February 2012 - mathfun@mailman.xmission.com

[math-fun] sets of integers with equal moments?
by Warren Smith 21 Feb '12

21 Feb '12

Your task is to find two disjoint sets (or multisets) {A1,A2,A3,...,Ak} and {B1,B2,B3,...,Bk} of integers with equal 0th, 1th, 2th,..., (k-1)th moments. k=1: {x} and {y} work, trivial. k=2: {0,0} and {-x,+x} have equal sums of Jth powers for J=0 and 1. k=4: {-5,0,0,+5} and {-4,-3,+3,+4} have equal sums of Jth powers for J=0,1,2,3. Also any pythagorean triple works instead of 3,4,5. k=6: ?I do not know of any example? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] sets of integers with equal moments?
by Warren Smith 21 Feb '12

21 Feb '12

an example I just found: {-241,-218,0,0,+218,+241}, {-266,-143,-120,+120,+143,+266} Are two multisets of 6 integers each with same sums of Jth powers for J=0,1,2,3,4,5. Note 0^4 + 218^4 + 241^4 = 5631933137 = 120^4 + 143^4 + 266^4 0^2 + 218^2 + 241^2 = 105605 = 120^2 + 143^2 + 266^2 and the sums of 1st and 3rd powers of course are 0 by odd symmetry. Prouhet–Tarry–Escott problem... thanks for that name... is the same problem except they demand sets not multisets ("multi"sets allow repeating an element): http://en.wikipedia.org/wiki/Prouhet–Tarry–Escott_problem

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[math-fun] N G. de Bruijn, 1918 - 2012
by Thane Plambeck 21 Feb '12

21 Feb '12

Apparently NG de Bruijn has died. The article I received isn't suitable for math-fun (it's a scan of a newspaper article). Maybe someone knows a good link. -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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Re: [math-fun] isosceles triangle dissections (Stuart Anderson)
by Stuart Anderson 21 Feb '12

21 Feb '12

> > ?Has anyone a picture of this dissection?>> > --rwg > > Nothing mysterious, I assumed (and made an ass of u and me) the > diagrams discussed in the text were all given in illustrations in the > following pages, but looks like he didnt give the 9 triangle > dissection. >> > I am advised (but cant verify); It's a little further along (page 258) in the answers section. ;)

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[math-fun] Next generation sorting algorithm
by quad 21 Feb '12

21 Feb '12

I am please to inform you, math-fun, that I have developed the next generation sorting algorithm. I call it QUOBOSORT. Let me describe it. Randomly permute a list X until the minimum is in the first position of X. When this occurs, QUOBOSORT the rest of X until we reach an empty list. Here is the more formal algorithm. ALGORITHM --------- QUOBOSORT(X) := -- INPUT: a list of integers X -- OUTPUT: sorted X Q0. Is X an empty list? Yes: Return X. No : Go to Q1. Q1. Find the minimum M of X. Q2. Randomly permute X. Q3. Is M in the first position of X? Yes: Concatenate M and QUOBOSORT(REST(X)) No : Go to Q2. After one sorting iteration, the algorithm speeds up about N times! I hope you all are impressed. Sincerely, Robert Smith Inventor of QUOBOSORT

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Re: [math-fun] (further) generalized Lambert series with Theta-convergence
by Bill Gosper 21 Feb '12

