Wikipedia:Reference desk/Archives/Mathematics/2020 November 23

Mathematics desk
< November 22 << Oct | November | Dec >> November 24 >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


November 23

edit

Area->Perimeter Sequence.

edit

Musing about another sequence defined as follows. if A(k) is a number A(k+1) is the length of the Perimeter of the rectangle (including squares) of area A(k) which has whole number sides and is closest to a square. So if A(1) = 5, since the rectangle closest to a square is 1x5 which has a perimeter of 12, so A(2) =12. Similarly

  • A(1) = 100, rectangle = 10x10, so A(2)=40
  • A(1) = 7, rectangle = 7x1, so A(2) = 16
  • A(1) = 27, rectangle = 9x3, A(2) = 24.

There are two stable points: 16 (rectangle = 4x4, so back to 16) and 18 (rectangle = 6x3). I think that it can be proved that any sequence eventually ends up there, but I'm not sure of a clean method of proving it.Naraht (talk) 15:02, 23 November 2020 (UTC)[reply]

May I suggest that you change your notation? Instead, define   to be the perimeter of the rectangle of area n that is closest to a square. So  , etc. And you are asking what happens to the iterates  .
Here is what you should do: first, check the result for all small values of n (up to 100 will certainly do) and then assume n > 100 (or whatever). Second, since   is always even, you may as well start from an even number. Third, show that if n is twice a composite number and larger than 100, then  . Fourth, suppose that n is twice a prime number p, so  . If you are lucky and   is not prime, then show  . If you are unlucky and p is a twin prime, then   but now   must be composite, so show that  . Finally, this shows that the trajectory   contains a decreasing subsequence that inevitably must include a number less than 100; declare victory. --JBL (talk) 16:44, 23 November 2020 (UTC)[reply]
There is a cycle 22 = 2×11 → 2×(2+11) = 26 = 2×13 → 2×(2+13) = 30 = 5×6 → 2×(5+6) = 22. For a similar problem (and appropriate terminology), see Collatz conjecture.  --Lambiam 01:28, 24 November 2020 (UTC)[reply]
Of course, the proof strategy I outlined would also suffice to prove that every point eventually goes to one of the fixed points or to that circuit. --JBL (talk) 04:12, 27 November 2020 (UTC)[reply]
Yes. Even stronger, it can be used to reveal without hardly any trial which circuits are possible. For example, assume   is twice a twin prime  , so   for some  . Then
 
A 3-cycle starting from twice a twin prime is therefore possible only if  , or  . Indeed, 22 is twice the twin prime 11: Bingo! The fixed points can likewise be found algebraically.  --Lambiam 08:17, 27 November 2020 (UTC)[reply]
Cute! --JBL (talk) 16:13, 27 November 2020 (UTC)[reply]