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Re: [math-fun] Add first diff of digits and repeat
by Michael Reid 02 Dec '07

02 Dec '07

> Let's say a(n) = 329441. How do we build a(n+1)? > First, sum, from left to right, the absolute differences of a(n)'s digits: > a(n) = 3 2 9 4 4 1 > abs. diff = 1+7+5+0+3 = 16 > Now add this sum to a(n) so to produce a(n+1): > 329441 + 16 = 329457 = a(n+1) > And repeat: [ ... ] > The sequence stops when all digits off a(n) are the same, of course: > If we start with 18, for instance, we have: > S = 18, 25, 28, 34, 35, 37, 41, 44, 44, 44, ... > Are there integers leading to infinite sequences? no, you can never increase the number of digits. for 1-digit numbers, this is clear. for 2-digit numbers, you add <= 9 at every step, so if you reach 3-digits, the previous number is 9a for some digit a , but this would make the next term 99 . for 3-digit numbers, you add <= 18 at every step, so if you reach 4-digits, the last 3-digit number is in the range 982 - 999 . but for these numbers, you add at most 10 . thus the last 3-digit number is at least 990 . for any number of the form 99a , the next term is 999 . for n >= 4 digits, at each step, you increase by at most 9 (n - 1) , which is less than 10^(n-2) . if you reach n + 1 digits, the last n-digit number has the form 99abc...x . for such a number, you increase it by the same amount as you increase 9abc...x , which means that the sequence is the same, except for putting an extra digit of 9 in front. since n-1 digit numbers can't reach n digits, the same holds for n digit numbers by induction. mike

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Re: [math-fun] what's special about 786433 ?
by Michael Reid 02 Dec '07

02 Dec '07

> what is the least prime p, such that mod(p, k)=1 for 18 values of k, k<= floor( > sqrt(n) ) ? > Answer: n=786433 > > Why would, among the first 100,000 primes, > it be so rare for p mod k to occur just 18 times? you are counting the number of divisors, d , of p-1 that are in the range 1 < d <= sqrt(p) . if you were to include 1 in the range, you would have half the divisors of p-1 , with the convention that if p-1 is a square, you include the divisor sqrt(p-1) . this means that your count is equivalent to floor((number_of_divisors(p-1) - 1)/ 2) . in order to get an 18 , you need p - 1 to have either 37 or 38 divisors. this means p - 1 has the form a^36 or a^18 b , where a and b are primes. since a^36 + 1 is divisible by a^4 + 1 , the first form cannot occur if p is prime. mike

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[math-fun] what's special about 786433 ?
by wouter meeussen 02 Dec '07

02 Dec '07

what is the least prime p, such that mod(p, k)=1 for 18 values of k, k<= floor( sqrt(n) ) ? Answer: n=786433 Why would, among the first 100,000 primes, it be so rare for p mod k to occur just 18 times? the count {k , frequency of mod(p,k)=1 } among the first 10^5 primes is {0,2},{1,5380},{2,2897},{3,15798},{4,823},{5,10186},{6,232}, {7,17325},{8,1398},{9,2239},{10,28},{11,12585},{12,9},{13,602}, {14,550},{15,9327},{16,3},{17,2828},{18,1},{19,2043},{20,149}, {21,36},{22,22},{23,6659},{24,42},{25,10},{26,236},{27,422}, {28,0},{29,766},{30,0},{31,2190},{32,13},{33,0}, ... does it mean anything special that 786433 mod k = 1 for k= {2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768} and each of these factors into 2's and 3's? (Pierpont primes A005109) This 786433 also occurs in A070806 (and on the 19th place). Hmmm. Wouter. ------------------------------------ for Mathematica Affectionado's: it2=Table[{Floor[Sqrt[Prime[n]]],Count[Mod[Prime[n],Range@Floor[Sqrt[Prime[n]]]],1]},{n,100000}]; {#,Count[it2,{_,#}]}&/@Range[0,Max[Last/@it2]] ------------------------------------

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[math-fun] Add first diff of digits and repeat
by Eric Angelini 01 Dec '07

01 Dec '07

Hello Math-fun, Let's say a(n) = 329441. How do we build a(n+1)? First, sum, from left to right, the absolute differences of a(n)'s digits: a(n) = 3 2 9 4 4 1 abs. diff = 1+7+5+0+3 = 16 Now add this sum to a(n) so to produce a(n+1): 329441 + 16 = 329457 = a(n+1) And repeat: 329457 + 16 = 329473 329473 + 20 = 329493 329493 + 24 = 329517 329517 + 22 = 329539 etc. The sequence stops when all digits off a(n) are the same, of course: If we start with 18, for instance, we have: S = 18, 25, 28, 34, 35, 37, 41, 44, 44, 44, ... Are there integers leading to infinite sequences? If yes, what would be the smallest one? Best, É. (hoping, as usual, that this is not old hat)

