradius = ---------, perimeter = 6 s, sagitta = ---------------, 2 2 side = s] But you can also supply equations specifying the other properties: (c4) solve_regular_polygon(name = octagon,area = 1) 1/4 2 (d4) [name = octagon, n = 8, area = 1, circumradius = ----, 2 sqrt(2) sqrt(sqrt(2) + 1) inradius = -------------------------, 4 perimeter = 4 sqrt(sqrt(2) - 1) sqrt(2), %pi sqrt(sqrt(2) - 1) sqrt(2) tan(---) 16 sagitta = ----------------------------------, 4 sqrt(sqrt(2) - 1) sqrt(2) side = -------------------------] 2 (c5) solve_regular_polygon(radius = r,sagitta = r/2) 2 3 sqrt(3) r (d5) [name = equilateral_triangle, n = 3, area = ------------, 4 r r circumradius = r, inradius = -, perimeter = 3 sqrt(3) r, sagitta = -, 2 2 side = sqrt(3) r] Some answers can be neat: (c6) solve_regular_polygon(area = s^2,perimeter = 4*s) 2 sqrt(2) s (d6) [name = square, n = 4, area = s , circumradius = ---------, 2 s (sqrt(2) - 1) s inradius = -, perimeter = 4 s, sagitta = ---------------, side = s] 2 2 and some can be messy: (c7) solve_regular_polygon(area = \a,perimeter = p) 2 2 4 %pi 2 sqrt(16 n A + p ) - tan(---) p 2 n (d7) [name = regular_polygon, n = ---------------------------------, 4 A area = A, circumradius = 2 2 4 %pi 2 2 4 2 A sqrt((sqrt(16 n A + p ) - tan(---) p ) + p ) 2 n ---------------------------------------------------, . . .] 2 2 4 %pi 2 p (sqrt(16 n A + p ) - tan(---) p ) 2 n Even when short, all the dimensions you weren't interested in are messy by definition. The trick is to substitute into what you want instead of letting all the junk print out: (c8) subst(solve_regular_polygon(area = \a,perimeter = p),inradius) 2 A (d8) --- p (c9) subst(solve_regular_polygon(radius = \r,apothem = r), [sides = n,\a/p = area/perimeter,sag = sagitta]) %pi A r (d9) [sides = ---------------------, - = -, sag = R - r] 2 2 p 2 sqrt(R - r ) 2 atan(-------------) R + r Tweak: (c10) block([assume_pos:true],rootscontract(substpart(factor(piece),%,1,2,2,2,1,1,1))) %pi A r (d10) [sides = -------------------, - = -, sag = R - r] R - r p 2 2 atan(sqrt(-----)) R + r --rwg

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[math-fun] Fibonacci periodicity; ORed testing problem
by Richard Schroeppel 29 Jun '05

29 Jun '05

You can find primes with really short Fibonacci periods by factoring Fibonacci numbers. Fib25 = 75025 = 5 * 5 * 3001 Fib26 = 121393 == 1353 (mod 3001) == I == sqrt(-1) (mod 3001) so the full period for P=3001 is 4*25 = 100. There's a fairly strong analogy between Fibonaccis and Mersennes, so there's likely a Wieferich phenomenon for Fibonaccis. (1093^2 | M1092, and 3511^2 | M3510, and 11^2 | 3^10-1.) ---- I recently read about a new test for AIDS. It detects something about the actual virus -- its DNA or proteins -- whereas older tests detect antibodies. The new test detects an infection earlier than the antibody tests. The drawback is that the new test is expensive. One use of the test is confirm negative results from the older tests: A group of samples, from people who are older-test-negative, is mixed together for a run of the new expensive test. If the new test reports negative, all is well; if positive, then a series of retests & eliminations follows to locate the true positives. I've oversimplified this into a math problem. Even listing the oversimplifications would take a page, so I won't. Here's the "ORed Testing Problem": Assume you have some number of people, each independently likely to have property A with probability P. Property A does not change with time. A perfectly accurate test is available for property A, and it can operate on any set of one or more combined samples, and report either Negative (none of the samples in the test run has property A) or Positive (at least one sample has property A). Each run of the test costs $1, no matter how many samples are pooled for the run. Each person will provide however many samples are needed, and all samples from the same person produce the same result. There is no other test for property A. The goal is to determine each person's property A status, with an (expected) minimum total cost. The best strategy depends on P. If P is near 1, there's no point in doing any combined tests, since they will almost always be positive. Each person must be tested individually. But if P is small, there's an advantage to combining samples. An entire group can be tested at once, and there's a good chance they are all negative. In this case we get a total cost of only $1, saving $N-1 over the cost of individual tests. For two people, it turns out that a merged test is optimal when P < .382 = 2-phi = phi^-2. I started working on the three-person case, but it turned into an algebra morass. The best strategy needs to be calculated for the situations where you know that 1) the merged group XYZ is positive; 2) the merged pair XY is positive, and nothing is known about Z; 3) the merged pairs XY and XZ are positive; 4) the merged pairs XY, XZ, and YZ are positive; 5) nothing is known; 6) the merged pair XY is positive and Z's status is settled. The available inferences are pretty simple. A negative test applies to everyone in the merged sample, and deletes them from any positive groups. An individual may be determined positive if they are they only one left in a positive group test, or in the limiting case of an individual test. A positive test for a group G superannuates any previous test for a supergroup of G, including the case where G is a singleton. Hilarie suggested a first cut at a strategy for small P and a large group: Calculate the number of samples to merge so as to make the probability for a positive test about 50:50, and test subgroups of that size. Anyone in a positive group has his probability adjusted upward for the next round. Randomize the positive subgroups for the next round, and retest with the smaller subgroups based on the revised probabilities. Etc. Rich rcs(a)cs.arizona.edu

