October 2020 - mathfun@mailman.xmission.com

Re: [math-fun] Factoring/normalizing quaternions
by Henry Baker 14 Oct '20

14 Oct '20

By 'complex number' A, I sometimes mean the quaternion A+0j, constructed from the pair of complex numbers (A,0). This subgroup of the quaternions is isomorphic to the standard complex numbers. Yes, I'm abusing the notation, but mathematicians often abuse notation. :-) At 05:33 PM 10/13/2020, Dan Asimov wrote: >Henry, Despite whatever else you say, you also said: >----- >We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. >----- >As you have defined then, a "complex number" in the quaternions is a quaternion of form a + bi. > >So maybe that's not what you mean, or maybe you don't mean that you have factored a quaternion into two complex numbers. > >But those are things you said, and that was my point. > >ÂDan > >>Hi Dan: >> >>There's a lot of subtlety in my post that a quick scan might miss. >> >>There's a *huge* difference between 'i' and 'j', so the eye might not >>pick up on this subtlety. In the quaternions, 'i' is perpendicular to >>'j', and 'i' doesn't commute with 'j'. BTW, 'k' is there: it is >>simply represented by 'ij' rather than by a separate symbol. >> >>My post is basically a rehash of the standard theorem about factoring >>SU(2) matrices into (diagonal complex)(real cos/sin)(diagonal complex) >>SU(2) matrices, but it does it completely within the quaternion >>framework. >> >>At 02:16 PM 10/13/2020, Dan Asimov wrote: >>>I skipped over sections of this post by Henry Baker to try to find the points that it included. >>> >>>One point seems to be this: >>> ----- We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. ----- >>>But a general quaternion is *not* a product of quaternions that have zero j components and zero k components. So I not sure what this statement means. >>> >>> ÂDan

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Re: [math-fun] Factoring/normalizing quaternions
by Dan Asimov 14 Oct '20

14 Oct '20

Henry, Despite whatever else you say, you also said: ----- We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. ----- As you have defined then, a "complex number" in the quaternions is a quaternion of form a + bi. So maybe that's not what you mean, or maybe you don't mean that you have factored a quaternion into two complex numbers. But those are things you said, and that was my point. —Dan >Hi Dan: > >There's a lot of subtlety in my post that a quick scan might miss. > >There's a *huge* difference between 'i' and 'j', so the eye might not >pick up on this subtlety. In the quaternions, 'i' is perpendicular to >'j', and 'i' doesn't commute with 'j'. BTW, 'k' is there: it is >simply represented by 'ij' rather than by a separate symbol. > >My post is basically a rehash of the standard theorem about factoring >SU(2) matrices into (diagonal complex)(real cos/sin)(diagonal complex) >SU(2) matrices, but it does it completely within the quaternion >framework. > >At 02:16 PM 10/13/2020, Dan Asimov wrote: >>I skipped over sections of this post by Henry Baker to try to find the points that it included. >> >>One point seems to be this: >> ----- We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. ----- >>But a general quaternion is *not* a product of quaternions that have zero j components and zero k components. So I not sure what this statement means. >> >>Dan

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Re: [math-fun] Factoring/normalizing quaternions
by Henry Baker 14 Oct '20

14 Oct '20

Hi Dan: There's a lot of subtlety in my post that a quick scan might miss. There's a *huge* difference between 'i' and 'j', so the eye might not pick up on this subtlety. In the quaternions, 'i' is perpendicular to 'j', and 'i' doesn't commute with 'j'. BTW, 'k' is there: it is simply represented by 'ij' rather than by a separate symbol. My post is basically a rehash of the standard theorem about factoring SU(2) matrices into (diagonal complex)(real cos/sin)(diagonal complex) SU(2) matrices, but it does it completely within the quaternion framework. At 02:16 PM 10/13/2020, Dan Asimov wrote: >I skipped over sections of this post by Henry Baker to try to find the points that it included. > >One point seems to be this: > >----- >We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. >----- >But a general quaternion is *not* a product of quaternions that have zero j components and zero k components. So I not sure what this statement means. > >ÂDan

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Re: [math-fun] Factoring/normalizing quaternions
by Dan Asimov 13 Oct '20

