November 2013 - mathfun@mailman.xmission.com

Re: [math-fun] The Case Against Algebra II
by Dave Dyer 15 Nov '13

15 Nov '13

The only plausible explanations for inflicting advanced algebra on unmotivated students are variations on "learning to think". Every tired old bit of curriculum that has died, did so screaming the same thing. As applied, this is pretty much like arguing that beatings are good because they toughen you up.

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[math-fun] multi-D hilbert curve
by Warren D Smith 13 Nov '13

13 Nov '13

> 3D (e.g.) Hilbert is easy with the finite state machine. Three bits of t > in, > one bit of x, y, and z out, plus state change. > --rwg --well... actually figure out the state machine for arbitrary D, and do it in a way that we do not need no stinking huge 2^D entry table to store the state-rules, and then I'm sold. There's some disgusting computer programs out there I found with google to do multi-D hilbert curves, but there really ought to be a nice way that takes under 1 page of code. And they haven't found it.

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[math-fun] Peano-Smith space-filling curve in D dimensions
by Warren D Smith 13 Nov '13

13 Nov '13

The "Peano-Smith" curve PCS(t) in D dimensions is defined thus. Let the real number t have representation in "balanced ternary" (i.e. radix 3 with digit set {-1, 0, +1}) beginning with A as its most-significant trit, which is in the 3^k place for some integer k. Define ROT(x), where x is a D-dimensional vector, as the rotation-permutation of x, for example with D=3 we have ROT(x,y,z) = (y,z,x). PSC(t) := { compute A and k from t; s = t - A * 3^k; //chop off most-signif trit of t if(A==0){ s = -s; } v = ROT(PSC(s)); //recursive call; v[0,1,...,D-1] is D-dimensional real vector if(A==0){ v[0] = -v[0]; } v[0] = v[0] + A * Q * 3^(k/D); return v; } where Q is a scaling constant (I do not care what Q is, provided nonzero). It seems to me it is possible to implement this in O(1) arithmetic ops per trit via an un-recursivization of this code, assuming 3^(1/D) is pre-computed. (You avoid ever doing a ROT by instead walking indices mod D and keeping track of the sign change for each coordinate.) The "Peano-Smith curve" is based on the fact that a hyper-rectangle with sides 3^(k/D) for the kth side (k=1,2,3,...,D) can be divided, by trisecting the longest side, into 3 equally-scaled copies of the original shape. If D>=3 there is no simplex which can be divided, by bisecting longest edge, into 2 equally-scaled copies (perhaps mirrored) of original shape; but if D=2 such a simplex exists, namely 45-45-90 triangle. This leads to the Sierpinski curve in the plane, which has no higher dimensional analogue. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Kontsevich periodicity conjecture
by Mike Stay 13 Nov '13

13 Nov '13

The conjecture was recently proved: http://golem.ph.utexas.edu/category/2013/10/who_ordered_that.html Start with a 3×3 matrix. Take its transpose, then take the reciprocal of each entry, then take the inverse of the whole matrix. Take the transpose of that, then take the reciprocal of each entry, then take the matrix inverse. Take the transpose of that, then take the reciprocal of each entry, and then, finally, take the matrix inverse. Theorem: Up to a bit of messing about, you’re back where you started. Does anyone know what motivated taking the element-wise inverse? I ask because it's the only nonlinear operation in AES. -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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[math-fun] Mike Stay's goofy identity
by Warren D Smith 13 Nov '13

13 Nov '13

> --I have no idea what motivated this, but it is trivial to prove if by > "prove" we > mean, you are willing to accept probabilistic proof. I.e. using the > Schwartz-Zippel lemma. > Andy Latto: I don't think so, because of the "up to a bit of messing around". In the commutative case, the "up to a bit of messing around" is "up to multiplication by some diagonal matrix. If the entries in this diagonal matrix were themselves polynomial or rational in the original 9 variables, then the Schwartz-Zippel lemma would apply, and the theorem would also be provable by any decent computer algebra system. But Schwartz-Zippel only applies to "are these two polynomials identical?", not to "are these polynomials related by a constant multiple?" WDS: the diagonals indeed have entries that are rational fns, and the MAXIMA proof below extracted from the web page Stay pointed to, proves this: M: matrix([a,b,c],[d,e,f],[g,h,i]); M1: invert(transpose(1/M)); M2: invert(transpose(1/M1)); M3: invert(transpose(1/M2)); D1: M[1,1]/M3[1,1]; D2: M[2,1]/M3[2,1]; D3: M[3,1]/M3[3,1]; D: matrix([D1,0,0],[0,D2,0],[0,0,D3]); DM3: D . M3; E1: 1; /* M[1,1] = DM3[1,1] by choice of D1 */ E2: M[1,2]/DM3[1,2]; E3: M[1,3]/DM3[1,3]; E: matrix([E1,0,0],[0,E2,0],[0,0,E3]); DM3E: DM3 . E; ratsimp(DM3E); at which point it outputs the original matrix to prove the theorem. I am just saying, Schwartz-ZIppel lemma would have done this way faster to get a probabilistic proof. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] good deterministic algorithmic point sets > without repeated distances?
by Bill Gosper 13 Nov '13

