February 2012 - mathfun@mailman.xmission.com

Re: [math-fun] Another puzzle
by N. J. A. Sloane 02 Feb '12

02 Feb '12

>> What n satisfy 2^n == 3n+2 (mod n^2 - 2n) > The clue is to observe that n^2 - 2n = n(n-2). First note that n is necessarily odd, which means that n and n-2 are relatively prime. By the Chinese Remainder Theorem, the equation is equivalent to the following system of equations: > 2^n == 2 (mod n) > 2^n == 8 (mod n-2) > Since n-2 is odd, the second equation is equivalent to > 2^{n-2} == 2 (mod n-2). > Thus n is a solution if and only if each of n and n-2 is a prime or a Sarrus/Poulet pseudoprime. I hope someone will submit this to the OEIS! Neil

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[math-fun] Score Four
by Allan Wechsler 02 Feb '12

02 Feb '12

Wikipedia says that Connect Four was shown to be a first-player win in 1988, but doesn't mention any similar analysis of the structurally similar game Score Four. It also feels like a first-player win to me, and I wonder if anybody knows anything about it for certain. Score Four is played on a square 4x4 array of posts; each post is tall enough to accommodate 4 playing beads. The board starts empty. Two players take turns; on your turn you slide a bead of your assigned color onto a post of your choosing; you can't choose a post that already has four beads on it. The bead will slide down to the lowest unoccupied place on the post. The object of the game is to establish a line of four beads of your color. This line may be horizontal, vertical, or diagonal; the winning lines are exactly the same as in Qubic. Score Four may be regarded as Qubic with the extra rule that one is required to play in the lowest unoccupied cell of a given column. Good players establish "traps", incomplete lines of three out of four beads, where the missing cell of the line has one void below it. The opponent can't play on that post without enabling the placement of the missing bead. In the endgame, the players alternate playing on un-booby-trapped posts until they run out of room; the player who plays on a booby-trapped post first loses. Surely if the exercise hasn't been done yet, this game would succumb to mere minutes of computer analysis.

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[math-fun] Fwd: Somos++
by rcs＠xmission.com 02 Feb '12

02 Feb '12

I don't know if this is the whole story, or page 1 of a longer paper. --Rich ----- Forwarded message from jamespropp(a)GMAIL.COM ----- Date: Thu, 2 Feb 2012 00:36:08 -0800 From: James Propp <jamespropp(a)GMAIL.COM> Reply-To: The Robbins Forum <ROBBINS(a)listserv.uml.edu> Subject: [ROBBINS] posted on behalf of Michael Somos To: ROBBINS(a)listserv.uml.edu ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Trilinear Recursion Michael Somos somos(a)cis.csuohio.edu It has been known for some time that the recursion A(n) = (A(n-1) + C1) / A(n-2) (of which C1=1 is the Lyness 5-cycle) has a solution in terms of a general Somos-5 sequence a(n) as follows A(n) = (a(n+2) * a(n-3)) / (a(n) * a(n-1) * C2) 1 + A(n) = (a(n+1) * a(n-2) * C3) / (a(n) * a(n-1) * C2) for some constansts C2, C3. There are similar kinds of examples. One of the simplest is the following. If A(n) = (A(n-1) + A(n-2)) / A(n-3), A(1) = A(2) = A(3) = 1, then A(n) = (a(n) * a(n+7)) / (a(n+3) * a(n+4)) where a(n)=(a(n-1)*a(n-6)*a(n-8) + a(n-2)*a(n-4)*a(n-9)) / (a(n=5)*a(n-10)), a(1) = a(2) = ... = a(10) = 1. which is now sequence A205303 in the OEIS at oeis.org. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ ----- End forwarded message ----- ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Trilinear Recursion Michael Somos somos(a)cis.csuohio.edu It has been known for some time that the recursion A(n) = (A(n-1) + C1) / A(n-2) (of which C1=1 is the Lyness 5-cycle) has a solution in terms of a general Somos-5 sequence a(n) as follows A(n) = (a(n+2) * a(n-3)) / (a(n) * a(n-1) * C2) 1 + A(n) = (a(n+1) * a(n-2) * C3) / (a(n) * a(n-1) * C2) for some constansts C2, C3. There are similar kinds of examples. One of the simplest is the following. If A(n) = (A(n-1) + A(n-2)) / A(n-3), A(1) = A(2) = A(3) = 1, then A(n) = (a(n) * a(n+7)) / (a(n+3) * a(n+4)) where a(n)=(a(n-1)*a(n-6)*a(n-8) + a(n-2)*a(n-4)*a(n-9)) / (a(n=5)*a(n-10)), a(1) = a(2) = ... = a(10) = 1. which is now sequence A205303 in the OEIS at oeis.org. ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

