October 2019 - mathfun@mailman.xmission.com

[math-fun] q-numbers are actually fun.
by Bill Gosper 29 Oct '19

29 Oct '19

(Kids: I highly recommend grokking this. Ask me anything.) Years ago I wrote WRI a package of q-utilities, e.g. QPochhammer, QFactorial, QBinomial, ... . One of the items in the spec was QNumber, which is just, e.g., In[120]:= qNumber[4, -q^2] Out[120]= 1 - q^2 + q^4 - q^6 I.e., replace the number n by n terms of the geometric series with the specified term ratio. It seemed sort of useless and trivial. I was a bit relieved when they didn't install it. I was wrong. Let's play. Exercise: Expand In[121]:= qNumber[5, q] qNumber[5, -q] Out[121]= (1 - q + q^2 - q^3 + q^4) (1 + q + q^2 + q^3 + q^4) in your head! SPOILER: StringReverse@"]2^q,5[rebmuNq" Why? Nice Mathematica exercise: Inverse qNumber. E.g., In[125]:= unqNumber /@ %121 Out[125]= Inactive[qNumber][5, -q] Inactive[qNumber][5, q] Try it! Here's a way, if you give up: unqNumber[p_] := Module[{L = List @@ Together@p}, Inactive[qNumber][Length@L, L[[2]]] /; PolynomialQ[p, Variables[p][[1]]]&&SameQ @@ (Drop[L, 1]/Drop[L, -1])] If you study the following factorizations, you should formulate enough theorems to mentally factorize any q-number whose q➝1 you can mentally factorize. True or False: the factors are always q-numbers. 0 0 1 1 2 1+q 3 1+q+q^2 2^2 (1+q) (1+q^2) 5 1+q+q^2+q^3+q^4 2 3 (1+q) (1-q+q^2) (1+q+q^2) 7 1+q+q^2+q^3+q^4+q^5+q^6 2^3 (1+q) (1+q^2) (1+q^4) 3^2 (1+q+q^2) (1+q^3+q^6) 2 5 (1+q) (1-q+q^2-q^3+q^4) (1+q+q^2+q^3+q^4) 11 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10 2^2 3 (1+q) (1+q^2) (1-q+q^2) (1+q+q^2) (1-q^2+q^4) 13 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12 2 7 (1+q) (1-q+q^2-q^3+q^4-q^5+q^6) (1+q+q^2+q^3+q^4+q^5+q^6) 3 5 (1+q+q^2) (1+q+q^2+q^3+q^4) (1-q+q^3-q^4+q^5-q^7+q^8) 2^4 (1+q) (1+q^2) (1+q^4) (1+q^8) 17 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12+q^13+q^14+q^15+q^16 2 3^2 (1+q) (1-q+q^2) (1+q+q^2) (1-q^3+q^6) (1+q^3+q^6) 19 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12+q^13+q^14+q^15+q^16+q^17+q^18 2^2 5 (1+q) (1+q^2) (1-q+q^2-q^3+q^4) (1+q+q^2+q^3+q^4) (1-q^2+q^4-q^6+q^8) 3 7 (1+q+q^2) (1+q+q^2+q^3+q^4+q^5+q^6) (1-q+q^3-q^4+q^6-q^8+q^9-q^11+q^12) 2 11 (1+q) (1-q+q^2-q^3+q^4-q^5+q^6-q^7+q^8-q^9+q^10) (1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10) 23 1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12+q^13+q^14+q^15+q^16+q^17+q^18+q^19+q^20+q^21+q^22 2^3 3 (1+q) (1+q^2) (1-q+q^2) (1+q+q^2) (1+q^4) (1-q^2+q^4) (1-q^4+q^8) 5^2 (1+q+q^2+q^3+q^4) (1+q^5+q^10+q^15+q^20) 2 13 (1+q) (1-q+q^2-q^3+q^4-q^5+q^6-q^7+q^8-q^9+q^10-q^11+q^12) (1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+q^9+q^10+q^11+q^12) 3^3 (1+q+q^2) (1+q^3+q^6) (1+q^9+q^18) —rwg