21 Feb '12

Aha! I just derived the Rogers-Fine identity as Sum[(t^k*QPochhammer[a, q, k])/QPochhammer[b, q, k], {k, 0, Infinity}] == Sum[(b^k*q^((-1 + k)*k)*t^k*(1 - a*q^(2*k)*t)*QPochhammer[a, q, k]* QPochhammer[(a*q*t)/b, q, k])/ (QPochhammer[b, q, k]*QPochhammer[t, q, 1 + k]), {k, 0, Infinity}] from a contour in the k,n plane, starting at i,j,k,n ={t,a,0,b}, with the unspecialized matrix system {{i, {{(q^(-1 + k + n)*(1 - q^(1 + i + j + k)))/(1 - q^(i + k)), (1 - q^(-1 + k + n))/(1 - q^(i + k))}, {0, 1}}}, {j, {{((1 - q^(1 + i + j + k))*(1 - q^(j + k + n)))/(1 - q^(1 + j)), -((q^(1 + j)*(1 - q^(-1 + k + n)))/(1 - q^(1 + j)))}, {0, 1}}}, {k, {{(q^(i + 2*k + n)*(1 - q^(1 + i + j + k))*(1 - q^(j + k + n)))/((1 - q^(i + k))*(1 - q^(k + n))), (1 - q^(i + j + 2*k + n))/(1 - q^(i + k))}, {0, 1}}}, {n, {{(q^(i + k)*(1 - q^(j + k + n)))/(1 - q^(k + n)), 1}, {0, 1}}}} I can generalize this out the wazoo, but mainly to matrix products which aren't sums. But I should have found this in AIM 304. --rwg n -> i + k j + k + n q (1 - q ) ----------------------- k + n 1 - q 1 0 1 k -> i + 2 k + n 1 + i + j + k j + k + n i + j + 2 k + n q (1 - q ) (1 - q ) 1 - q -------------------------------------------------- -------------------- i + k k + n i + k (1 - q ) (1 - q ) 1 - q 0 1 j -> 1 + i + j + k j + k + n 1 + j -1 + k + n (1 - q ) (1 - q ) q (1 - q ) ------------------------------------- -(------------------------) 1 + j 1 + j 1 - q 1 - q 0 1 i -> -1 + k + n 1 + i + j + k -1 + k + n q (1 - q ) 1 - q -------------------------------- --------------- i + k i + k 1 - q 1 - q 0 1 Although that product of Lucas numbers in my fib formula is just a q-pochhammmer, it provides theta-convergence because |q|>1. And since you'd sum both it and yours effectively via matrix product, it's more competitive than it looks. (E.g, you don't win using an addition chain on q^k^2 because you need (k-1)^2 etc. as well, so you just maintain q^(2k+1). * Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> [Feb 07. 2012 11:40]:> [jj:]> without splitting into even and odd part (yes, RWG, AIM304 exists).> > YOW, I'd forgotten that stunt! joerg>That one needed a small correction, see http://www.jjj.de/lambert-paper/ (Warning: very first iteration, just spend today over it; certainly contains typos (tell me if you spot one!)). What I referred to as "Fine's versatile relation"> Now this one turns out to be easily obtained by setting a := -b> in Fine's "versatile" relation (14-1) (p.15 in "Basic Hypergeometric> Series and Applications"). appears to be known as "Rogers-Fine identity", this term is used in my draft. Later I'll process your other mails (big THANKS btw.!) > [...]

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Re: [math-fun] isosceles triangle dissections
by Stuart Anderson 21 Feb '12

21 Feb '12

> > Message: 5 > Date: Mon, 20 Feb 2012 19:47:59 -0800 > From: Bill Gosper <billgosper(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] isosceles triangle dissections Has anyone a picture of this dissection?>> > --rwg > > SA>Martin Gardner gave the 9 triangle dissection in "The Last > Recreations" p253 (but Google books and Amazon wont show me that page) > Stuart Anderson > > [...] > This is mysterious. Neil just sent me some screenshots of his electronic > edition. P253 is a pure text description of a map-coloring game. Triangular > dissections are discussed on pp250 and 252, with illustrations of three > dissected squares filling p251, and no mention in the text of nonasecting the > 45-45-90 into acute isosceles. Is it likely Neil's edition isn't the latest? > --rwg Nothing mysterious, I assumed (and made an ass of u and me) the diagrams discussed in the text were all given in illustrations in the following pages, but looks like he didnt give the 9 triangle dissection. >

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[math-fun] Mma sum
by Bill Gosper 20 Feb '12

