February 2005 - mathfun@mailman.xmission.com

[math-fun] new Mesenne prime
by Richard Schroeppel 26 Feb '05

26 Feb '05

Goerge Woltman reports: > Tony Reix has verified the new Mersenne Prime! I'll have more details once the > press release and web site changes are ready. The exponent isn't released yet, but the new number is < 10M digits. Rich rcs(a)cs.arizona.edu

1 0

[math-fun] Re: Bob Dylan
by James Propp 25 Feb '05

25 Feb '05

>Are there any songs that mention mathematicians other than Dylan's "Tangled >up in blue"?: There's Christine Lavin's song "Doris and Edwin: The Movie" (http://www.christinelavin.com/00081507dorisandedwin.html) "Now I'm not much of a mathematician But even I can see In a building of 37 floors And 4000 employees Her chances of running into him Were really not too good Even though she hung out at the canteen And rode the elevator every spare minute she could." (The word "mathematician" makes a repeat appearance in the mock-tragic denoument.) Jim

2 1

[math-fun] True So Far
by Eric Angelini 25 Feb '05

25 Feb '05

Hello math-fun and seqfan, I've just sent this to the OEIS : 10 12 13 14 15 16 17 18 19 20 23 24 25 26 27 28 29 30 34 35 36 37 38 39 40 45 46 47 48 49 50 56 57 58 59 60 67 68 69 70 78 79 80 89 90 90 102 103 104 105 106 107 108 109 112 113 114 115 116 117 118 119 123 124 125 126 127 128 129 134 135 136 137 138 139 145 146 147 148 149 156 157 158 159 167 168 169 178 179 180 189... [more hand calculated terms here (hope no errors)]: http://www.cetteadressecomportecinquantesignes.com/TrueSoFar.htm Description : "True so far"-sequence. Last digit of a(n) must be seen as a glyph and preceding digits as a quantity. So "10" reads [one "0"] and "12" [one "2"] -- which are both true statements: there is only one "0" glyph so far in the sequence when [10] is read, and there is only one "2" glyph when [12] is read. The sequence is built with [a(n+1)-a(n)] being minimal and a(n+1) always "true so far". This explains why integers [11], [21], [22], [31], etc. are not in: their statements are false. The nice substring ...1112,1113,1114,1115,1116,1117 1118... appears in the sequence -- which means that so far the whole sequence has used 111 "2", 111 "3", 111 "4", 111 "5", 111 "6", 111 "7" and 111 "8"... Question which ruined my sleep tonight: « Will the sequence ever stop? » ... my intuition says yes... ... could someone compute this and check for some more integers? Thanks, É.

6 7

[math-fun] squares
by Gary McGuire 25 Feb '05

25 Feb '05

Question: You have 9 squares with side lengths 1,2,3,4,5,6,7,8,9. What is the smallest rectangle you can put them in? (smallest meaning smallest in area, and you put them in without overlap) A rectangle with area 1^2+...+9^2 won't work. Is there a standard technique for going at this type of question? I can make a pretty good stab at it, just by trial and error, and I'm fairly convinced I have the best possible, but how does one prove such a thing? Is there a general way to go about this with side lengths 1,2,3,....,n, or even with some random set of squares? Gary McGuire

4 3

[math-fun] Re: Bob Dylan
by Dean Hickerson 25 Feb '05

25 Feb '05

Marc LeBrun wrote: > "I wouldn't even think about playing music if I was born in these times... > I'd probably turn to something like mathematics. That would interest me." > --Bob Dylan Are there any songs that mention mathematicians other than Dylan's "Tangled up in blue"?: http://bobdylan.com/songs/tangled.html Dean Hickerson dean(a)math.ucdavis.edu

