i wrote
if 0 is not in D , then it's possible to get elements of S that end with 0 . this may not always occur, but it seems likely if the set S is reasonably large. the rightmost 0's double with each squaring and do not affect the other digits. thus, keeping the rightmost part of the splitting, we get arbitrarily large numbers, unless a splitting digit occurs as the rightmost non-zero digit.
the rightmost non-zero digit converges to an idempotent modulo b (the base). so if all of these idempotent are in the splitting set D , that prevents this process from producing arbitrarily large numbers. the first interesting case appears to be b = 3 , D = { 1 } . i've tried several choices for the initial number in S , and so far all have yielded only finitely many elements in S . in this case, elements in S have only 0's and 2's in their ternary representations ("cantor numbers"??), except perhaps the starting number. from the squares of cantor numbers, we excise 0-2 strings, and can safely eliminate any rightmost 0's because they'll get removed at the next iteration. most of the resulting numbers seem to be smaller than the previous number, but there are exceptions, such as 2 0^n 2 0^{n+1} 2 (where exponents indicate that many repetitions) whose square produces this palindromic substring: 22 0^{n-1} 202 0^{n-1} 22 is there an initial number for which S is infinite in this case (b = 3, D = {1}) ? rich: ) > do you know of any examples where 0 is not in D , and you can prove ) > that S is infinite? ) Sure. Base 4, D = {3}, starting value {11001}. ) The right ends of successive squarings will be xxx11[0*]1 and xxx22[0*]1. ) This is an infinite set, as the string of internal 0s grows by one 0 with ) each 4th-powering. this generalizes to any base that's a power of 2 . any other examples where we can show that S is infinite? (a trivial one is to take the set of splitting digits to be empty.) another question: is there an example of an initial number A , and a base b , and set of splitting digits D , such that the set S is finite, but for the same A and b , but for a larger set D , the set S is infinite? too many questions, not enough answers! mike
Michael Reid wrote:
the rightmost non-zero digit converges to an idempotent modulo b (the base).
Can't there be cycles?... sure, 2*2=4 and 4*4=22, mod 7. (The only idempotents mod any prime p are 0 and 1, of course.) --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On Thu, 12 Jan 2006, Michael Reid wrote:
too many questions, not enough answers!
How about this one? Does every positive integer K appear in some square? That is, given K are there integers x,y and N such that N^2 = xKy? Integers may be expressed in any base and xKy is concatenation of x, K and y. --Edwin
Thu, 12 Jan 2006 23:15:26 -0500 (EST) Edwin Clark <eclark@math.usf.edu> On Thu, 12 Jan 2006, Michael Reid wrote:
too many questions, not enough answers!
How about this one? Does every positive integer K appear in some square? That is, given K are there integers x,y and N such that N^2 = xKy? Yes, unless my early-morning-brain is muddled. In any base b, let x be ceiling(log_b (K)), and let d = ceil(fractional_part(K/2)) [in other words if K is odd, then d is a base b digit > b/2, else d < b/2]. Now, let N = 1 + floor(K/2)*b^{x+2} + d*b^{x+1} + 1. N^2 will be (K/2)^2 shifted 2x+4, K will appear at the {x+2} position, possibly followed by a single digit, then there will be a long string of zeros ending in 1. Integers may be expressed in any base and xKy is concatenation of x, K and y. --Edwin _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
[Minor typo, but in case it's not obvious: N should have only had +1 once.] Fri, 13 Jan 2006 08:11:30 EST Michael B Greenwald <mbgreen@central.cis.upenn.edu> Thu, 12 Jan 2006 23:15:26 -0500 (EST) Edwin Clark <eclark@math.usf.edu> On Thu, 12 Jan 2006, Michael Reid wrote: > too many questions, not enough answers! How about this one? Does every positive integer K appear in some square? That is, given K are there integers x,y and N such that N^2 = xKy? Yes, unless my early-morning-brain is muddled. In any base b, let x be ceiling(log_b (K)), and let d = ceil(fractional_part(K/2)) [in other words if K is odd, then d is a base b digit > b/2, else d < b/2]. Now, let N = 1 + floor(K/2)*b^{x+2} + d*b^{x+1} + 1. N^2 will be (K/2)^2 shifted 2x+4, K will appear at the {x+2} position, possibly followed by a single digit, then there will be a long string of zeros ending in 1. Integers may be expressed in any base and xKy is concatenation of x, K an d y. --Edwin _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Fri, 13 Jan 2006, Michael B Greenwald wrote:
Does every positive integer K appear in some square? That is, given K are there integers x,y and N such that N^2 = xKy?
Yes, unless my early-morning-brain is muddled. In any base b, let x be ceiling(log_b (K)), and let d = ceil(fractional_part(K/2)) [in other words if K is odd, then d is a base b digit > b/2, else d < b/2]. Now, let N = 1 + floor(K/2)*b^{x+2} + d*b^{x+1}
That's nice! I figured there must be a simple way to do it. And if b is even, it is even simpler since then we can replace K by Kb and and let d = 0 so N is simply (K/2)*b^{x+2} + 1. At any rate, I think this indicates that solving David's problem may not be easy. --Edwin
participants (4)
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Edwin Clark -
Michael B Greenwald -
Michael Kleber -
Michael Reid