20 Feb '12

VE> How does one report bugs to Wolfram? In:=Sum[(-1)^k Binomial[m, k] k, {k, 0, m}] Out:=0 It gets the correct answer when m is given explicitly. I'm guessing it knows Sum[z^k Binomial[m, k] k, {k, 0, m}] and then chokes when it evaluates for z=-1. Veit As far as I'm concerned, the real bug is that it did anything at all before you typed FunctionExpand or the like. http://www.wolfram.com/support/contact/email/ should work. They used to have support.wolfram.com/submitabug.cgi . I can send you a copy of the web page and associated files if you want to try it. Speaking of Wolfram.com, http://reference.wolfram.com/mathematica/ref/QPochhammer.html (Neat examples) gives "Hirschhorn's modular identity ": In[673]:= Series[ QPochhammer[q, q]^5 - QPochhammer[q^5, q^5], {q, 0, 99}] Out[673]= SeriesData[q, 0, {-5, 5, 10, -15, -5, -5, 25, 15, -20, 10, -45, -5, 25, 20, 10, 15, 20, -50, -35, -30, 55, -50, 15, 80, 0, 50, -35, -45, -15, 5, -50, -25, -55, 85, 50, 50, 10, -40, 65, 10, -10, -115, 50, -115, -100, 85, 80, -30, 5, 20, 45, 70, 65, 45, -55, -100, -45, 10, -115, 110, -160, 55, -20, -70, 110, -20, 105, 185, 70, -130, -85, 65, 60, -50, 0, -130, 10, -55, -165, -110, 165, 15, 55, -30, 160, 45, -60, 90, -5, 10, 100, 165, -155, 60, -230, -145, -55, -45, 35}, 1, 100, 1] In fact this seems to hold for *all* primes: In[716]:= Table[Collect[Series[QPochhammer[q]^p - QPochhammer[q^p], {q, 0, 999}], q, Modulus -> p], {p, Prime@Range[99]}] Out[716]= {0, 0, 0, 0,..., 0} For p=5, the coefficients match for a peculiar (infinite?) power sequence: In[679]:=Flatten[ Position[CoefficientList[Normal[Series[QPochhammer[q, q]^5 - QPochhammer[q^5, q^5], {q, 0, 4076}]], q], 0]]-1 Out[679]= {0, 25, 75, 175, 350, 1560, 1802, 1838, 2318, 2690, 3174, 3742, 3925} QPochhammer[q, q]^3 = QPochhammer[q^3, q^3] *except* at q^binomial(n,2), n>1. QPochhammer[q, q]^2 matches QPochhammer[q^2, q^2] at q powers 0, 2, 7, 10, 11, 12, 17, 18, 21, 22, 24, 25, 32, 37, 39, 41, 42, 43, 44, 46, 47, 49, 52, 54, 57, 58, 60, 62, 65, 67, 68, 70, ... QPochhammer[q, q]^15 matches QPochhammer[q^15, q^15] at powers 0, 53, 482, 1340, ... QPochhammer[q, q]^n appears to match QPochhammer[q^n, q^n] only at q^0 for n in {7,9,11,13,17,...}! For n=4, the matches are 0, 9, 14, 19, 24, 31, 34, 39, 42, 44, 49, 53, 59, 64, 65, 69, 74, 75,... For n=6, 0, 5, 8, 14, 17, 19, 23, 26, 32, 33, 35, 40, 41, 44, 47, 50, 52, 53, 54 For n=8, 0, 3, 7, 11, 13, 15, 18, 19, 23, 27, 28, 29, 31, 35, 38, 39 Lotsa sequences here. --rwg

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Re: [math-fun] isosceles triangle dissections
by Bill Gosper 20 Feb '12

20 Feb '12

>> On Wed, Feb 15, 2012 at 4:11 PM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:>> > Neil has just uncovered fq.math.ca/Scanned/6-6/hoggatt.pdf>> > predating my http://mathworld.wolfram.com/SquareDissection.html>> > of a square into ten acute isosceles triangles.>> >>> > Neil also found in a Gardner book a reference to a Monthly paper>> > (June-July 1962, pp550-552) claiming that *any* obtuse triangle can>> > be cut into eight acute isosceles triangles, implying at most nine for>> > dissecting a right isosceles, in contradiction of my round(tan(69))>> > solutions assertion. Has anyone a picture of this dissection?>> > --rwg SA>Martin Gardner gave the 9 triangle dissection in "The Last Recreations" p253 (but Google books and Amazon wont show me that page) Stuart Anderson [...] This is mysterious. Neil just sent me some screenshots of his electronic edition. P253 is a pure text description of a map-coloring game. Triangular dissections are discussed on pp250 and 252, with illustrations of three dissected squares filling p251, and no mention in the text of nonasecting the 45-45-90 into acute isosceles. Is it likely Neil's edition isn't the latest? --rwg

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Re: [math-fun] Waring-type question
by Dan Asimov 20 Feb '12

20 Feb '12

If I'm understanding: What ensures that n^3 - (n-1)^3 won't require another (n-1)^3 in its sum of cubes? --Dan << Sure -- just let n^3 - (n-1)^3 be sufficiently large as well, no? --Michael On Feb 19, 2012 10:48 AM, "Dan Asimov" <dasimov(a)earthlink.net> wrote: In what follows, "cube" means the cube of a positive integer. It is known that every sufficiently large integer is the sum of distinct cubes. See < http://oeis.org/A001476 >. QUESTION: Is every sufficiently large integer the sum of *more than one* distinct cubes? --Dan >> ________________________________________________________________________________________ It goes without saying that .

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