October 2020 - mathfun@mailman.xmission.com

[math-fun] Taxicab 5.2.2 / A046881
by Frank Stevenson 10 Oct '20

10 Oct '20

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. This is perhaps not a surprise, since the density of possible solutions falls precipitously. I ran the search over a couple of days on 8 x Tesla V100 GPUs, and can give details if anyone is interested. But my main point of curiosity now is what the current thinking about this open problem is: Do any taxicab numbers exist for powers >= 5 ? Is it even something that is tackled seriously ? Frank

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[math-fun] Fibo-quaternion sequences
by Henry Baker 09 Oct '20

09 Oct '20

Given 2 quaternions q0,q1, we can define the traditional vector cross product operation: q0 x q1 = (q0 q1 - q1 q0)/2 [Note that this result is *independent* of the scalar parts of q0,q1, so q0 x q1 = V(q0) x V(q1) = (V(q0)V(q1) - V(q0)V(q1))/2] Given two arbitrary quaternions a,b, we can form the infinite sequence: q0=a, q1=b, q2=q0xq1, q3=q1xq2, ..., q(n)=q(n-2)xq(n-1). >From q2 forward, the qi are all 'pure vector' quaternions having no scalar part. For such pure vector quaternions qi, qi^2=-(qi.qi)=-|qi|^2, so we can normalize all non-zero pure vectors into pure 'unit' vectors: u(v) = v/|v| = v/sqrt(-v^2) Unless the vector part of our arbitrary quaternion b is a scalar multiple of the vector part of our arbitrary quaternion a, then our qi sequence describes a spiral from the origin about the axis vector u(q2)+u(q3)+u(q4) and q(n)=M*q(n-3), for some real scalar M. If we consider the normalized sequence u(q2),u(q3),u(q4),u(q5),..., then this sequence cycles through 3 pure unit vectors: u(q2)->u(q3)->u(q4)->u(q1). Note that u(q2)^2=u(q3)^2=u(q4)^2=u(q2)u(q3)u(q4)=-1, so I=u(q2),J=u(q3),K=u(q4) make perfectly good basis vectors for our quaternion space. Finally, we can define q' = -(q+IqI+JqJ+KqK)/2 which gives us a perfectly good 'conjugate' operation for our quaternion space. Using our conjugate operation, we can now extract the scalar and vector parts of an arbitrary quaternion: s(q) = (q+q')/2 V(q) = q-s(q) Thus, our quaternion sequence can construct a 'coordinate-free' representation of quaternions from the minimal raw materials of the scalar field, the non- commutative product operation, and two given arbitrary quaternions with linearly independent vector parts. ---- We could also consider other 'fibo-quaternion' sequences, e.g., qn=q(n-2)q(n-1), as well as all such sequences over other fields -- e.g., GF(q^k).

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[math-fun] The surprise
by Hans Havermann 08 Oct '20

08 Oct '20

A week ago I was verifying and extending a sequence of numbers/assertions that Claudio Meller had published in Futility Closet nine years ago: https://www.futilitycloset.com/2011/10/01/on-target-2/ I soon enough realized that for the letter 'w' a purely numerical proxy could be employed. Consider the list of positive base-ten integers: 1, 2, 3, ... The second occurrence of the digit 2 in the list is in 12; the fifth and sixth, in 22. A search for the n-th occurrence of 2 happening in the number n came up with three pairs: 28263827 & 28263828, 242463827 & 242463828, and 528263827 & 528263828. Note that the third pair is exactly 500 million more than the first. On Sunday I found the next one: The 2222222222nd occurrence of 2 in the list is the second 2 in 2222222222. I was of course mildly surprised. Is this an amazing coincidence or is there a way of framing the observation so that it appears somewhat less remarkable?

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[math-fun] rank of a symbolic algebra matrix
by Henry Baker 07 Oct '20

07 Oct '20

Suppose you're using a symbolic algebra system & want to estimate the rank of a matrix whose entries are symbolic expressions involving many different variables. My intuition tells me that I should be able to randomly guess values for all of the variables, numerically compute the rank for that assignment of variables, and do it again several more times. The *maximum* rank achieved should be close to the actual symbolic rank of the matrix. 2 cases: integers and floating point. If the symbolic expressions enable exact integer values, then the rank could be computed using arbitrary-precision arithmetic. However, it might be more efficient to do the computation modulo a large number of different primes. If we work in floating point, the expressions could be incredibly ill-conditioned, in which case telling the difference between very small numbers and actual zeros could be impossible. Other than such ill-conditioning, is it possible that the rank could be *over* estimated by such random sampling?