13 Oct '20

PS What I find a useful way to express quaternions is to distinguish between real and "pure imaginary" portions, where "pure imaginary" means of form bi + cj + dk for real numbers b,c,d. Any non-real quaternion q = a + bi +cj +dk can clearly be expressed as q = a * 1 + s * u where u is the unit pure quaternion u = (bi +cj +dk) / s. and where s is the real number s = √(b^2+c^2+d^2). The subalgebra <1, u> of the quaternions generated over the reals by 1 and u is isomorphic to the complex field C. The map h : C —> <1, u> given by h(1) = 1 and h(i) = u, and extending linearly, is a field isomorphism. Hence anything true in C is correspondingly true of <1, u>. Hence the quaternion q above may be expressed as q = r exp(𝛉u) where r = √(a^2+b^2+c^2+d^2) and tan(𝛉) = s/a (s as above) and exp is the usual function given by exp(t) = 𝝨 t^n / n! (n = 0,1,2, ...). —Dan

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[math-fun] Factoring/normalizing quaternions
by Henry Baker 13 Oct '20

13 Oct '20

[The following exposition covers some familiar territory, but with perhaps a more perspicuous narrative.] Factoring/normalizing quaternions Henry Baker Posted 10/13/2020 to math-fun Let Q=A+Bj be an arbitrary quaternion, for arbitrary complex numbers A,B. I.e., quaternions can be represented as pairs of ordinary complex numbers; j is a symbol which does NOT commute with the ordinary complex 'i'. Complex numbers are commutative and always have square roots. Note that i^2=j^2=-1, and ij=k, and ijk=-1, as is usual for quaternions. (We won't be needing or using 'k' in this exposition.) Let A' be the complex conjugate of A, B' be the complex conjugate of B. Let N(A)=N(A')=AA'=|A|^2. N() is a multiplicative homomorphism. Now jB = B'j, so we can commute complex numbers with the j quaternion, so long as we conjugate them when they hop over j in either direction. We can 'normalize' the A component of Q into a non-negative real number by pre and post multiplication: (1/sqrt(A))Q(sqrt(A))' = (1/sqrt(A)) (A+Bj) (sqrt(A))' = (1/sqrt(A)) (A+Bj) sqrt(A') = (1/sqrt(A)) A(sqrt(A')) + (1/sqrt(A)) Bj (sqrt(A')) = sqrt(A) sqrt(A') + (1/sqrt(A)) B (sqrt(A))j = sqrt(AA') + Bj = sqrt(|A|^2) + Bj = |A| + Bj So Q = A+Bj = sqrt(A) (|A|+Bj) (1/sqrt(A))' Note that this works even when A is already a negative real number, e.g., A=-1, because sqrt(A)=sqrt(-1)=i, and (1/sqrt(A))'=(1/i)'=(-i)'=i, and A+Bj = -1+Bj = i (1+Bj) i = i^2 + iBji = -1 + i(-i)Bj = -1+Bj. Similarly, we an 'normalize' the B component of Q into a non-negative real number by pre and post multiplication: (1/sqrt(B))Q(sqrt(B)) = (1/sqrt(B)) (A+Bj) sqrt(B) = (1/sqrt(B)) A (sqrt(B)) + (1/sqrt(B)) Bj (sqrt(B)) = A + sqrt(B) j (sqrt(B)) = A + sqrt(B) sqrt(B') j = A + sqrt(BB') j = A + sqrt(|B|^2) j = A + |B|j So Q = A+Bj = sqrt(B) (A+|B|j) (1/sqrt(B)) Combining these two operations into one: (1/sqrt(AB))Q(sqrt(A'B) = (1/sqrt(AB)) (A+Bj) (sqrt(A'B)) = (1/sqrt(AB)) A (sqrt(A'B)) + (1/sqrt(AB)) Bj (sqrt(A'B)) = (1/sqrt(B)) sqrt(A) (sqrt(A'B)) + (1/sqrt(A)) sqrt(B) j (sqrt(A'B)) = sqrt(A) sqrt(A') + (1/sqrt(A)) sqrt(B) sqrt(B'A) j = sqrt(|A|^2) + sqrt(B) sqrt(B') j = |A| + sqrt(BB') j = |A| + |B|j