August 2005 - mathfun@mailman.xmission.com

[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

3 2

[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

3 2

[math-fun] one-to-one relation (odd partitions) <-> (non-repeat partitions)
by wouter meeussen 22 Aug '05

22 Aug '05

It is well known that both are equal in count, but can anyone point me to a one-to-one transformation rule between both sets? example: {{7}, {6,1}, {5,2}, {4,3}, {4,2,1}} to & from {{7}, {5,1,1}, {3,3,1}, {3,1,1,1,1}, {1,1,1,1,1,1,1}} or is this a well known *dumb* question? W.

2 3

[math-fun] Re: Asymptotic for A058797
by Alec Mihailovs 20 Aug '05

20 Aug '05

"Franklin T. Adams-Watters" <franktaw(a)netscape.net> wrote: > The entry for A058797 states that it is asymptotic to c*n!, with c=.224... > I conjecture that c = BesselJ(0,2) = Sum (-1)^k/(k!)^2 = 0.223890779... > (A091681). I found a formula for A007754: a(n,k) = Pi*(BesselJ(n+k+1,2)*BesselY(k,2) - BesselY(n+k+1,2)*BesselJ(k,2)) In particular, for k=0, we get a formula for A058797, a(n) = Pi (BesselJ(n + 1, 2)*BesselY(0, 2) - BesselY(n + 1, 2) BesselJ(0, 2)) Now, BesselJ(n+1,2) ~ 1/(n+1)! and BesselY(n+1,2) ~ -n!/Pi That gives exactly that asymptotics, a(n) ~ BesselJ(0,2)*n! Similarly, the asymptotics for the k-th column of A007754 is a(n,k) ~ BesselJ(k,2)*(n+k)! (for fixed k and n -> infinity). Alec Mihailovs http://math.tntech.edu/alec/

1 0

[math-fun] orbit trivia
by R. William Gosper 19 Aug '05

19 Aug '05

Responding to a highschooler's Shuttle questions, some rough calculations turned up a) at roughly 49.2 degrees north and south lattitude, there are countours where g = 79000 mph/hr, exactly. b) although the equation of motion of an elliptical orbit requires higher transcendental functions, the shape, in Earth centered polar coordinates, is simply 2 r = ------------------------------, (k0 - k1) cos(theta) + k1 + k0 where k0 = 1/perigee and k1 = 1/apogee. (The hard part is theta(t), which apparently even transcends Lambert_W.) One question was what happens if you shoot a 1000mph bullet straight down from ISS orbit (225 mi). The line extrapolating the velocity vector sum comes nowhere near the Earth, so I was a little surprised to find the orbit grazing the surface about 1/4 the way around, recrossing the ISS orbit about halfway around, reaching an apogee of 484 mi at about the 3/4 point. So there ought to be a continuum of slightly lower muzzle velocities and slightly rearward aimpoints wherein you shoot yourself an orbit later. These unfortunate orbits also cross yours about halfway around, but "safely" ahead. Each provides an additional self-shooting solution--the vector difference between the shooter's orbital velocity and that of the first outbound crossing of the bullet, which will turn out to be slightly backward from straight "up" (away from the Earth.) This will cross once behind you and then hit you on the upswing after one full orbit. A couple of other shuttle trivia: c) the deorbit dv is only about 158 mph. d) reentry is a lot more like the old capsule/heat-shield scheme than I realized. The Shuttle comes in almost belly-first (rolling alternately almost + and - 90 degrees to avoid regaining altitude with the very high angle of attack), with the black undersurface acting as heat shield. (I always wondered why the windows didn't melt.) Similarly, the capsules had a deliberately eccentric center of mass so that the heat shield maintained a significant angle of attack. An alternative to the careening, serpentine descent would be multiple skips (with roll ~0), which might increase the number of orbit counts in reach of KSC, and reduce maximal tile heating. I'm guessing this is dangerously over-reliant on the rather feeble attitude thrusters, which in the current scheme are used once to get the attitude exactly right well in advance of atmospheric effects, after which most of the hypergolic fuel is dumped. The Earth turns about 22 degrees per Shuttle orbit, so they get a shot at KSC when the ground track passes just east of Florida, and another when it passes just west on the succeeding orbit. After the second waveoff Monday morning, NASA told the public that the next opportunity would be about 24 hr after the first one. However, the groundtracks passed Florida again after only 10 hrs or so, heading SE instead of NE. I asked some NASA people why nobody was considering this and got a rather defensive "I can't answer that." Could it be that they calculated that another reentry accident would only kill the program if the debris landed in the Midwest instead of the Caribbean? Maybe it was just a circadian or sonic boom decision. Lastly, the NASA PR types who showed the reentry live in the Ames visitor center theater also mentioned that the Mars rovers recently got life-extended when a friendly dust devil cleaned off their solar panels! --rwg PS, about 200 people (including children) stayed up both nights to watch. There were also scads of TV newscrews to tape the horror-stricken faces in case of an accident. PPS: The center has a moon rock, some meteoritic martian basalt, and what they claim is a piece of Vesta. How the he** do they know which asteroid it came from? Nobody taped the horror-stricken faces of the PR types I asked.