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[math-fun] primoids. An elementary primes lesson ends in a bit of a surprise.
by Bill Gosper 28 Oct '19

28 Oct '19

primoids <http://gosper.org/primoids.pdf> —rwg

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[math-fun] Chunk & Sum (minimal rule)
by Éric Angelini 27 Oct '19

27 Oct '19

Hello Allan, I have found -- thanks to a remark made to me by Frank Stevenson in a private mail -- the "minimal pieces" rule I was trying to explain. Alice always wants to check if there is a sum that is sound (according to the first rule -- increasing chunks) _and_ inferior to the sum claimed by Bob -- when considering Bob's arrangement. This is sometimes difficult to see. Best, and thanks to Frank, É.

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[math-fun] "Self-describing" integer tuples.
by Allan Wechsler 26 Oct '19

26 Oct '19

I am sure this has been discussed here before, so if somebody just has a pointer to any previous discussion, let me know. Suppose we have a finite sequence of non-negative integers. We can create a new sequence that is a "census" of the first. For example, the census of (1,4,1,4,2,1,3) is (3,1,1,2,1,3,2,4) -- that is, three 1's, one 2, one 3, and two 4's. I am interested in any tuple that is its own census. A simple example is (2,2); a more complicated example is (2,1,3,2,2,3,1,K). The K can be replaced with any number bigger than 3. I found another class of examples of the form (K,1,3,2,2,3,2,K,1,A,1,B...) where K,A,B... are distinct from each other and bigger than 3, and there are exactly K-1 terms in the A,B... set. Let's declare that the choice of the value of singletons doesn't matter. I have an intuition that the number of types of self-censusing sequences is quite small -- maybe only the three types I just mentioned? Or did I miss a few? I would not be surprised if I did, but I would be surprised if there were a _lot_ more types.

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[math-fun] The "bottom row" of q-Pascal's triangle
by Bill Gosper 26 Oct '19

26 Oct '19

makes sense because successive rows approach a limit, which is experimentally Inactive[Table][1/QPochhammer[q,q,k],{k,0,∞}], and shouldn't be hard to prove. For finite segments, you should just be able to say Table[QBinomial[∞,k],{k,0,...}], but Mathematica returns nonsense. —rwg

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[math-fun] Mathematician Solves 48-Year-Old Problem, Finds New Way To Multiply
by Ray Tayek 23 Oct '19

23 Oct '19

https://science.slashdot.org/story/19/10/20/164235/mathematician-solves-48-… An anonymous reader quotes Popular Mechanics: /An assistant professor from the University of New South Wales Sydney in Australia has developed a new method for multiplying giant numbers together that's more efficient <https://www.popularmechanics.com/science/a29514208/faster-way-multiply/> than the "long multiplication" so many are taught at an early age. "More technically, we have proved a 1971 conjecture of Schönhage and Strassen about the complexity of integer multiplication," associate professor David Harvey says in this video <https://youtu.be/2w-lDOQfI5s>... Schönhage and Strassen predicted that an algorithm multiplying /n/-digit numbers using /n/ * log(/n/) basic operations should exist, Harvey says. His paper is the first known proof that it does... The /[original 1971]/ Schönhage-Strassen method <https://en.wikipedia.org/wiki/Sch%C3%B6nhage%E2%80%93Strassen_algorithm> is very fast, Harvey says. If a computer were to use the squared method taught in school on a problem where two numbers had a billion digits each, it would take months. A computer using the Schönhage-Strassen method could do so in 30 seconds. But if the numbers keep rising into the trillions and beyond, the algorithm developed by Harvey and collaborator Joris van der Hoeven at École Polytechnique in France could find solutions faster than the 1971 Schönhage-Strassen algorithm. "It means you can do all sorts of arithmetic more efficiently, for example division and square roots," he says. "You could also calculate digits of pi more efficiently than before. It even has applications to problems involving huge prime numbers./ "The question is, how deep does /n/ have to be for this algorithm to actually be faster than the previous algorithms?" the assistant professor says in the video. "The answer is we don't know. "It could be billions of digits. It could be trillions. It could be much bigger than that. We really have no idea at this point." -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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[math-fun] Pick any 4 successive terms a, b, c, d and have a/b = c/d.
by Éric Angelini 22 Oct '19