13 Nov '13

For "self avoiding frequency" I should have said "self-avoiding frequency and phase". I.e. Hilbert[f+wt] repeats no value as t steps through {0,1/n,2/n,...}. If you self contact using the golden ratio, I want to hear about it. 3D (e.g.) Hilbert is easy with the finite state machine. Three bits of t in, one bit of x, y, and z out, plus state change. --rwg On 2013-11-13 10:41, Warren D Smith wrote: >>rwg: > If that's bad, you can choose a beating frequency close to a self-avoiding > frequency: gosper.org/hilbeat.png > > --well, I'm not sure what you mean by "self avoiding frequency" but an > obvious candidate > stepsize is the golden ratio as "the most irrational number" (and friends). > > I was not complaining your construction yielding some arbitrarily > small distances > was "bad"; I was merely "interested" to know whether its distribution of > nearest neighbor distances, differed from that of random points. (A > question which > likely could be answered computationally.) > >>rwg: > Here (again) is the exact Hilbert fcn > > Clear[Hilbert]; > Hilbert[t_, a1_: 1, a0_: 0] := > Hilbert[t, b1_: 1, > b0_: 0] = (Hilbert[t, s1_: 1, s0_: 0] = ((a0 - s0)/(s1 - a1)); > Module[{t4 = 4*t, n}, n = Floor[t4]; t4 -= n; Switch[n, > 0, I*(1 - Hilbert[1 - t4, -I*a1/2, a0 + a1*I/2])/2, > 1, (I + Hilbert[t4, a1/2, a0 + a1*I/2])/2, > 2, (1 + I + Hilbert[t4, a1/2, a0 + a1*(1 + I)/2])/2, > 3, 1 + I*Hilbert[1 - t4, a1*I/2, a0 + a1]/2, > 4, 1]]) > > --elegant! > Generalizing to D dimensions using gray code makes my head hurt, but > it is pretty easy to go to D dimensions if you instead of Hilbert, employ > Peano space filling curve idea. This is a curve from 0000..000 to 1111...1111 > in the hypercube [0,1]^D. > Slice your hyper-rectangle into 3 > equal-width slabs splitting coordinate 1. > Inside each slab, employ Peano with the D coordinates > rotated 1 slot, frobulate, and then the first coordinate is scaled by 1/3. > Here "frobulate" is to do nothing for the outer two slabs, but for the > middle slab it > is to apply the map x --> 1-x to each coordinate other than the first. > > Helps if you own a ternary computer. >

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[math-fun] Mike Stay's goofy identity
by Warren D Smith 13 Nov '13

13 Nov '13

> The conjecture was recently proved: > http://golem.ph.utexas.edu/category/2013/10/who_ordered_that.html > > Start with a 3?3 matrix. > Take its transpose, then take the reciprocal of each entry, then take > the inverse of the whole matrix. > Take the transpose of that, then take the reciprocal of each entry, > then take the matrix inverse. > Take the transpose of that, then take the reciprocal of each entry, > and then, finally, take the matrix inverse. > Theorem: Up to a bit of messing about, you?re back where you started. --I have no idea what motivated this, but it is trivial to prove if by "prove" we mean, you are willing to accept probabilistic proof. I.e. using the Schwartz-Zippel lemma.

