For which numbers $n$ can the sequence $1$ to $n$ be rearranged such that each pair of consecutive terms adds up to a perfect square?
Can this be done on the set of natural numbers as well? Integers? Rationals?
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1Related. – vadim123 Feb 28 '16 at 23:42
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2Could you put these numbers on a circle? ( I mean can these numbers be put so that the sum of the first and last numbers are a square as well?) – S.C.B. Feb 28 '16 at 23:45
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5Am I the only one who finds the animation in the graphic annoying? – Barry Cipra Feb 29 '16 at 00:14
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1@TitoPiezasIII My understanding of the question was that the pairs would work like this, starting from the blue end: [1,3],[3,13],[13,23],[23,26], etc., so that the second number in pair N is the first number in pair N+1. – nhgrif Feb 29 '16 at 00:38
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related or duplicate – Elaqqad Feb 29 '16 at 00:38
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3it is conjectured that there is a solution for all $n>24$ – Elaqqad Feb 29 '16 at 00:41
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@nhgrif: I see. – Tito Piezas III Feb 29 '16 at 00:41
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3@BarryCipra: I think the graphic helped the post get more than 10 upvotes, while this post from *exactly* a year ago and which is very similar got only 3. I guess some readers are quite visual. :) – Tito Piezas III Feb 29 '16 at 00:59
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2https://oeis.org/A090460 – Feb 29 '16 at 09:19
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3Speaking of all natural numbers, see https://oeis.org/A034175. – Ivan Neretin Feb 29 '16 at 12:15
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The animated part seems quite pointless as it is it seems it illustrates strictly nothing. Would it run through the pairs and display the sum there'd be some point to it. – quid Feb 29 '16 at 16:15
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@MXYMXY: Yes, and there are $57$ different ways to put them in such a circle. Kindly see answer below. – Tito Piezas III Feb 29 '16 at 18:00
1 Answers
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(Just to summarize things so people don't have to jump from MSE, MO, OEIS, SO.)
This is a rather interesting question, but there are two previous MSE posts that have already covered it. Post 1 (MSE) asks for which $n$ we can arrange {$1,2,\dots n$} so that the sum $S^k$ of every two adjacent numbers is a square (or $k=2$). A commenter pointed A090461 hence,
$$n = 15,16,17,23,25,26,27,\dots,\infty$$
so it is conjectured for all $n>24$. That, in turn, was inspired by Post 2 (MSE) which was the general case, but focused on sums $S^k$ for $k>2$. For $k=3$, the OP gave an example as $n=305$.
Post 3 (MO) gives an example for $k=4$ as $n=9641$. It was also a cyclic arrangement; that is, the first and last entries also have a sum $S^k$.
P.S. Re MYXMYX's question here if there is a cyclic arrangement for $n=35$ for squares, MJD found there are a whopping $17175$ possible arrangements, so chances are good. By the update below, OEIS says there are $57$ ways to do it.)
Tito Piezas III
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(Update.) Let $n$ have all adjacent pair-sums as $S^2$. Then the number of circular arrangements $m_c$ starting with $n=32$ is given by A071984 as, $$\begin{array}{|c|c|}\hline n&\text{# of};m_c\32&1\33&1\34&11\35&57\ \hline\end{array}$$ – Tito Piezas III Feb 29 '16 at 17:58