22 Oct '19

Hello Math-Fun, I've sent this 5 days ago to SeqFans but it seems that it never reached the list. I try here, Best, É. ---------------------------------------------------- Hello (SeqFans) Math-Fun, I was looking for a seq S (of distinct integers) where one could pick any 4 successive terms a, b, c, d and have a/b = c/d. I quickly found https://oeis.org/A038754 [Original definition of the seq is between brackets] [a(2n) = 3^n, a(2n+1) = 2*3^n]. S = 1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486,... The above seq starts with 1,2,3. The start 1,2,4 produces the powers of 2. The start 1,2,5 produces https://oeis.org/A026383 [a(n) = 5a(n-2), starting 1,2]. The start 1,2,6, produces https://oeis.org/A026549 [Ratios of successive terms are 2,3,2,3,2,3,2,3...] The start 1,2,7 produces https://oeis.org/A123752 [a(n) = 7*a(n-2), a(0) = 1, a(1) = 2]. The start 1,2,8 produces https://oeis.org/A098232 [Largest power of 2 <= 3^n]. The start 1,2,9 produces https://oeis.org/A083423 [a(n) = (5*3^n + (-3)^n)/6]. The start 1,2,10 produces https://oeis.org/A004643 [Powers of 2 written in base 4]. etc. (I love the rainbow of definitions!) The start 1,3,2 produces https://oeis.org/A164073 [a(n) = 2*a(n-2) for n > 2; a(1) = 1, a(2) = 3]. The start 1,3,4 produces https://oeis.org/A084221 [a(n+2) = 4*a(n), with a(0)=1, a(1)=3]. The start 1,3,5 produces https://oeis.org/A056487 [a(n) = 5^(n/2) for n even, a(n) = 3*5^((n-1)/2) for n odd]. ....(!?) and produces also https://oeis.org/A111386 [a(1) = 1, a(2) = 3; for n >= 3, take a(n) to be the smallest odd number not occurring earlier such that a(n-1) divides the concatenation a(n-2)a(n)]. The start 1,3,6 produces https://oeis.org/A026532 [Ratios of successive terms are 3,2,3,2,3,2,3,2...] The start 1,3,7 produces 1,3,7,21,49,147,343,1029,... _not in the OEIS_. The start 1,3,8 produces https://oeis.org/A096886 [Expansion of (1+3*x)/(1-8*x^2)]. The start 1,3,9 produces the powers of 3 The start 1,3,10 produces https://oeis.org/A004663 [Powers of 3 written in base 9]. etc. The start 1,4,2 produces https://oeis.org/A143095 [(1, 2, 4, 8,...) interleaved with (4, 8, 16, 32,...)]. But we have repeated terms there. The start 1,4,3 produces https://oeis.org/A166552 [a(n) = 3*a(n-2) for n > 2; a(1) = 1; a(2) = 4] The start 1,4,5 produces https://oeis.org/A133632 [a(1)=1, a(n)=(p-1)*a(n-1), if n is even, else a(n)=p*a(n-2), where p=5]. The start 1,4,6 produces https://oeis.org/A164532 [a(n) = 6*a(n-2) for n > 2; a(1) = 1, a(2) = 4]. The start 1,4,7 produces 1,4,7,28,49,196,343,1372,... _nt in the OEIS_ The start 1,4,8 produces https://oeis.org/A094015 [Expansion of (1+4x)/(1-8x^2)]. The start 1,4,9 produces https://oeis.org/A133125 [A133080 * A000244]. The start 1,4,10 produces https://oeis.org/A136859 [Numbers n such that n and the square of n use only the digits 0, 1, 4 and 6]. etc. (another rainbow!) The start 1,5,2 produces 1,5,2,10,4,20,8,40,16,80,32,160,... _not in the OEIS_. The start 1,5,3 produces https://oeis.org/A166465 [a(n) = 3*a(n-2) for n > 2; a(1) = 1, a(2) = 5]. The start 1,5,4 produces 1,5,4,20,16,80,64,320,256,... _not in the OEIS_. The start 1,5,6 produces https://oeis.org/A166023 [a(n) = 6*a(n-2) for n > 2; a(1) = 1, a(2) = 5] The start 1,5,7 produces 1,5,7,35,49,245,343,... _not in the OEIS_. The start 1,5,8 produces 1,5,8,40,64,320,512,... _not in the OEIS_. The start 1,5,9 produces 1,5,9,45,81,405,729,... _not in the OEIS_. The start 1,5,10 produces https://oeis.org/A268100 [a(n) = 2^((n-1) mod 2)*5*10^floor((n-1)/2)] etc. The start 1,6,2 produces https://oeis.org/A163864 [a(n) = 2*a(n-2) for