1 0

[math-fun] RE: True So Far
by Lior Manor 25 Feb '05

25 Feb '05

Yes, I had a problem with the indices, it's 2024. I just saw your comments in the OEIS and I got the same numbers. ________________________________ From: PlasteredDragon(a)comcast.net [mailto:PlasteredDragon@comcast.net] Sent: Thu 24-Feb-05 23:28 To: Lior Manor; Eric Angelini Cc: math-fun Subject: RE: True So Far Hi Lior, Your list of last terms looks good to me, but I'm curious about your indices. By my count True So Far (in base 10 anyway) only has 2024 terms in it. -- Chuck -------------- Original message -------------- > Hi Eric, > > Looks like a(6078) = 8945 is the last term of your sequence. > According to my calculations the last terms ending with 0-9 digits are: > a(5914) = 5890 > a(6060) = 8201 > a(6064) = 8302 > a(6073) = 8613 > a(6075) = 8734 > a(6078) = 8945 > a(6059) = 7756 > a(6067) = 8563 > a(6051) = 6778 > a(5918) = 5979 > It is easy to see that there will be no additional terms. > You can view the full sequence in sci.misc > (http://groups-beta.google.com/group/sci.misc/msg/aa57ea84b421d2a2) > > Lior Manor > >

1 0

[math-fun] Zeroless powers of 2 and Wilson's conjecture.
by David Wilson 25 Feb '05

25 Feb '05

Since the discussion has again rolled around to the perennial chestnut of whether there are a finite number of zeroless powers of 2, I thought I would again state my perennial umbrella conjecture. To start, if we are given two arbitrary integer sequences each exhibiting exponential growth, a probabilistic argument says that we should expect the sequences to share only a finite number of elements. There are cases where we can prove the number of shared elements is finite (e.g, {2^n} and {3^n}) and others where we know the number of shared elements is infinite (e.g, {Fib(n)} and {Fib(2n)}), but in the absence of evidence either way, the safer assumption is a finite intersection. A b-automatic (b >= 2) set is a set of nonnegative integers whose base-b representations form a regular language, that is, are accepted by a finite automaton and representable by a regular expression. For example, the powers of 2 form a 2-automatic set with the regular expression 10*. Similarly, the zeroless numbers form a 10-automatic set with regular expression (1+2+3+4+5+6+7+8+9)(1+2+3+4+5+6+7+8+9)*. Call two bases b and c compatible if b and c share a common power > 1. Thus binary, octal and hexadecimal are compatible bases because 2, 8 and 16 have the common power 4096. If b and c are compatible, then the b-automatic and c-automatic sets coincide. If b and c are incompatible, the sets that are both b-automatic and c-automatic sets include precisely the finite sets and sets whose first differences are eventually periodic (Cobham 1964). ---------------------------------------------------------- Let B be a b-automatic set with density 0 over the integers. It can be proved that B is either finite or exhibits exponential growth (as a sorted sequence). Let B be a zero-density b-automatic set and C a zero-density c-automatic set. If b and c are compatible, it is easy enough to find B and C having either finite or infinite intersection. For finite intersection, let B = C = S = {1,2,3} (or any other finite set). By virtue of its finiteness, B = C = S is both b-automatic and c-automatic, as well as zero-density, and the intersection of B and C is S, which is finite. For infinite intersection, let B = C = S = {b^n}. B = C = S is zero-density over the positive integers. B is b-automatic, having base-b regular expression 10*. Because c is compatible with b, C = B is also c-automatic. The intersection of B and C is S, which is infinite. If b and c are incompatible, it is easy enough to find B and C with finite intersection. For example, b = 2 and c = 3 are incompatible. Let B = {2^n} and C = {3^n}. B and C are 2-automatic and 3-automatic respectively, having regular expression 10* in their respective bases, and their intersection is {1}, which is finite. On the other hand, I have not been able to find such a B and C with infinite intersection, which leads to the conjecture: Wilson's Conjecture: If b and c are incompatible bases, then zero-density b-automatic set B and zero-density c-automatic set C have finite intersection. (I adopt this conjecture because I have not seen it elsewhere). Since B and C are zero-density automatic sets, they are either finite or exhibit exponential growth. If either is finite, their intersection is cleary so. If both have exponential growth, applying our statistical argument above, we would expect them to have finite intersection. So what Wilson's conjecture really says is that the incompatibility of bases b and c can be taken as evidence that B and C are "relatively random" enough to obey the probabilistic argument that indicate finite intersection. ---------------------------------------------------------- Admittedly, Wilson's conjecture is a yet another generalization of certain problems that are themselves notoriously difficult to prove. However, I think that it is a useful generalization in that (1) it tells us what to expect in a large number of similar problems, and (2) it has number theoretical implications beyond mere digit twiddling. The finity of the zeroless powers of 2 is a straightforward consequence of Wilson's conjecture. The generalization is easy enough to show: Let b and c be incompatible bases, and let d be a base-c digit (0 <= d < c). The powers of b form a zero-density b-automatic set B, while the base-c numbers without digit d form a zero-density c-automatic set C. Wilson's conjecture then implies the intersection of B and C is finite, that is, there are finitely many base-c powers of b witout digit d. Letting b = 2 and c = 10 are incompatible, and we get a finite number of zeroless base-10 powers of 2. If instead of powers of 2, we had chosen a sequence S to be a random sequence of base-10 numbers such that S(n) has the same number of digits as 2^n, S would have had a finite number of zeroless elements with probability 1. So in this case, Wilson's conjecture tells us that in one respect the powers of 2 act like similarly sized random numbers with respect to base 10. More generally, Wilson's theorem says something about the "relative randomness" of incompatible bases in certain contexts. Anyone who has programmed arbitrary precision arithmetic knows that converting between compatible bases is computationally easy while converting between incompatible bases is difficult. To convert between compatible octal and hexadecimal bases, one can simply start at the right, converting each block of 4 octal digits to an equivalent block of 3 hexadecimal digits, and O(n) process. Converting between incompatible binary and decimal is much harder. Changing a single base-2 digit potentially affects arbitrarily many base-10 digits. I'm not even sure that a sub-O(n^2) algorithm exists for binary-to-decimal conversion. This is one aspect of the "relative randomness" of incompatible bases. I am taxed to remember the details of a certain prize problem concerning the divisibility of choice numbers by 4 and 9 (any help finding this problem would be appreciated). The problem asked whether there were a finite number of choice numbers with a certain property, and the conjectured solution was implied by Wilson's theorem. So here was an instance of an independent number theoretical (as opposed to digit twiddling) problem where Wilson's theorem gave the expected result. Another purely number theoretical implication of Wilson's theorem is that for incompatible b and c, |b^x - c^y| <= d has a finite number of solutions. This result, whether true or false, probably has a goodly amount of number theoretical background, which might be reversed engineered to shed light on Wilson's theorem. -- No virus found in this outgoing message. Checked by AVG Anti-Virus. Version: 7.0.300 / Virus Database: 266.4.0 - Release Date: 2/22/2005