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Re: [math-fun] good deterministic algorithmic point sets > without repeated distances?
by Warren D Smith 13 Nov '13

13 Nov '13

>rwg: If that's bad, you can choose a beating frequency close to a self-avoiding frequency: gosper.org/hilbeat.png --well, I'm not sure what you mean by "self avoiding frequency" but an obvious candidate stepsize is the golden ratio as "the most irrational number" (and friends). I was not complaining your construction yielding some arbitrarily small distances was "bad"; I was merely "interested" to know whether its distribution of nearest neighbor distances, differed from that of random points. (A question which likely could be answered computationally.) >rwg: Here (again) is the exact Hilbert fcn Clear[Hilbert]; Hilbert[t_, a1_: 1, a0_: 0] := Hilbert[t, b1_: 1, b0_: 0] = (Hilbert[t, s1_: 1, s0_: 0] = ((a0 - s0)/(s1 - a1)); Module[{t4 = 4*t, n}, n = Floor[t4]; t4 -= n; Switch[n, 0, I*(1 - Hilbert[1 - t4, -I*a1/2, a0 + a1*I/2])/2, 1, (I + Hilbert[t4, a1/2, a0 + a1*I/2])/2, 2, (1 + I + Hilbert[t4, a1/2, a0 + a1*(1 + I)/2])/2, 3, 1 + I*Hilbert[1 - t4, a1*I/2, a0 + a1]/2, 4, 1]]) --elegant! Generalizing to D dimensions using gray code makes my head hurt, but it is pretty easy to go to D dimensions if you instead of Hilbert, employ Peano space filling curve idea. This is a curve from 0000..000 to 1111...1111 in the hypercube [0,1]^D. Slice your hyper-rectangle into 3 equal-width slabs splitting coordinate 1. Inside each slab, employ Peano with the D coordinates rotated 1 slot, frobulate, and then the first coordinate is scaled by 1/3. Here "frobulate" is to do nothing for the outer two slabs, but for the middle slab it is to apply the map x --> 1-x to each coordinate other than the first. Helps if you own a ternary computer.

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[math-fun] good deterministic algorithmic point sets without repeated distances?
by Bill Gosper 13 Nov '13

13 Nov '13

How about Hilbert's squarefill evaluated with a fixed, irrational step size?gosper.org/hilbrand.png --rwg --I like that. I had the idea (did I post it?) another time of integrating high-dimensional functions by regarding them as 1D integrals along a spacefilling curve... which hopefully would work better than plain monte-carlo integration. One could also "adapt" rather easily using that approach. Is the nearest-neighbor distance-distribution different for this kind of point set, than for random points? (I presume yes, but can see that arbitrarily small distances still do occur.) --wds If that's bad, you can choose a beating frequency close to a self-avoiding frequency: gosper.org/hilbeat.png Here (again) is the exact Hilbert fcn Clear[Hilbert]; Hilbert[t_, a1_: 1, a0_: 0] := Hilbert[t, b1_: 1, b0_: 0] = (Hilbert[t, s1_: 1, s0_: 0] = ((a0 - s0)/(s1 - a1)); Module[{t4 = 4*t, n}, n = Floor[t4]; t4 -= n; Switch[n, 0, I*(1 - Hilbert[1 - t4, -I*a1/2, a0 + a1*I/2])/2, 1, (I + Hilbert[t4, a1/2, a0 + a1*I/2])/2, 2, (1 + I + Hilbert[t4, a1/2, a0 + a1*(1 + I)/2])/2, 3, 1 + I*Hilbert[1 - t4, a1*I/2, a0 + a1]/2, 4, 1]]) which is way too slow. You probably want the finite state machine version using dyadic rational approximations. --rwg As expected, that hourlong PrimeQ of SylvesterSequence[19] came up False: In[25]:= Timing[PrimeQ[Nest[#^2 - # + 1 &, 2, 19]]] Out[25]= {2737.828416, False}

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[math-fun] good deterministic algorithmic point sets without repeated distances?
by Warren D Smith 13 Nov '13