n > 2; a(1) = 1, a(2) = 6]. The start 1,6,3 produces https://oeis.org/A166450 [a(n) = 3*a(n-2) for n > 2; a(1) = 1, a(2) = 6]. The start 1,6,4 produces https://oeis.org/A081631 [2*2^n-(-2)^n]. The start 1,6,5 produces 1,6,5,30,25,150,125,750,... _not in the OEIS_. The start 1,6,7 produces 1,6,7,42,49,... _not in the OEIS_. The start 1,6,8 produces https://oeis.org/A164640 [a(n) = 8*a(n-2) for n > 2; a(1) = 1, a(2) = 6]. The start 1,6,9 produces 1,6,9,54,81,... _not in the OEIS_. The start 1,6,10 produces 1,6,10,60,100,600,1000,6000,10000 _not in the OEIS_. etc. The start 1,7,2 produces 1,7,2,14,4,28,8,56,16,112,32,224,... _not in the OEIS_. The start 1,7,3 produces https://oeis.org/A166481 [a(n) = 3*a(n-2) for n > 2; a(1) = 1; a(2) = 7]. The start 1,7,4 produces 1,7,4,28,16,96,64,... _not in the OEIS_. The start 1,7,5 produces 1,7,5,35,25,175,... _not in the OEIS_. The start 1,7,6 produces 1,7,6,42,36,252,... _not in the OEIS_. The start 1,7,8 produces 1,7,8,56,64,... _not in the OEIS_. The start 1,7,9 produces 1,7,9,63,81,... _not in the OEIS_. The start 1,7,10 produces 1,7,10,70,100,700,1000,7000,... _not in the OEIS_. etc. The start 1,8,2 produces https://oeis.org/A164587 [a(n) = 2*a(n - 2) for n > 2; a(1) = 1, a(2) = 8]. The start 1,8,3 produces 1,8,3,24,9,72,27,216,81,648,243,... _not in the OEIS_. The start 1,8,4 produces almost https://oeis.org/A135520 [erase the 1st term, then a(n) = 4*a(n-2)]. The start 1,8,5 produces 1,8,5,40,25,200,125,... _not in the OEIS_. The start 1,8,6 produces 1,8,6,48,36,... _not in the OEIS_. The start 1,8,7 produces 1,8,7,56,49,... _not in the OEIS_. The start 1,8,9 produces 1,8,9,72,81,... _not in the OEIS_. The start 1,8,10 produces 1,8,10,80,100,800,1000,... _not in the OEIS_. etc. The start 1,9,2 produces 1,9,2,18,4,36,8,72,16,144,32,288,... _not in the OEIS_. The start 1,9,3 produces almost https://oeis.org/A162852 [a(n) = 3*a(n-2) for n > 2; a(1) = 3, a(2) = -1]. But, again, a lot of not-distinct terms. The start 1,9,4 produces 1,9,4,36,16,144,... _not in the OEIS_. The start 1,9,5 produces 1,9,5,45,25,225,... _not in the OEIS_. The start 1,9 6 produces 1,9,6,54,36,... _notin the OEIS_. The start 1,9,7 produces 1,9,7,63,49,... _not in the OEIS_. The start 1,9,8 produces 1,9,8,72,64,... _not in the OEIS_. The start 1,9,10 produces 1,9,10,90,100,900,1000,... _not in the OEIS_. etc. The start 1,10,2 produces 1,10,2,20,4,40,8,80,16,160,32,320,... _not in the OEIS_. The start 1,10,3 produces 1,10,3,30,9,90,27,270,... _not in the OEIS_. etc. The start 2,1,6 produces 2,1,6,3,18,9,54,27,... _not in the OEIS_. As 19 is prime, the start 20,19 produces 20,19,40,38,80,72,... _not in the OEIS_ (and never in the OEIS, I guess). etc. Note that the remark here https://oeis.org/A164073, by a friend of mine ("Absolute second differences are the sequence itself"), is always true for some seqs above that follow the pattern: 1, a, 2, 2a, 4, 4a, 8, 8a, 16, 16a... for a = 0 to k. Oh, and what about picking 5 (instead of 4) successive integers a, b, c, d, e in a seq of distinct terms such that a/b is always equal to d/e? Never mind, this has been already submitted by Paul Curtz: https://oeis.org/A133464 Too good, the OEIS'authors!-)) My two,three, four cents. É. P.-S.#1 Many thanks to Klaus Brockhaus who computed 10 years ago a lot of the above, and forgive the errors I left in this mail, despite my carefull rerereading). P.-S.#2 Let me pick again (in a similar seq T) 5 successive integers a, b, c, d, e -- but such that a/c is now equal to d/e: T = 1, 2, 3, 4, 12, 24, 96, 576,... Ah ah, _T is not in the OEIS_!-)) Best, É.