2 2

[math-fun] Re: Describe and add
by Eric Angelini 19 Aug '05

19 Aug '05

Thanks so much, Hugo I'll add this to the OEIS with both yr name and Perl pgm. Best, E. ------------------------ (...) > Here's some perl code: > perl -Mbigint -wle '$n=$a=1; while (1) { print "$n: $a"; ++$n; $a = desc($a) + $a } sub desc { my %a; ++$a{$_} for split //, shift; join "", map +($a{$_}, $_), sort { $a <=> $b } keys %a }' > Compute: > 1: 1 > 2: 12 > 3: 1124 > 4: 212338 > 5: 11434656 > 6: 2124676182 > 7: 215338937900 > 8: 2011338654109729 > 9: 20314234480170281558 > 10: 20317265802504533296 > 11: 50431498946030705115 > 12: 50834622289546876944 > 13: 50937835732083050773 > 14: 50937875857418523592 > 15: 51038998088873897421 > 16: 51241110320289174260 > 17: 81755323561805346079 > 18: 101966556706331618898 > 19: 101968597938887792727 > 20: 102070720070404367576 > [...] > 25000: 209208887438850939711053 > [...] > 1718184: 18417097587111721307162018 > > :Questions: > : > :- has this seq. a fixed point (or loop)? > :- what about other beginnings? > :(start with a(1) = 0 for instance) > > No: a(n+1) > a(n) is guaranteed by the definition. > > See references to "self-describing" in OEIS for some sequences defined > along similar lines. > > Hugo > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

1 0

[math-fun] Factorization of 1 + n + n^2 + . . . + n^k
by dasimov＠earthlink.net 17 Aug '05

17 Aug '05

My car's odometer recently passed 111,111 -- which out of curiosity I quickly factored as 3*7*11*13*37. Which got me wondering, What is known in general about the factorization of S(n,k) = 1 + n + n^2 + . . . + n^(k-1) ??? Of course S(n,k) = (n^k - 1) / (n - 1) = Prod r=1 to k-1 of (n - w^r), where w is a primitive kth root of unity. And conjugate pairs can be grouped together to get the product of just real factors: S(n,k) = fac(n,k) * Prod r=1 to [(k-1)/2] of (n^2 - 2 cos(2pi*r/k) + 1) where fac(n,k) = n+1 if k is even, = 1 if k is odd. But this doesn't seem to help much. Is this problem more solvable for n a prime? Are there asymptotics for the number of prime factors, their size, or their exponents? --Dan

2 1

[math-fun] A016142
by Richard Guy 17 Aug '05

17 Aug '05

This (also) appears to be 3^(n-1) * (3^n - 1)/2. VERY WILD surmise: It's the number of integer-sided Heron triangles whose circumdiameter is the product of n distinct primes of shape 4k + 1. (3^n - 1)/2 of these are (believed to be) right triangles, so that the number of non-right ones is (3^(n-1) - 1) * (3^n - 1)/2 A003462 and a new(?) sequence 0, 8, 104, 1040, 9680, ... R.