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[math-fun] Another puzzle
by David Wilson 02 Feb '12

02 Feb '12

What n satisfy 2^n == 3n+2 (mod n^2 - 2n)

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[math-fun] (further) generalized Lambert series with Theta-convergence
by Joerg Arndt 02 Feb '12

02 Feb '12

>From the department of formulas that could possibly be found in some paper from 200 years ago. Semi-recently I made the observation (and posted it here) that sum(n>=0, x*q^n/(1-x*q^n) can be computed as sum(n>=0, q^(n^2+n)*x^(n+1)*(1 - x*q^(2*n+1)) / ((1 - x*q^n)*(1 - q*q^n))) and this allows the fast computation of sums of inverse Fibs (and infinitely more sums of reciprocal order-two linrecs) without splitting into even and odd part (yes, RWG, AIM304 exists). After much effort I found that sum(n>=0, t^n / (1-x*q^n) ) can be computed as sum(n=0, S, (1-x*t*q^(2*n))*(x*t)^n*q^(n^2) / ( (1-x*q^n)*(1-t*q^n)) ) (which btw. is symmetric in x and t). Now this one turns out to be easily obtained by setting a := -b in Fine's "versatile" relation (14-1) (p.15 in "Basic Hypergeometric Series and Applications"). Anybody with one bit of interest in q-things: Buy this book, it's a marvel.

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[math-fun] z contiguity in basic series
by Bill Gosper 02 Feb '12

02 Feb '12

rwg>[...] Is there some way to trick the *N(h,i,j,k/3,(k+1)/3,(k+2)/3,n)* system to give * N(j/3,(j+1)/3,(j+2)/3,k/3,(k+1)/3,(k+2)/3,n)*, even though k is in the numerator and j in the denominator? Yes, you can reciprocate the term ratio by inverting all the matrices with the transformation h->-h-1, i->-1-i, j->-j-1, k->-k-1, n->-n-1, and then cobble together the two systems by renaming variables. Interestingly, for q matrices, (where you have 1-q^n instead of n), you also need to reciprocate q. This reflects the existence of a seventh dimension (unfortunately, the next letter is e) in the q-Rosetta grid providing four-term recurrences relating 3φ2[...,q^e], 3φ2[...,q^(e+1)], etc. I.e., we also need e->-e-1. (That was surprise #4.) For the (generalized, d instead of q) 2φ1[z], the recurrences are merely three-term. The z-bumping recurrence is Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]* z^n)/(QPochhammer[c, q, n]* QPochhammer[d, q, n]), {n, 0, Infinity}] == ((-(c*d - a*b*q^2*z))* Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]*q^(2*n)*z^n)/ (QPochhammer[c, q, n]*QPochhammer[d, q, n]), {n, 0,Infinity}] +q*((-b)*q*z - a*q*z + d + c)* Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]*q^n*z^n)/ (QPochhammer[c, q, n]*QPochhammer[d, q, n]), {n, 0, Infinity}] + (q - c)*(q - d))/(q^2*(1 - z)) (Note the inhomogeneity.) It seems to me the d=q case of this could be derived more traditionally by a limit process on a known 3φ2[1] three-term recurrence, but I've never seen it noted. A corollary is the contiguous (note the q^(-2n)) Heine (q-Gauss) identity Sum[(QPochhammer[a, q, n]*QPochhammer[b, q, n]*c^n)/ (a^n*b^n*QPochhammer[c, q, n]*q^(2*n)*QPochhammer[q, q, n]), {n, 0, Infinity}] == (QPochhammer[c/a, q]*QPochhammer[c/b, q]* (a^2*(b^2*q^3 + (b - 1)*c^2*(b*q - 1) + (b - 1)*b*c*q*(q + 1)) - a*b*c*(q + 1)*(b*q + (b - 1)*c) + b^2*c^2))/ (a^2*b^2*QPochhammer[c, q]*QPochhammer[c/(a*b*q^2), q]*q^3) Special thanks to Carol Cronin @ WRI for her after hours efforts enabling Neil and me to derive these this evening. --rwg

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[math-fun] Artin's conjecture
by Dan Asimov 01 Feb '12

01 Feb '12

The absence of concrete results on Artin's conjecture is amazing. According to Wikipedia there is not even one integer N for which it is known that N is a primitive root modulo p for infinitely many primes p. Not even N = 2. Amazing. --Dan << And that is devastating because (pathetically!) it is not even known that 10 is a prim-root modulo more than a finite set of primes: http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots >> ________________________________________________________________________________________ It goes without saying that .

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