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[math-fun] Dean Hickerson
by James Propp 22 Oct '19

22 Oct '19

Anyone have an email address for Dean Hickerson? Jim Propp

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[math-fun] Dean Hickerson
by James Propp 22 Oct '19

22 Oct '19

Anyone have a current email address for Dean Hickerson? "dean at math daht ucdavis daht edu" and "dean at ucdmath daht ucdavis daht edu" aren't working anymore. Thanks, Jim Propp

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[math-fun] "LHS Sometimes returns precision errors? What the bleep is it doing numerics for an identity map?"
by Bill Gosper 21 Oct '19

21 Oct '19

Brad, there's a really interesting answer. E.g., In[68]:= Sin[π/34] // FunctionExpand Out[68]= 1/(4 Sqrt[2/( 15 + Sqrt[17] - Sqrt[2 (17 - Sqrt[17])] - Sqrt[ 2 (34 + 6 Sqrt[17] + Sqrt[2 (17 - Sqrt[17])] - Sqrt[ 34 (17 - Sqrt[17])] + 8 Sqrt[2 (17 + Sqrt[17])])])]) (Rather cumbrous, Now undo it.) In[69]:= ArcSin@% // FullSimplify // Timing Out[69]= {0.312125, π/34} How could it take .3 sec to look that up? Alternatively, how could it get π/34 so quickly? Or at all? Here Mathematica recognizes Macsyma's nicer rendition: In[70]:= ArcSin[(-\[Sqrt](\[Sqrt]17 + 3) \[Sqrt](4 \[Sqrt]17 - 2 \[Sqrt](34 - 2 \[Sqrt]17)) + \[Sqrt](34 - 2 \[Sqrt]17) + \[Sqrt]17 - 1)/16] // FullSimplify // Timing Out[70]= {0.948516, π/34} even though it wasn't able to simplify its version to Macsyma's. So where did it get the notion of π/34? It guessed it numerically! It is then fairly routine to show that the two surds are identical, or that Macsyma's satisfies In[72]:= MinimalPolynomial[Sin[π/34], x] // Timing Out[72]= {0.002874, 1 - 8 x - 40 x^2 + 80 x^3 + 240 x^4 - 192 x^5 - 448 x^6 + 128 x^7 + 256 x^8} At least this is how Macsyma did it. Without even requiring a ratsimp or radcan. —rwg

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