So, Q = A+Bj = sqrt(AB) (|A|+|B|j) (1/sqrt(A'B)) We have thus factored our quaternion into two complex numbers sqrt(AB) and 1/sqrt(A'B), and a quaternion having only real (rather than complex) components. If Q is a 'unit' quaternion -- i.e., QQ'=1 -- then QQ' = N(sqrt(AB)) N(|A|+|B|j) N(1/sqrt(A'B) = sqrt(N(A)) sqrt(N(B)) N(|A|+|B|j) 1/(sqrt(N(A')) sqrt(N(B))) = sqrt(N(A)) N(|A|+|B|j) 1/sqrt(N(A')) = sqrt(N(A)) N(|A|+|B|j) 1/sqrt(N(A))) = N(|A|+|B|j) = N(A)+N(B) = |A|^2+|B|^2 = 1 So in this unit quaternion case, we can set |A|=cos(theta), |B|=sin(theta), for a suitable real angle theta in the first quadrant, and Q = A+Bj = sqrt(AB) (|A|+|B|j) (1/sqrt(A'B)) = sqrt(AB) (cos(theta)+sin(theta)j) (1/sqrt(A'B)) Note for the record that sqrt(AB)/sqrt(A'B) = sqrt(A)/sqrt(A') = sqrt(A/A') = sqrt(AA/AA') = sqrt(AA)/sqrt(AA') = A/sqrt(|A|^2) = A/|A| = phase(A) However, our factorization is NOT the same as the more usual factorization in terms of phases, since |sqrt(AB)| /= 1, and |1/sqrt(A'B)| /= 1. We can always recover the more traditional factorization by taking phases: Q = A+Bj = phase(sqrt(AB)) (|A|+|B|j) phase(1/sqrt(A'B)) since |sqrt(AB)|/|sqrt(A'B)| = |phase(A)| = 1. and if QQ'=1, we get the classical factorization Q = A+Bj = phase(sqrt(AB)) (cos(theta)+sin(theta)j) phase(1/sqrt(A'B)) = exp(i*gamma) (cos(theta)+sin(theta)j) exp(i*delta) for suitable real angles gamma, delta, theta. We have one last factorization to make. If m,n are *real* numbers then m+nj is a quaternion both of whose complex components are missing their imaginary parts. We can convert such a quaternion m+nj to the complex number m+ni by means of following product: m+ni = -(i+j)(m+nj)(i+j)/2 or -2m-2ni = (i+j)(m+nj)(i+j) = m(i+j)^2 + n(i+j)j(i+j) = m(i^2+j^2) + n(i+j)(ji+jj) = -2m + n(i+j)(-ij-1) = -2m - n(i+j)(1+ij) = -2m - n(i+i^2j+j+jij) = -2m - n(i-j+j-ij^2) = -2m - n(i+i) = -2m - 2ni But the same transformation also works in reverse! -2m-2nj = (i+j)(m+ni)(i+j) = m(i+j)^2 + n(i+j)i(i+j) = -2m + n(i+j)(i^2+ij) = -2m + n(i+j)(-1+ij) = -2m + n(-i -j +i^2j + jij) = -2m + n(-i -j -j - ij^2) = -2m + n(-2j -i + i) = -2m -2nj This transformation thus allows us to convert cos(theta)+sin(theta)j into cos(theta)+sin(theta)i, and then into exp(i*theta). -2*exp(i*theta) = -2*(cos(theta)+sin(theta)i) = (i+j) (cos(theta)+sin(theta)j) (i+j) So our original arbitrary quaternion Q can be factored as Q = A+Bj = sqrt(AB) (|A|+|B|j) (1/sqrt(A'B)) = -(sqrt(AB) (i+j) (|A|+|B|i) (1/sqrt(A'B)))/2 where |A|+|B|i is now an ordinary complex number in the first quadrant. We note that |A|,|B| supply only 2 degrees of freedom, so the other two degrees of freedom are supplied by phase(sqrt(AB)) and phase(1/sqrt(A'B)). Once again, if Q is a unit quaternion, QQ'=1, then Q = A+Bj = phase(sqrt(AB)) (cos(theta)+sin(theta)j) phase(1/sqrt(A'B)) = exp(i*gamma) (cos(theta)+sin(theta)j) exp(i*delta) = -exp(i*gamma) (i+j) (cos(theta)+sin(theta)i) (i+j) exp(i*delta)/2 = -exp(i*gamma) (i+j) exp(i*theta) (i+j) exp(i*delta)/2 for suitably chosen angles gamma, theta, delta. A unit quaternion has 3 real degrees of freedom, and this 3-parameter factorization provides one convenient representation.