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RE: [math-fun] factoring formal power series
by dasimov＠earthlink.net 28 Jun '05

28 Jun '05

Neil asked (excerpt): << [W]hat about [factoring] power series with integer coefficients? There are two versions, depending on whether you work to order n, or with the full power series. Version 1: Given 2 3 4 5 6 7 8 t3 := 1 + 3 x + 7 x + 17 x + 42 x + 113 x + 321 x + 976 x + 3109 x + 9 10 11 12 10258 x + 34698 x + 119554 x + O(x ), find t1 and t2 such that t1*t2 = t3 (mod x^12). Answer: > t1; 2 3 4 5 6 7 8 9 1 + x + 2 x + 5 x + 14 x + 42 x + 132 x + 429 x + 1430 x + 4862 x 10 11 + 16796 x + 58786 x + ... > t2; 2 3 4 5 6 7 8 9 10 11 1 + 2 x + 3 x + 5 x + 7 x + 11 x + 13 x + 17 x + 19 x + 23 x + 29 x + 31 x + ... ___________________________________________________________ Version 2 (harder): Factor: 1 + 2*Sum_{m >= 1} x^(m^2) = 1 + 2x + 2x^4 + 2x^9 + 2x^16 + ... Answer: Product_{n >= 1} (1-x^2n)(1+x^(2n-1))^2 >> --------------------------------------------------------------------------- Since the group of units is exactly the set of power series beginning with constant term +-1, finding the equivalence classes up to multiplication by a unit seems worthwhile. Suppose the lowest term of a power series S(x) in Z[[x]] is +-N*(x^d), with d >= 0 and N > 0. Then there is a unique unit U_S(x) such that (*) U_S(x)S(x) = x^d * (N + Sum_{k >= 1} (L_k x^k)), where for all k >= 1 we have 0 <= L_k < N. Thus each equivalence class up to multiplication by a unit has a unique representative of the form [RHS of (*)]. (In fact the cofficients L_k are just the original coefficients of x^(d+k) reduced mod N.) This last fact allows us to multiply such representatives of each equivalence class and figure out the corresponding representative of the product. But still unanswered is the crucial question: What are the representatives of the irreducible elements of Z[[x]] ? --Dan

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[math-fun] Fibs mod n
by Gary McGuire 28 Jun '05

28 Jun '05

Just for the record: In a previous post I gave the following link http://www.earlham.edu/~rodrimi/Periods%20of%20Fibonacci%20Sequences%20Mod%… to a page with information about Fibonacci numbers mod n. (No authorship on that page.) I now see exactly the same web page at Kevin Brown's mathpages: http://www.mathpages.com/home/kmath078.htm Gary McGuire

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[math-fun] Almost-Langford
by Eric Angelini 27 Jun '05

27 Jun '05

Hello, This is a Langford number : 231213 Between the two 1's there is 1 digit Between the two 2's there are 2 digits Between the two 3's there are 3 digits. This is an almost-Langford number : 62312136 Between the two 1's there is 1 digit Between the two 2's there are 2 digits Between the two 3's there are 3 digits Between the two 6's there are 6 digits The difference between a Langford and an almost- Langford number is that the first one shows all pairs of integers from 1 to n (here 1 to 3) -- which is not the case for the second one. What could be the biggest almost-Langford number? This number cannot have more than 20 digits as pairs of digits must be unique -- your theoretical alphabet to write this number is: 00112233445566778899 Best, É. (hope this is not old-hat)