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Re: [math-fun] Integer determinant question
by Dan Asimov 07 Oct '20

07 Oct '20

I recently asked the question at bottom. This is an update. >From googling I learned that an n x n matrix whose rows are successive cyclic rotated versions of each other are called "circulant" matrices (a term I'd heard but never bothered to learn about before) — and that these matrices are actually very interesting!!! (See for instance this article: <https://www.ams.org/notices/201203/rtx120300368p.pdf>. It is apparently an open problem to determine which integers can be the determinants of integer circulant matrices — see for instance this article: <https://projecteuclid.org/download/pdf_1/euclid.ijm/1256047802>. Perhaps the additional condition I somewhat arbitrarily imposed, that the row sum a(0) + ... + a(n-1) = 1, might allow solution of the problem in this special case; maybe not. A subproblem might be to reduce entries of a circulant matrix modulo a prime p and then ask what the determinant can be modulo p. Maybe that question is more tractable. —Dan ----- The lattice in R^3 generated by the symmetrical vectors { (2,-1,0), (0,2,-1), (-1,0,2) } (whose sum is (1,1,1)) has a fundamental cell whose volume = 7. I was surprised that a 3-fold symmetrical set of vectors would generate a parallelepiped whose volume had volume = 7. So I'm curious: Suppose n integers a(0), ..., a(n-1) have sum a(0) + ... + a(n-1) = 1. Then what integers can be the determinant |M| of the n x n matrix M = (m(i,j)) if m(i,j) = a(i+j) where i+j is calculated modulo n ??? -----

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Re: [math-fun] Lucas, Fibo, Phi, capital O
by Steve Witham 07 Oct '20

07 Oct '20

> My first thought seeing this was that >Â Â Â Â Lucas(n) / Fibo(n) --> sqrt(5) > and the int sequences can be done by repeated squaring, > so you could get O(2^n) digits of sqrt(5) > with O(n) int multiplications and only one division!! Ugh, using "n" for two different things.Â Lucas(n) and Fibo(n) do have O(n) digits, but the following is wrong: > And the work that Newton's method does is also about > the same because the divisions need only 1, 2, 4... > 2^(n-1), 2^n digits of precision. Sorry, instead say 2^1, 2^2, 2^4... 2^(k/2), 2^k digits in k steps. The point is that the order of work is that of the last division. Â --Steve

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[math-fun] Lucas, Fibo, Phi, capital O
by Steve Witham 07 Oct '20

07 Oct '20

Lucas(n=0, 1, 2...) = 2, 1, 3, 4, 7, 11, 18... = phi^n + (-phi)^(-n) Fibonacci(n=0, 1, 2...) = 0, 1, 1, 2, 3, 5, 8... = (phi^n - (-phi)^(-n)) / sqrt(5) (True for n < 0 also.) My first thought seeing this was that Lucas(n) / Fibo(n) --> sqrt(5) and the int sequences can be done by repeated squaring, so you could get O(2^n) digits of sqrt(5) with O(n) int multiplications and only one division!! But then I thought, continued fractions probably solve the general sqrt(int) problem just as fast. And the work that Newton's method does is also about the same because the divisions only need 1, 2, 4... 2^(n-1), 2^n digits of precision. This is math. This is my brain on math. --Steve

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[math-fun] Integer determinant question
by Dan Asimov 07 Oct '20

07 Oct '20

The lattice in R^3 generated by the symmetrical vectors { (2,-1,0), (0,2,-1), (-1,0,2) } (whose sum is (1,1,1)) has a fundamental cell whose volume = 7. I was surprised that a 3-fold symmetrical set of vectors would generate a parallelepiped whose volume had volume = 7. So I'm curious: Suppose n integers a(0), ..., a(n-1) have sum a(0) + ... + a(n-1) = 1. Then what integers can be the determinant |M| of the n x n matrix M = (m(i,j)) if m(i,j) = a(i+j) where i+j is calculated modulo n ??? —Dan

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[math-fun] binary roulette wheels
by Veit Elser 06 Oct '20

06 Oct '20

Is there a name for cyclic sequences (necklaces) of length 2^n that contain all the integers 0, … , 2^n-1 expressed in binary in the 2^n subsequences of length n? For example, for n=4 the sequence 0000100110101111 contains 0, 1, 2, 4, 9, 3, 6, 13, 10, 5, 11, 7, 15, 14, 12, 8. -Veit

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[math-fun] Using AI to help Superseeker guess formulas for sequences
by Neil Sloane 05 Oct '20

05 Oct '20

Dear Math Fun: I keep coming across sequences that badly need a formula, possibly expressing them in terms of some known sequence. A337641 and A337655 are two recent examples. Why not use AI to strengthen Superseeker?, I keep saying There is a talk on Monday that might be relevant: This announcement just arrived in email: Springer Nature <springernature(a)newsletter.springernature.com> George Em Karniadakis (Brown University and MIT) >From PINNs to DeepOnets: Approximating functions, functionals, and operators using deep neural networks Throughout October, authors from the new journal SN Partial Differential Equations and Applications will be hosting a series of free webinars on topics related to partial differential equations. Up next is a presentation by George Em Karniadakis of Brown University and MIT on approximating functions using deep neural networks. Join us for a lively discussion on October 5th at 9am EST, 3pm CET. I signed up, but AI and PDEs are not my field. If anyone is interested, please watch too

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