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Re: [math-fun] [seqfan] Re: Re: Squares packings
by Fred Lunnon 13 Oct '20

13 Oct '20

Looks good to me. WFL On 10/13/20, Leonardo Costa <leonardocostalesage(a)gmail.com> wrote: > Part (A) can be settled by using the same argument as in the original 23x23 > problem. The idea is to paint the squares of the grid white and black. > To be more precise, all squares in odd columns will be black and all even > ones will be white. Observe that there are as many black and > white squares in any 2x2 square you place and any 3x3 square you place will > have 3 more white squares than black ones or vice versa. > > Now suppose we can cover a nxn square (n = 6k+1, 6k-1) with 2x2 and 3x3 > pieces. Then, the difference between the amount of black and white squares > will be n, but it will also be a multiple of 3, because only 3x3 squares > cover a different amount of black and white squares. But 3 doesn't divide > n, contradiction. > > So if I haven't made any mistakes, I think this problem is settled. Good > job everyone! > > Leonardo Costa > > On Tue, 13 Oct 2020 at 03:39, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > >> Doh --- what's worse, JG had just beaten me to making the the exact >> same remark myself! >> >> WFL >> >> >> >> On 10/13/20, Rob Pratt <robert.william.pratt(a)gmail.com> wrote: >> > (B) is already settled by Jack Grahl’s argument. >> > >> >> On Oct 12, 2020, at 9:11 PM, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: >> >> >> >> I don't know --- I was hoping that we have some seasoned packer of >> >> squares >> >> on hand who can tell us what might be known. >> >> >> >> In the meantime, just to set up a formal aunt Sally: >> >> Conjecture: For natural n mod 6 in {1, 5} : >> >> (A) no packing of a n x n box by 1 x 1, 2 x 2, 3 x 3 tiles >> >> exists with no 1 x 1 tile; >> >> (B) there exists some packing with only a single 1 x 1 tile. >> >> >> >> WFL >> >> >> >> >> >> >> >>> On 10/13/20, rcs(a)xmission.com <rcs(a)xmission.com> wrote: >> >>> Is it established that large 6k+-1 squares require any 1x1s? --Rich >> >>> >> >>> ----- >> >>> Quoting Fred Lunnon <fred.lunnon(a)gmail.com>: >> >>>> << Sounds good for the upper bound. I have also confirmed the lower >> >>>> bound >> >>>> up >> >>>> through n = 100. >> >> >>>> >> >>>> Unclear: does your program establish that _NO_ solution exists >> without >> >>>> some 1x1 tile for edge length n mod 6 in {1, 5} & n < 100 ? >> >>>> >> >>>> If so, this suggests an interesting conjecture ... WFL >> >>>> >> >>>> >> >>>> >> >>>> On 10/12/20, Rob Pratt <robert.william.pratt(a)gmail.com> wrote: >> >>>>> Sounds good for the upper bound. I have also confirmed the lower >> bound >> >>>>> up >> >>>>> through n = 100. >> >>>>> >> >>>>> On Mon, Oct 12, 2020 at 3:06 AM Jack Grahl <jack.grahl(a)gmail.com> >> >>>>> wrote: >> >>>>> >> >>>>>> If we have a solution with 1 small square for n=11 and n=13, >> >>>>>> doesn't >> >>>>>> that >> >>>>>> imply a solution with 1 small square for all larger numbers 6k+1 >> >>>>>> and >> >>>>>> 6k-1 >> >>>>>> (ie all for which the solution is greater than zero)? >> >>>>>> >> >>>>>> Simply form an L-shape with width 6, to extend a solution for 6k+1 >> to >> >>>>>> a >> >>>>>> solution for 6(k+1)+1. The L shape is made up of a 6x6 square, and >> two >> >>>>>> 6x(6k+1) strips. Since any number >1 can be formed by a sum of 2's >> and >> >>>>>> 3's, >> >>>>>> we can always make the strips. Same for 6k-1 of course. >> >>>>>> >> >>>>>> Jack Grahl >> >>>>>> >> >>>>>> On Sun, 11 Oct 2020, 20:25 Rob Pratt, < >> robert.william.pratt(a)gmail.com> >> >>>>>> wrote: >> >>>>>> >> >>>>>>> Via