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[math-fun] Totient Summing Tidbit
by wouter meeussen 26 Jun '05

26 Jun '05

given the nice entry in http://mathworld.wolfram.com/TotientSummatoryFunction.html isn't it cute to guess that Sum[Sum[EulerPhi[k],{k,1,n}],{n,1,w}] = Sum[EulerPhi[k]*(w+1-k),{k,1,w}] = (2*w^3+3*w^2+4*w)/(2*Pi^2) numerical test: Table[w=2^ww; {Sum[EulerPhi[k]*(w+1-k),{k,1,w}], Floor[(w*(4+w*(3+2*w)))/(2*Pi^2)]},{ww,16}] gives: { {3,1}, {13,9}, {75,63}, {497,457}, {3649,3482}, {27833,27196}, {217521,215001}, {1719835,1709899}, {13679291,13639043}, {109111251,108952364}, {871617967,870980213}, {6967839977,6965289383}, {55722310869,55712110761}, {445696933143,445656078836}, {3565248463365,3565085421581}, {28520683400617,28520030576058} } not too bad, no? W.

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[math-fun] an almost trivial telescopy
by R. William Gosper 26 Jun '05

26 Jun '05

(Spun off from the Sierpinski triangle-squarefree string business.) n /===\ | | [ f(k) g(k) ] | | [ ] = | | [ g(k) f(k) ] k = 1 n n /===\ /===\ [ 1 1 ] | | [ 1 - 1 ] | | [ ] | | (g(k) + f(k)) + [ ] | | (f(k) - g(k)) [ 1 1 ] | | [ - 1 1 ] | | k = 1 k = 1 ---------------------------------------------------------------. 2 ("Proof": Replace the lhs matrix with the prod from k to k of the rhs, then note how the crossterms vanish.) E.g., [ k ] [ 1 q ] [ inf inf ] inf [ -------- -------- ] [ ==== ==== ] /===\ [ 2 k 2 k ] [ \ n \ n ] | | [ 1 - q 1 - q ] [ > p (n) q > p (n) q ] | | [ ] = [ / even / odd ], | | [ k ] [ ==== ==== ] k = 1 [ q 1 ] [ n = 0 n = 0 ] [ -------- -------- ] [ ] [ 2 k 2 k ] [ vice versa ] [ 1 - q 1 - q ] where p_even(n) and p_odd(n) are the partitions of n into an even and odd number of parts: inf inf /===\ /===\ | | 1 | | 1 inf inf | | ------ + | | ------ ==== ==== 2 n | | k | | k \ n \ q k = 1 1 - q k = 1 1 + q > p (n) q = > --------- = ---------------------------, / even / (q; q) 2 ==== ==== 2 n n = 0 n = 0 2 3 4 5 6 7 8 9 = 1 + q + q + 3 q + 3 q + 6 q + 7 q + 12 q + 14 q + . . . (A027187) and inf inf /===\ /===\ | | 1 | | 1 inf inf | | ------ - | | ------ ==== ==== 2 n + 1 | | k | | k \ n \ q k = 1 1 - q k = 1 1 + q > p (n) q = > ------------- = ---------------------------. / odd / (q; q) 2 ==== ==== 2 n + 1 n = 0 n = 0 2 3 4 5 6 7 8 9 = q + q + 2 q + 2 q + 4 q + 5 q + 8 q + 10 q + 16 q + . . . (A027193). --rwg PS, RKG mentioned the open problem of packing the 1/n by 1/(n+1) rectangles, n>0, into the unit square. What's the greatest n that have been crammed?

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[math-fun] Re: Formal power series Z[[x]]
by dasimov＠earthlink.net 23 Jun '05

23 Jun '05

Especially to Neil: T.Y. Lam has confirmed that the ring Z[[x]] of formal power series in one variable over the integers is a unique factorization domain. I was told this is based on this paper by Pierre Samuel: On unique factorization domains. Illinois J. Math. 5 1961 1--17. MR 22#12121 George Bergman mentioned that there is also another paper by Samuel about UFD's in the Monthly: 75(1968)945-952, "which apparently includes discussion of power series". --Dan

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