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[math-fun] The 3 terms after "3" cannot be 4, 5 or 6.
by Eric Angelini 29 Nov '07

29 Nov '07

Hello Math-fun, (thanks for the private answers to my earlier posts on other topics) --- Start the sequence S with a(1)=1. The next term is the smallest one not yet present in S with the rule: the a(n) terms immediately after a(n) must be < a(n) or > 2*a(n). Example: The 7 terms immediately after 7 cannot be 8, 9, 10, ... 14. The 15 termrs immediately after 15 cannot be 16, 17, 18, ..., 30. ... The term immediately after 1 cannot be 2. S starts like this (I think): s = 1, 3, 2, 7, 15, 4, 31, 63, 127, 255, 5, 8, ... Could please someone interested compute a hundred or so terms and find a formula giving the n-th term? I hope this is not old hat, Best, É.

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Re: [math-fun] Re: math-fun Digest, Vol 57, Issue 17
by Mike Stay 28 Nov '07

28 Nov '07

On Nov 28, 2007 11:10 AM, Steve Witham <sw(a)tiac.net> wrote: > >From: "Fred lunnon" <fred.lunnon(a)gmail.com> > > > > > http://math.ucr.edu/home/baez/square.root.html > > > >I've read this webpage three times, and I still don't understand it. > > I didn't make much sense of it... I feel compelled to say that this is, perhaps, the least illuminating of all Dr. Baez' writings. He's actually well-known for the clarity of his presentations, and works very hard to explain things so well that what once was "mysterious" looks almost trivial in retrospect. -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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RE: [math-fun] The Axiom of Choice for roots of z^2 + 1
by Dan Asimov 28 Nov '07

28 Nov '07

<< If i and -i were treated as indistinguishable, wouldn't this destroy the map of the complexes onto the XY plane? >> << I think . . . Rather than addition, start with . . . i=-i . . . >> -------------------------------------------------- My point about the complex roots of X^2 + 1 being indistinguishable was not to say that in any sense they are equal*. Just that there is no way of specifying one rather than the other. --Dan __________________________________________________ * This might be a feasible approach, but it would, I agree, sure make a mess of the complex plane.

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[math-fun] smallest "non-Web" positive integer?
by James Propp 28 Nov '07

28 Nov '07

A Google search indicates that the number 184362536 (chosen more or less at random, after a half dozen failed attempts) does not occur anywhere on the Web. What is the smallest positive integer with this property as of right now? Jim Propp

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Re: [math-fun] Re: mirrors (was The Axiom of Choice for roots of z^2 + 1)
by Fred lunnon 28 Nov '07

28 Nov '07

On 11/26/07, Steve Witham <sw(a)tiac.net> wrote: > ... > Mirrored pairs are called "right-handed" and "left-handed"-- > is that the "badly-formulated linguistic trope" you mean, Fred? Yes indeed. That's what I had in mind when I queried to association of left-handed writing with left-handed coordinates. > Or is it the whole treatment of "swap" as a verb, or...? This could be queried as well, I suppose --- there's an implicit "modulo proper isometries", as several people pointed out. Of course, with a little surgery, the mirror could be implanted along the observer's plan of (approximate) bilateral symmetry, when it really does "swap left with right" ... WFL > >From: Bill Thurston <wpt4(a)cornell.edu> > > > > Ultimately, I mellowed. I realized that people were unconvinced by > >each other's explanations because they weren't actually confused > >about what a mirror does. There's not an actual question here without > >words. Sure --- but understanding takes place at several levels. If you are caught out by the question --- as most people are, unless they've made a specific study of the Euclidean group --- then I claim you have something to gain by grasping exactly why. WFL

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[math-fun] Re: math-fun Digest, Vol 57, Issue 17
by Steve Witham 28 Nov '07

28 Nov '07

>From: "Fred lunnon" <fred.lunnon(a)gmail.com> > > > http://math.ucr.edu/home/baez/square.root.html > >I've read this webpage three times, and I still don't understand it. I didn't make much sense of it, but maybe he's leading up to the idea that while you can flip the number line with two rotations, trying to build a complex conjugate out of rotations is a blind alley? --Steve

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