integer linear programming, I get instead >> >>>>>>> 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 >> >>>>>>> >> >>>>>>> Here's an optimal solution for n = 17, with the only 1x1 appearing >> in >> >>>>>> cell >> >>>>>>> (12,6): >> >>>>>>> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 >> >>>>>>> 1 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8 >> >>>>>>> 2 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8 >> >>>>>>> 3 9 9 10 10 38 38 38 11 11 12 12 13 13 14 14 15 15 >> >>>>>>> 4 9 9 10 10 39 39 39 11 11 12 12 13 13 14 14 15 15 >> >>>>>>> 5 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41 >> >>>>>>> 6 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41 >> >>>>>>> 7 42 42 42 20 20 21 21 22 22 23 23 40 40 40 41 41 41 >> >>>>>>> 8 42 42 42 20 20 21 21 22 22 23 23 24 24 25 25 26 26 >> >>>>>>> 9 42 42 42 27 27 43 43 43 44 44 44 24 24 25 25 26 26 >> >>>>>>> 10 45 45 45 27 27 43 43 43 44 44 44 28 28 29 29 30 30 >> >>>>>>> 11 45 45 45 31 31 43 43 43 44 44 44 28 28 29 29 30 30 >> >>>>>>> 12 45 45 45 31 31 1 46 46 46 47 47 47 32 32 48 48 48 >> >>>>>>> 13 33 33 34 34 35 35 46 46 46 47 47 47 32 32 48 48 48 >> >>>>>>> 14 33 33 34 34 35 35 46 46 46 47 47 47 36 36 48 48 48 >> >>>>>>> 15 49 49 49 50 50 50 51 51 51 52 52 52 36 36 53 53 53 >> >>>>>>> 16 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53 >> >>>>>>> 17 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53 >> >>>>>>> >> >>>>>>> Clearly, even 2n can be partitioned into 2x2 only, and 3n can be >> >>>>>>> partitioned into 3x3 only. Otherwise, it appears that the minimum >> is >> >>>>>>> 1 >> >>>>>> for >> >>>>>>> n >= 11. >> >>>>>>> >> >>>>>>> On Sun, Oct 11, 2020 at 3:11 AM <michel.marcus(a)free.fr> wrote: >> >>>>>>> >> >>>>>>>> >> >>>>>>>> Hello Seqfans, >> >>>>>>>> >> >>>>>>>> >> >>>>>>>> "Images du CNRS" French site 4th problem for September 2020 is: >> >>>>>>>> . >> >>>>>>>> What is the minimum number of 1X1 pieces to fill a 23X23 square >> >>>>>>>> with >> >>>>>> 1X1, >> >>>>>>>> 2X2, and 3X3 pieces. >> >>>>>>>> Problem is at >> >>>>>>>> http://images.math.cnrs.fr/Septembre-2020-4e-defi.html. >> >>>>>>>> Solution is at >> >>>>>>>> http://images.math.cnrs.fr/Octobre-2020-1er-defi.html >> >>>>>>>> (click in Solution du 4e défi de septembre) >> >>>>>>>> >> >>>>>>>> >> >>>>>>>> I wondered what we get for other squares and painstakingly >> obtained >> >>>>>>>> for >> >>>>>>>> n=1 up to n=23: >> >>>>>>>> 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 4 0 4 0 0 0 1 >> >>>>>>>> >> >>>>>>>> >> >>>>>>>> Do you see some mistakes? Is it possible to extend it? >> >>>>>>>> >> >>>>>>>> >> >>>>>>>> Thanks. Best. >> >>>>>>>> MM >> >>>>>>>> >> >>>>>>>> -- >> >>>>>>>> Seqfan Mailing list - http://list.seqfan.eu/ >> >>>>>>>> >> >>>>>>> >> >>>>>>> -- >> >>>>>>> Seqfan Mailing list - http://list.seqfan.eu/ >> >>>>>>> >> >>>>>> >> >>>>>> -- >> >>>>>> Seqfan Mailing list - http://list.seqfan.eu/ >> >>>>>> >> >>>>> >> >>>>> -- >> >>>>> Seqfan Mailing list - http://list.seqfan.eu/ >> >>>>> >> >>>> >> >>>> _______________________________________________ >> >>>> math-fun mailing list >> >>>> math-fun(a)mailman.xmission.com >> >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >>>> >> >>> >> >>> >> >>> >> >>> _______________________________________________ >> >>> math-fun mailing list >> >>> math-fun(a)mailman.xmission.com >> >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >>> >> >> >> >> -- >> >> Seqfan Mailing list - http://list.seqfan.eu/ >> > >> > -- >> > Seqfan Mailing list - http://list.seqfan.eu/ >> > >> >> -- >> Seqfan Mailing list - http://list.seqfan.eu/ >> > > -- > Seqfan Mailing list - http://list.seqfan.eu/ >

