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

= November 23 =

Area->Perimeter Sequence.
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)


 * May I suggest that you change your notation? Instead, define $$A(n)$$ to be the perimeter of the rectangle of area n that is closest to a square.  So $$A(1) = 4, A(5)  = 12, A(100) = 40$$, etc.  And you are asking what happens to the iterates $$A(n), A(A(n)), \ldots$$. 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 $$A(n)$$ 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 $$A(n)<n$$.  Fourth, suppose that n is twice a prime number p, so $$A(p) = 2(p + 2)$$.  If you are lucky and $$p + 2$$ is not prime, then show $$A(A(n)) < n$$.  If you are unlucky and p is a twin prime, then $$A(A(n)) = 2(p + 4)$$ but now $$p + 4$$ must be composite, so show that $$A(A(A(n))) < n$$.  Finally, this shows that the trajectory $$A(n), A(A(n)), \ldots$$ contains a decreasing subsequence that inevitably must include a number less than 100; declare victory. --JBL (talk) 16:44, 23 November 2020 (UTC)


 * 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)


 * 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)
 * Yes. Even stronger, it can be used to reveal without hardly any trial which circuits are possible. For example, assume $$n$$ is twice a twin prime $$p \ge 5$$, so $$p = 6q+5$$ for some $$q$$. Then
 * $$A(A(A(n))) = A(2(p+4)) = A(6(2q+3)) \le 2(2q+9) = \frac{n+44}{3}.$$
 * A 3-cycle starting from twice a twin prime is therefore possible only if $$n \le \frac{n+44}{3}$$, or $$n \le 22$$. Indeed, 22 is twice the twin prime 11: Bingo! The fixed points can likewise be found algebraically. --Lambiam 08:17, 27 November 2020 (UTC)
 * Cute! --JBL (talk) 16:13, 27 November 2020 (UTC)