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[math-fun] Re: Describe and add
by Eric Angelini 06 Sep '05

06 Sep '05

Hello to all, back from the Lido, [Hugo]: > Check Eric's original series of examples again: > this is not Conway's "look and say" type of des- > cription. My guess (as encoded in my perl) was > that the digits are sorted before describing, > so that the description of 212338 is 11222318. ... yes, indeed -- this is exactly it -- bravo Hugo, and apologizes for not beeing clear enough (happens to me far too often, I admit) Could I kindly ask to Hugo to express in good english and mathematical terms how A109229 works -- and add this comment to the OEIS? And, Michele, why wouldn't you submit your (equi- valently sound) sequence to the database? Best to both of you, É.

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[math-fun] more on bijections on partitions
by wouter meeussen 04 Sep '05

04 Sep '05

Odd-Strict partitions <--> Self Conjugate partitions OEIS : A000700 the Ferrer's plots are a dead give-away: start from a self-conjugate partition with its symmetric Ferrer's plot, peel the onion by taking away its first row and column, repeat on the rest until nothing left. The lengths of the peels (obviously different, obviously odd) are the corresponding odd-strict partition. Mma: peel[par:{__Integer}]:=-1+2*Map[First,Drop[FixedPointList[DeleteCases[Rest[#]-1,0|{}]&,par,SameTest- >( #2==={}&)],-1]] joinpeels[rinds:{__Integer}]:=Flatten[{#,Drop[TransposePartition[#],Length[#]]}]&[(1+rinds)/2+Range[ 0,Length[rinds]-1] ] Partitions in more than one part, you *got* to love them. W.

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Re: [math-fun] orbit problem -> segment thickness problem
by R. William Gosper 03 Sep '05

03 Sep '05

hgb>This is very cool! Do you have a reference to an article for amateurs? A couple of years ago, the Car Talk radio show and website featured the puzzle of how to quarter-fill a horizontal cylindrical tank, so a lot of math teachers may have been embarrassed already. Unfortunately, I didn't see their solution, and it's no longer obvious on their site. I also haven't seen the area problem phrased as a "parabolic" (epsilon = 1) case of Kepler's, but it's fairly obvious. An efficient area solution might be useful as the lead term of an expansion at epsilon = 1 of the high eccentricity Kepler case. Watson gives the solution in terms of Bessel functions as a standard Fourier expansion involving a slick integration by parts which turns Kepler's equation into one of the usual definite integral definitions of the Bessel function. I'll send it along if there's any interest. Twice I said >b) although the equation of motion of an elliptical orbit requires higher >transcendental functions,[...] (The hard part is theta(t), which apparently even transcends Lambert_W.) But just barely. t = time ~ M, the "mean anomaly", and the problem is to solve E - eps sin(E) = M for E, the "eccentric anomaly". In a (comfortable) neighborhood of M = 0, Kepler's eqn is (nonobviously) equivalent to | i M - eps cos(E) | |W(- e eps)| = eps. >Also, do the transcendental functions show up for "even" numbers of >dimensions? For dimensions 0..5, thickness x, the volume formula gives sqrt(x) 2 asin(-------) sqrt(2) v(0) = ---------------, v(1) = x, %pi sqrt(x) sqrt(2) asin(-------) (x - 2) (x - 1) sqrt(2) v(2) = - sqrt(2) (--------------- - ---------------------) sqrt(x), x sqrt(x) 2 sqrt(1 - -) 2 x 2 v(3) = %pi (1 - -) x , v(4) = sqrt(2) %pi sqrt(x) 3 sqrt(x) 2 3 sqrt(2) asin(-------) (x - 2) (x - 1) (2 x - 4 x - 3) sqrt(2) (-------------------------------- + -----------------------)/6, x sqrt(x) 2 sqrt(1 - -) 2 x 3 (1 - -) x 2 3 5 2 %pi x (1 - -----------) 4 v(5) = ---------------------------, 3 and the hypergeometric three-term relations give the recurrence v(n) = 2 %pi (2 %pi v(n - 4) (x - 2) x 2 - v(n - 2) ((n - 2) (x - 2 x - 1) - 1))/((n - 1) n), which proves transcendence iff n even. --rwg SELF-DEVOURING GIRDLE-OF-VENUS STREAMLINES RAIMENTLESS ETERNALISMS

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[math-fun] Inverse factorial?
by Michael Kleber 29 Aug '05