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Re: [math-fun] [seqfan] Re: Squares packings
by Fred Lunnon 13 Oct '20

13 Oct '20

<< Sounds good for the upper bound. I have also confirmed the lower bound up through n = 100. >> Unclear: does your program establish that _NO_ solution exists without some 1x1 tile for edge length n mod 6 in {1, 5} & n < 100 ? If so, this suggests an interesting conjecture ... WFL On 10/12/20, Rob Pratt <robert.william.pratt(a)gmail.com> wrote: > Sounds good for the upper bound. I have also confirmed the lower bound up > through n = 100. > > On Mon, Oct 12, 2020 at 3:06 AM Jack Grahl <jack.grahl(a)gmail.com> wrote: > >> If we have a solution with 1 small square for n=11 and n=13, doesn't that >> imply a solution with 1 small square for all larger numbers 6k+1 and 6k-1 >> (ie all for which the solution is greater than zero)? >> >> Simply form an L-shape with width 6, to extend a solution for 6k+1 to a >> solution for 6(k+1)+1. The L shape is made up of a 6x6 square, and two >> 6x(6k+1) strips. Since any number >1 can be formed by a sum of 2's and >> 3's, >> we can always make the strips. Same for 6k-1 of course. >> >> Jack Grahl >> >> On Sun, 11 Oct 2020, 20:25 Rob Pratt, <robert.william.pratt(a)gmail.com> >> wrote: >> >> > Via integer linear programming, I get instead >> > 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 >> > >> > Here's an optimal solution for n = 17, with the only 1x1 appearing in >> cell >> > (12,6): >> > 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 >> > 1 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8 >> > 2 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8 >> > 3 9 9 10 10 38 38 38 11 11 12 12 13 13 14 14 15 15 >> > 4 9 9 10 10 39 39 39 11 11 12 12 13 13 14 14 15 15 >> > 5 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41 >> > 6 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41 >> > 7 42 42 42 20 20 21 21 22 22 23 23 40 40 40 41 41 41 >> > 8 42 42 42 20 20 21 21 22 22 23 23 24 24 25 25 26 26 >> > 9 42 42 42 27 27 43 43 43 44 44 44 24 24 25 25 26 26 >> > 10 45 45 45 27 27 43 43 43 44 44 44 28 28 29 29 30 30 >> > 11 45 45 45 31 31 43 43 43 44 44 44 28 28 29 29 30 30 >> > 12 45 45 45 31 31 1 46 46 46 47 47 47 32 32 48 48 48 >> > 13 33 33 34 34 35 35 46 46 46 47 47 47 32 32 48 48 48 >> > 14 33 33 34 34 35 35 46 46 46 47 47 47 36 36 48 48 48 >> > 15 49 49 49 50 50 50 51 51 51 52 52 52 36 36 53 53 53 >> > 16 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53 >> > 17 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53 >> > >> > Clearly, even 2n can be partitioned into 2x2 only, and 3n can be >> > partitioned into 3x3 only. Otherwise, it appears that the minimum is 1 >> for >> > n >= 11. >> > >> > On Sun, Oct 11, 2020 at 3:11 AM <michel.marcus(a)free.fr> wrote: >> > >> > > >> > > Hello Seqfans, >> > > >> > > >> > > "Images du CNRS" French site 4th problem for September 2020 is: >> > > . >> > > What is the minimum number of 1X1 pieces to fill a 23X23 square with >> 1X1, >> > > 2X2, and 3X3 pieces. >> > > Problem is at http://images.math.cnrs.fr/Septembre-2020-4e-defi.html. >> > > Solution is at http://images.math.cnrs.fr/Octobre-2020-1er-defi.html >> > > (click in Solution du 4e défi de septembre) >> > > >> > > >> > > I wondered what we get for other squares and painstakingly obtained >> > > for >> > > n=1 up to n=23: >> > > 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 4 0 4 0 0 0 1 >> > > >> > > >> > > Do you see some mistakes? Is it possible to extend it? >> > > >> > > >> > > Thanks. Best. >> > > MM >> > > >> > > -- >> > > Seqfan Mailing list - http://list.seqfan.eu/ >> > > >> > >> > -- >> > Seqfan Mailing list - http://list.seqfan.eu/ >> > >> >> -- >> Seqfan Mailing list - http://list.seqfan.eu/ >> > > -- > Seqfan Mailing list - http://list.seqfan.eu/ >