13 Nov '13

How about Hilbert's squarefill evaluated with a fixed, irrational step size? gosper.org/hilbrand.png --rwg --I like that. I had the idea (did I post it?) another time of integrating high-dimensional functions by regarding them as 1D integrals along a spacefilling curve... which hopefully would work better than plain monte-carlo integration. One could also "adapt" rather easily using that approach. Is the nearest-neighbor distance-distribution different for this kind of point set, than for random points? (I presume yes, but can see that arbitrarily small distances still do occur.) On 11/13/13, math-fun-request(a)mailman.xmission.com <math-fun-request(a)mailman.xmission.com> wrote: > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > Today's Topics: > > 1. Re: math-fun Digest, Vol 129, Issue 12 (Warren D Smith) > 2. Re: density of primorial+1 primes (Bill Gosper) > 3. Re: density of primorial+1 primes (rcs(a)xmission.com) > 4. Re: density of primorial+1 primes (Bill Gosper) > 5. good deterministic algorithmic point sets without repeated > distances? (Warren D Smith) > 6. Re: good deterministic algorithmic point sets without > repeated distances? (meekerdb) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Mon, 11 Nov 2013 22:09:36 -0500 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] math-fun Digest, Vol 129, Issue 12 > Message-ID: > <CAAJP7Y3LB8Rb=hFMqwdqi8myvLzwm+kPzn1mDrMr8u+bCPPqrw(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > >>> I'm curious: Can one state in closed form what is the asymptotic density >>> of prime numbers among numbers Q(n) of the form >>> >>> Q(n) := primorial(n) + 1 >>> >>> i.e., >>> >>> Q(n) = 2*3*5*7*...*p_n + 1 >>> >>> (where p_n is the nth prime) ??? > > --I would think that Q(n) is prime with "probability" about > loglog(Q(n)) times greater > than the usual 1/log(Q(n)) probability. > > > > ------------------------------ > > Message: 2 > Date: Tue, 12 Nov 2013 11:56:48 -0800 > From: Bill Gosper <billgosper(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] density of primorial+1 primes > Message-ID: > <CAA-4O0FvpLthgQeVHFkWsB86iHYrsMMRhn_pfLZoogR5E4uAhw(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > DanA> > > I'm curious: Can one state in closed form what is the asymptotic density > of prime numbers among numbers Q(n) of the form > > Q(n) := primorial(n) + 1 > > i.e., > > Q(n) = 2*3*5*7*...*p_n + 1 > > (where p_n is the nth prime) ??? > > CG>It's not even known that there are infinitely many. Caldwell & Gallot > conjecture that there are about e^gamma * log x prime Q(n) with n <= x, > that is, that the usual heuristic gives the right answer. > > I think sieve theory can give an upper bound if that's enough for your > purposes. > > Charles Greathouse > > WDS>--I would think that Q(n) is prime with "probability" about > loglog(Q(n)) times greater > than the usual 1/log(Q(n)) probability. > [Warren] > > Only slightly less infeasible might be the coprime sequence A000058 > (interesting comments). > In[10]:= NestList[#^2 - # + 1 &, 2, 9] > > Out[10]= {2, 3, 7, 43, 1807, 3263443, 10650056950807, > 113423713055421844361000443, > 12864938683278671740537145998360961546653259485195807, > 165506647324519964198468195444439180017513152706377497841851388766535868639572406808911988131737645185443} > > In[8]:= N[Timing[Nest[#^2 - # + 1 &, 2, 21]]] > > Out[8]= {0.007302, 1.853733657356281*10^426880} > > (Half a million digits in 7ms?) > > In[6]:= TimeConstrained[PrimeQ[#], 1] & /@ > NestList[#^2 - # + 1 &, 2, 21] > > Out[6]= {True, True, True, True, False, True, False, False, False, > False, False, False, False, False, False, False, False, False, > $Aborted, $Aborted, False, $Aborted} > > where $Aborted almost certainly means True. Changing the timeout > from 1 to 288 sec did zilch. Decertifying a_20 > In[9]:= Timing[PrimeQ[Nest[#^2 - # + 1 &, 2, 20]]] > > Out[9]= {0.150525, False} > > took < 1/6 sec. Can we show there are no (strong?) pseudoprimes of this > form? > > --rwg > > SUPERIMPOSED > > PSEUDOPRIMES > > > ------------------------------ > > Message: 3 > Date: Tue, 12 Nov 2013 13:17:42 -0700 > From: rcs(a)xmission.com > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] density of primorial+1 primes > Message-ID: <20131112131742.s0v082qagck4s4wg(a)webmail.xmission.com> > Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes"; > format="flowed" > > I'll guess that the False value for a_20 is from some