29 Aug '05

Suppose you place N things in N buckets. How many do expect the fullest bucket to contain? I know that the number of buckets containing k things, on average, is N/(k! e). So I can pretend to get the fullest bucket by setting this equal to 1 and solving for k: you expect around one bucket with "(N/e) un-factorial" items. (Surely this function has a name?) Any recommendations on a smarter approach? --Michael -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

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[math-fun] Re: Confused questions about finding bijections
by James Propp 29 Aug '05

29 Aug '05

>The partition bijection question originally asked by Wouter >left open the possibility that might be no "natural" bijection >between these two kinds of partitions. > >I guess the quintessential example is, given an arbitrary >vector space V (e.g., say V is a 1-dimensional v.s. over R), >there is no "natural" bijection between V and V* (= linear >maps from V to R). I can think of two examples from combinatorics in which one can demonstrate that there is no "natural" bijection between two sets of the same size. #1: The set of linear orderings of an n-element set S has n! elements, as does the set of invertible self-maps from S to itself. And one can put the two sets in bijection if one specifies a special linear ordering of the elements of S. But there is no "natural" bijection, if by a natural bijection one means a G-equivariant bijection, where G is the group of permutations of S. After all, there is a very special self-map from S to itself, namely the identity map, but there is no privileged linear ordering of a set S that comes with no extra structure. (Putting things in the language of G-equivariance: the G-set of self-maps has a fixed point under the action of G, while the G-set of linear orderings does not.) #2: If S is an n-element set with n>0 even, S has 2^{n-1} subsets of even cardinality and it also has 2^{n-1} subsets of odd cardinality. And one can put the two collections of subsets in bijection (indeed, if one views the power set of S as an n-dimensional vector space over the field with 2 elements, then a linear map will do). But there is again no G-equivariant bijection. After all, there is a very special even-cardinality-subset of S, namely S itself (or, if you prefer, the empty set), but there is no privileged odd-cardinality-subset of S. If you like thinking about things this way, you might enjoy the article "Producing New Bijections from Old" by David Feldman and myself (published in Advances in Mathematics, volume 113 (1995), pages 1-44), which unfortunately was written just before I started putting all my work on the web. Jim

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[math-fun] two bijections odd <-> strict partitions
by wouter meeussen 27 Aug '05

27 Aug '05

upto now, I collected two (odd - strict) bijections. The first, thanks to Neil (njas) & Bill (rwg) for explaining the (Hardy and Wright) -way: odd-> strict : keep pairwise adding repeated integers until no more repeats, strict -> odd: keep splitting evens in two repeating halfs until no more evens. These are cute Mathematica one-liners: odd2strict[li:{__Integer}]:= Reverse[Sort[li //. {a___,b_,b_,c___} :> Sort[{a,2b,c}] ]] strict2odd[li:{__Integer}]:= Reverse[Sort[li //. q_?EvenQ -> Sequence[q/2,q/2] ]] The second, thanks to Rich (rkg) for explaining the Franklin (?)-way. This one is more tricky, and gives not-so-nice Mma-functions: rkg2strict[par:{__Integer}] := Block[{rr = Length[par], cc = Max[par], t1, t2}, t1 = Table[If[c<cc/2, Min[2*r,cc+1-2*c],0] + If[c>=cc/2,Min[2*r-1, 2*c-cc],0], {r,rr}, {c,cc}]; t2 = Table[Take[t1[[k]], {cc/2- par[[k]]/2 +1, cc/2+ par[[k]]/2}], {k, rr}]; Length /@ Split[Sort[Flatten[t2]]] ] rkg2odd[par:{__Integer}] := Block[{zero, one, two, three, four, five, argu, p1, p2}, zero = par - Reverse[Range[Length[par]]]; one = zero - Append[Rest[zero], 0]; two = Partition[one, 2, 2, {1, 1}, 0]; three = Reverse /@ Transpose[two]; width = 1 + 2*Floor[Length[par]/2]; hight = Floor[Length[par]/2 + 1/2]; four= Rest[FoldList[Plus, 0, Last[three]]]; five= Rest[FoldList[Plus, 0, First[three]]]; p1 = width + 2*Reverse[four]; argu = hight + five; p2 = (-1 + 2*Count[argu, q_/; q >= #1]&)/@ Range[Length[p1]+1, Last[argu]]; Join[p1, p2] ] the Pretty Bit in all this of course graphical: take the strict 24,18,13,9,2,1 as test, it has 6 parts, so 6,5,4,3,2,1 is its 'minor'. Plot the (not quite Ferrer of the) minor as one 6, two 5's, three 4's, ... 2 2 2 1 1 1 1 4 4 2 1 3 3 3 6 4 2 1 3 5 5 where . . 2 1 . . . . 4 . . 3 . . 6 . . . . 5 . is the frame with two diagonals, and the empty spaces are filled from the left when right of the right diagonal, and from the right when left of the left diagonal, and from above otherwise. Now complete the 'minor' for the 'excess' of the test: excess :: 24-6=18 extra 1's, 18-5=13 extra 2's, 13-4= 9 extra 3's, 9-3=6 extra 4's, 2-2=0 extra 5's and also no extra 6's. 2 2 2 2 2 2 2 2 2 2 x x x x x x x 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 x x x x x x x 3 3 3 3 3 3 x x x x x x x 2 1 3 2 1 3 2 1 3 1 1 1 1 1 by first completing the 6 extra 4's to the left (where the evens are in the minor), then completing the 9 extra 3's, from right to left (keeping symmetry) and extending below whatever sticks out; This bit sticking out determines where the 13 extra 2's go, upward, and the 10 stragglers peeping out to the top left ... Finally, simply read off the lengths of the rows: 27,19,7,3,3,3,1,1,1,1,1 Wow, this was fun! Any more such bijections? W.

