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

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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 $p\geq 5$, so $p=6q+5$ for some $q$. Then

$A(A(A(n)))=A(2(p+4))=A(6(2q+3))\leq 2(2q+9)={\frac {n+44}{3}}.$

A 3-cycle starting from twice a twin prime is therefore possible only if $n\leq {\frac {n+44}{3}}$, or $n\leq 22$. 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)