small factor, > and that the Aborted values are because the test ran out of time > before completing even one superimposed test. Try asking for a > few digits of 2^(a_12) (mod a_12), and then see how far you can > get with a_13 ... before running out of gas. > > Rich > > --------- > Quoting Bill Gosper <billgosper(a)gmail.com>: > <eliding earlier comments --rich> >> Only slightly less infeasible might be the coprime sequence A000058 >> (interesting comments). >> In[10]:= NestList[#^2 - # + 1 &, 2, 9] >> >> Out[10]= {2, 3, 7, 43, 1807, 3263443, 10650056950807, >> 113423713055421844361000443, >> 12864938683278671740537145998360961546653259485195807, >> 165506647324519964198468195444439180017513152706377497841851388766535868639572406808911988131737645185443} >> >> In[8]:= N[Timing[Nest[#^2 - # + 1 &, 2, 21]]] >> >> Out[8]= {0.007302, 1.853733657356281*10^426880} >> >> (Half a million digits in 7ms?) >> >> In[6]:= TimeConstrained[PrimeQ[#], 1] & /@ >> NestList[#^2 - # + 1 &, 2, 21] >> >> Out[6]= {True, True, True, True, False, True, False, False, False, >> False, False, False, False, False, False, False, False, False, >> $Aborted, $Aborted, False, $Aborted} >> >> where $Aborted almost certainly means True. Changing the timeout >> from 1 to 288 sec did zilch. Decertifying a_20 >> In[9]:= Timing[PrimeQ[Nest[#^2 - # + 1 &, 2, 20]]] >> >> Out[9]= {0.150525, False} >> >> took < 1/6 sec. Can we show there are no (strong?) pseudoprimes of >> this form? >> >> --rwg >> >> SUPERIMPOSED >> >> PSEUDOPRIMES >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > > > > ------------------------------ > > Message: 4 > Date: Tue, 12 Nov 2013 16:21:29 -0800 > From: Bill Gosper <billgosper(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] density of primorial+1 primes > Message-ID: > <CAA-4O0H1FR1Ubvj8Fn=tQcs3faNfq9OEB3r57YbfBens7VRzbw(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > RCS> > > I'll guess that the False value for a_20 is from some small factor, > and that the Aborted values are because the test ran out of time > before completing even one superimposed test. Try asking for a > few digits of 2^(a_12) (mod a_12), and then see how far you can > get with a_13 ... before running out of gas. > > Rich > > --------- > > I lose; you win: > In[11]:= PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 12] > > Out[11]= 3366937412175501591717367777101460820640224690509113100163277\ > 6539067634240985481258893779792671452248127456725155260727280357834657\ > 6076738692420210769091266874710826881662611178920169012031118635505607\ > 4310982786125119105965083788824019629431580952133240830246962982908185\ > 9746496798634116562354568928408751636377986351524509214103104546223261\ > 2396128362132130189597314029272968730759961945103880790832377276137373\ > 9360515945586462465327970571098621823207535551580401784234812938313843\ > 6459926693612089970288170337138209267562435148320741084111322614415098\ > 1287168301351535936792810718614137830956322091599451719533327897728719\ > 2022822793035110202282332960887794400493112644185559756058506634629406\ > 8091290492788783650041163770989366969890285626688901808066658454549560\ > 7443533211603079106672889774858913012113951640476740718704380411099010\ > 204 > > In[13]:= PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 6] > > Out[13]= 4457343787994 > > (Note: a_0:=2 .) > > In[14]:= PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 5] > > Out[14]= 2 > > In[16]:= PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 4] > > Out[16]= 1103 > > In[17]:= PowerMod[3, #, #] &@Nest[#^2 - # + 1 &, 2, 5] > > Out[17]= 3 > > In[18]:= PowerMod[8, #, #] &@Nest[#^2 - # + 1 &, 2, 5] > > Out[18]= 8 > > In[12]:= Timing[N[PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 13]]] > > Out[12]= {0.105467, 2.415419924381151*10^1667} > > In[19]:= Timing[N[PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 14]]] > > Out[19]= {0.647739, 7.344539857481122*10^3334} > > In[20]:= Timing[N[PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 15]]] > > Out[20]= {3.835341, 6.597333832436021*10^6669} > > In[21]:= Timing[N[PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 16]]] > > Out[21]= {21.430643, 1.412908137918137*10^13339} > > In[22]:= Timing[N[PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 17]]] > > Out[22]= {119.182441, 8.856579016766857*10^26679} > > In[23]:= Timing[N[PowerMod[2, #, #] &@Nest[#^2 - # + 1 &, 2, 18]]] > > Out[23]= {576.777472, 