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[math-fun] Casey's sequel to Euclid
by Richard Guy 25 Aug '05

25 Aug '05

If anyone has access to Casey, would they be willing to give a brief account of the extent he mentions the Malfatti problem (3 circles, mutually touching and touchingthe sides of a triangle in pairs)? [include edition #, date, & pages -- in the 5th edition, it may be on pp.154 & 155.] Better still, would anyone like to donate a copy of any edition to the Strens Collection? Or I will pay a reasonable price. R.

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[math-fun] Max dimension of a lattice
by N. J. A. Sloane 23 Aug '05

23 Aug '05

In his 1992 book "Combinatorics and Partially Ordered Sets", Tom Trotter mentions a 1984 question from Sands, What is the max dimension of a lattice having n elements? I won't repeat the definition since it is complicated. But does anyone know if the first few terms have been computed? He gives references to papers by Ganter et al., Furedi and Kahn, Sali, so there is a fair amount of literature. NJAS PS A week or two ago I asked if someone could compute the next term of %S A001289 1,2,3,8,48,150357 %N A001289 Number of equivalence classes of Boolean functions modulo linear functions Since then I have heard that a(7) was computed 10 years ago, so the sequence now reads: %S A001289 1,2,3,8,48,150357,63379147320777408548 ... %D A001289 Xiang-Dong Hou, AGL(m,2) acting on R(r,m)/R(s,m), J. Algebra, 171 (1995), 921-938. ...

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[math-fun] re one-to-one relation (odd partitions) <-> (non-repeat partitions)
by N. J. A. Sloane 23 Aug '05

23 Aug '05

Answer: (from Hardy and Wright) given n = l1 * 1 + l2 * 3 + l3 * 5 + l7 * 7 + ..., a ptn into odd parts, write each li in binary, e.g. li = 2^a1 + 2^a2 + 2^a3 + ... where the aj's are all different then expand n = (2^a1 * 1 + ...)*1 + ... by removing the brackets and we get a partition into distinct parts NJAS

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[math-fun] Confused questions about finding bijections
by dasimov＠earthlink.net 22 Aug '05

22 Aug '05

The partition bijection question originally asked by Wouter left open the possibility that might be no "natural" bijection between these two kinds of partitions. I guess the quintessential example is, given an arbitrary vector space V (e.g., say V is a 1-dimensional v.s. over R), there is no "natural" bijection between V and V* (= linear maps from V to R). Although this must be true, I've never seen it expressed as a theorem. Can it be? E.g., "Every vector space has a basis" is easily equivalent to the Axiom of Choice via Zorn's Lemma. But is it accurate to say that without AC, there is no bijection from V to V* ? (I suspect one can prove that w/o AC, there is no "natural transformation from the category of all vector spaces to the category of all dual vector spaces" -- or something like that. But what about merely one bijection from V to V* where V is a 1-dim v.s. over R ?) AND, are there countable examples of cases like the two kinds of partitions that don't -- in some sense -- have a natural bijection (more accurately a family of bijections, one for each N) ??? (Imitating the v.s. case, I suppose it's fair to say that w/o AC there's no bijection between G and Hom(G, H), where G is a gp. isomorphic to Z+Z and H is a gp. isomorphic to Z. True? And what about Hom(H,H) ?) --Dan

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