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Re: [math-fun] Do "Hawking Points" Actually Exist?
by Dan Asimov 11 Oct '20

11 Oct '20

Penrose has been insisting that human consciousness is transmitted/generated/*something* by features of neurons called microtubules. I'm convinced this is utter nonsense. —Dan > Whenever it surfaces, I do wonder whether theoretical physics is "math-fun"; >--- anyway, till we get slapped down, here's my (totally amateur) >two-penn'orth ... >I've been fascinated by physics since childhood, but somehow never managed to >focus motivation sufficiently to learn to _do_ any: so I can usually >understand the >questions well enough, even if the answers often go over my head. > > Not this time round. "behind (or before) the CMB" puts the problem >in a nutshell. >Both general relativity and quantum theory deal with observations made by an >observer in a specific frame of reference located in (Minkwski) spacetime. >Penrose in the video kicks off by admitting this, then cheerfully proceeds to >speculate about some enveloping notional spacetime without any attempt at >defining what behind (or before) might mean. It's turtles all the way down ... > > I sniff a fundamental cognitive trap lurking. Our notion of time is >inextricably >connected with our consciousness, which in turn we remain unable to relate to >any physical framework --- despite strenuous efforts (by Penrose and others) >over many millennia. The possibility of progress seems to demand that such >questions must be stripped of their dependence on a nonexistent frame of >reference even to make sense. > >Fred Lunnon > > >On 10/9/20, Brad Klee <bradklee(a)gmail.com> wrote: >> I was wondering about this question recently, so I was >> happy to get a dissenting opinion via Forbes: >> >> https://www.forbes.com/sites/startswithabang/2020/10/08/no-roger-penrose-we… >> >> But the tone of the Forbes article is too businesslike for >> my taste, and does it mention that Penrose himself >> describes the idea as "crazy"? See also: >> >> https://www.youtube.com/watch?v=YTttUigXulk >> >> When you watch this video, how can you help but wonder >> about the incredible mystery of what exactly is going on in >> the background out there? Maybe universal reincarnation >> is a plausible cosmogony? We don't know. >> >> Fifty or one hundred years from now we could have better >> gravity detection, and a better idea of what is going on >> behind (or before) the CMB. Until then... >> >> Anyways, congratulations Penrose! More and more cheers >> and congratulations for unexcelled creativity! >> >> --Brad

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Re: [math-fun] Taxicab 5.2.2 / A046881
by Dan Asimov 10 Oct '20

10 Oct '20

1) What are the arguments for/against the existence of generalized taxicab numbers A^n + B^n = C^n + D^n ? 2) Is there a reason to believe that these should exist only for finitely many values of n ≥ 1 and positive A, B, C, D ??? 3) And what if (for odd n) A, B, C, D are allowed to be negative??? —Dan ----- On Fri, Oct 9, 2020 at 12:06 PM Frank Stevenson < frankstevensonmobile(a)gmail.com> wrote: As a programming exercise I just finished a search for a solution A⁵ + B⁵ = C⁵ + D⁵ for numbers up to 4.25e37 ( 2 ^ 125 ) but did not find anything. ... ... -----

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[math-fun] Pseudo-Fibonacci numbers?
by James Propp 10 Oct '20

10 Oct '20

Does there exist a positive integer n that isn’t a Fibonacci number, such that for every modulus m there is a Fibonacci number congruent to n mod m? Jim Propp

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