7.425006686698617*10^53359} > > In[24]:= Timing[PrimeQ[Nest[#^2 - # + 1 &, 2, 18]]] > > Out[24]= {577.039477, False} > > In[25]:= Timing[PrimeQ[Nest[#^2 - # + 1 &, 2, 19]]] > should take about an hour. Meanwhile, we can now conjecture, as with > Fermat #s, that there are no more of these primes! > --rwg > Quoting Bill Gosper <billgosper(a)gmail.com>: <eliding earlier comments > --rich> > > Only slightly less infeasible might be the coprime sequence A000058 > (interesting comments). In[10]:= NestList[#^2 - # + 1 &, 2, 9] Out[10]= {2, > 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443, > 12864938683278671740537145998360961546653259485195807, > 165506647324519964198468195444439180017513152706377497841851388766535868639572406808911988131737645185443} > In[8]:= N[Timing[Nest[#^2 - # + 1 &, 2, 21]]] Out[8]= {0.007302, > 1.853733657356281*10^426880} (Half a million digits in 7ms?) In[6]:= > TimeConstrained[PrimeQ[#], 1] & /@ NestList[#^2 - # + 1 &, 2, 21] Out[6]= > {True, True, True, True, False, True, False, False, False, False, False, > False, False, False, False, False, False, False, $Aborted, $Aborted, False, > $Aborted} where $Aborted almost certainly means True. Changing the timeout > from 1 to 288 sec did zilch. Decertifying a_20 In[9]:= > Timing[PrimeQ[Nest[#^2 - # + 1 &, 2, 20]]] Out[9]= {0.150525, False} took < > 1/6 sec. Can we show there are no (strong?) pseudoprimes of this form? > --rwg > > SUPERIMPOSED PSEUDOPRIMES > <disformatting earlier comments --Firefox> > > > ------------------------------ > > Message: 5 > Date: Wed, 13 Nov 2013 00:16:49 -0500 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] good deterministic algorithmic point sets without > repeated distances? > Message-ID: > <CAAJP7Y2mMwA54-adLHnnUs-R8z6Ju4octFC8_XjxpaSHWyuVfg(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > In the plane (or some other low dimensional space) I want a good > infinite point set that has "density 1." > > What does "good" mean? Well, "more uniform than random" in some vague > sense, such as > sphere packing, sphere covering, and providing good estimates of > integrals via sums. > > BUT, what I also want, is the set has no repeated distances. So > lattices are undesirable. > Random point sets will not repeat a distance (with probability 1) so > they are great, except they fail to be good in the sense of the > previous paragraph, while lattices can be quite good in that sense. > > Another thing I want is an easy, algorithmic description of the point set, > and > it should be deterministic. > > The question is how to get the best of both worlds. > > I admit, I am being vague here. Ideas are solicited. > > > > ------------------------------ > > Message: 6 > Date: Tue, 12 Nov 2013 21:45:25 -0800 > From: meekerdb <meekerdb(a)verizon.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] good deterministic algorithmic point sets > without repeated distances? > Message-ID: <528311F5.3070308(a)verizon.net> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > How about lattice points plus small random vectors. > > Brent > > On 11/12/2013 9:16 PM, Warren D Smith wrote: >> In the plane (or some other low dimensional space) I want a good >> infinite point set that has "density 1." >> >> What does "good" mean? Well, "more uniform than random" in some vague >> sense, such as >> sphere packing, sphere covering, and providing good estimates of >> integrals via sums. >> >> BUT, what I also want, is the set has no repeated distances. So >> lattices are undesirable. >> Random point sets will not repeat a distance (with probability 1) so >> they are great, except they fail to be good in the sense of the >> previous paragraph, while lattices can be quite good in that sense. >> >> Another thing I want is an easy, algorithmic description of the point set, >> and >> it should be deterministic. >> >> The question is how to get the best of both worlds. >> >> I admit, I am being vague here. Ideas are solicited. >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> ----- >> No virus found in this message. >> Checked by AVG - www.avg.com >> Version: 2014.0.4158 / Virus Database: 3629/6830 - Release Date: 11/12/13 >> >> > > > > > ------------------------------ > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > End of math-fun Digest, Vol 129, Issue 13 > ***************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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