3 5

[math-fun] a hard sequence worthy of attention
by N. J. A. Sloane 16 Aug '05

16 Aug '05

There was some discussion on this list a while back about sequences that would make good projects. Here is an old one where nothing has been done for many years: %I A001289 %S A001289 1,2,3,8,48,150357 %N A001289 Number of equivalence classes of Boolean functions modulo linear functions. %C A001289 Number of equivalence classes of maps from GF(2)^n to GF(2), where maps f and g are equivalent iff there exists an invertible n X n binary matrix M, two n-dimensional binary vectors a and b, and a binary scalar c such that g(x) = f(Mx+a) + b.x + c. %D A001289 Berlekamp, Elwyn R. and Welch, Lloyd R., Weight distributions of the cosets of the (32,6) Reed-Muller code, IEEE Trans. Information Theory, IT-18 (1972), 203-207. %D A001289 R. J. Lechner, Harmonic Analysis of Switching Functions, in A. Mukhopadhyay, ed., Recent Developments in Switching Theory, Acad. Press, 1971, pp. 121-254, esp. p. 186. %D A001289 F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1977, p. 431. %D A001289 I. Strazdins, Universal affine classification of Boolean functions, Acta Applic. Math. 46 (1997), 147-167. %H A001289 <a href="http://www.research.att.com/~njas/sequences/Sindx_Bo.html#Boolean">Index entries for sequences related to Boolean functions</a> %Y A001289 Adjacent sequences: A001286 A001287 A001288 this_sequence A001290 A001291 A001292 %Y A001289 Sequence in context: A013208 A094370 A066084 this_sequence A103045 A041979 A001686 %K A001289 nonn,hard %O A001289 1,2 %A A001289 njas It is important in the study of Reed-Muller codes. NJAS

2 3

[math-fun] No. of inequivalent sudokus
by N. J. A. Sloane 13 Aug '05

13 Aug '05

I'm sending this to both the math fun and seq fan lists, becuse even though it is about a sequence, most of the sudoku discussions i've seen have been on the former list. The number of (completed) n^2 X n^2 sudokus is: %I A107739 %S A107739 1,1,288,6670903752021072936960 %N A107739 Number of (completed) sudokus of size n^2 X n^2. based on %H A107739 Bertram Felgenhauer and Frazer Jarvis, http://www.shef.ac.uk/~pm1afj/sudoku/ There are 6670903752021072936960 Sudoku grids Ed Russell and Frazer Jarvis also compute the number of inequivalent 9x9's, see "There are 5472730538 essentially different Sudoku grids" http://www.shef.ac.uk/~pm1afj/sudoku/sudgroup.html allowing (i quote) Relabelling entries; Reflection; Rotation; Permutation of blocks of columns 1-3, 4-6 and 7-9; Permutation of blocks of rows 1-3, 4-6 and 7-9; Permutation of columns 1-3; Permutation of rows 1-3; Permutation of columns 4-6; Permutation of rows 4-6; Permutation of columns 7-9; Permutation of rows 7-9. So there's a new sequence here: 0 1 2 3 1 1 x 5472730538 where x is some small number! My question is, what is x? Using analogues of the above transformations, how many inequivalent 4 X 4 grids are there? Hugo van der Sanden sent me an upper bound x <= 6, since any 4X4 grid is equivalent to one of: 1234 1234 1234 1234 1234 1234 3412 3412 3421 4312 4321 4321 2143 2341 2143 2143 2143 2413 4321 4123 4312 3421 3412 3142 but this can surely be reduced. So is x 1, 2 or 3? NJAS

2 1