1 1

Re: [math-fun] Bob Dylan
by Paul Reiners 25 Feb '05

25 Feb '05

Like Schillinger, Copland, and Stravinsky, Bob has alway been a math/music hacker: http://www.therestisnoise.com/2004/10/bob.html On 2/24/05 2:53 PM, "Marc LeBrun" <mlb(a)fxpt.com> wrote: > "I wouldn't even think about playing music if I was born in these times... > I'd probably turn to something like mathematics. That would interest me." > --Bob Dylan > > > (Either that, or maybe a Victoria's Secret spokesmodel?)

1 0

[math-fun] RE: True So Far
by PlasteredDragon＠comcast.net 24 Feb '05

24 Feb '05

Hi Lior, Your list of last terms looks good to me, but I'm curious about your indices. By my count True So Far (in base 10 anyway) only has 2024 terms in it. -- Chuck -------------- Original message -------------- > Hi Eric, > > Looks like a(6078) = 8945 is the last term of your sequence. > According to my calculations the last terms ending with 0-9 digits are: > a(5914) = 5890 > a(6060) = 8201 > a(6064) = 8302 > a(6073) = 8613 > a(6075) = 8734 > a(6078) = 8945 > a(6059) = 7756 > a(6067) = 8563 > a(6051) = 6778 > a(5918) = 5979 > It is easy to see that there will be no additional terms. > You can view the full sequence in sci.misc > (http://groups-beta.google.com/group/sci.misc/msg/aa57ea84b421d2a2) > > Lior Manor > >

1 0

[math-fun] Bob Dylan
by Marc LeBrun 24 Feb '05

24 Feb '05

"I wouldn't even think about playing music if I was born in these times... I'd probably turn to something like mathematics. That would interest me." --Bob Dylan (Either that, or maybe a Victoria's Secret spokesmodel?)

1 0