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I read a book called Things to Make and Do in the Fourth Dimension, written by Australian mathe-matician and comedian, Matt Parker. In one part of the book, I remember him explaining about a conjecture such that for all $n\geqslant 89$, you can arrange the elements of the set $\{1, 2,\ldots,n\}$ in a certain way where each adjacent pair of elements sums to a squared number. For example, let $n = 17$. Al-though $17<89$, below is a good example to demonstrate what I mean:

We have the set $\{1, 2, 3,\ldots, 17\}$ which can also be written as $\mathbb{N}_{\leqslant 17}$. How can we order each number in a certain way such that every adjacent pair of numbers in the ordered sequence add up to a squa-red number? Well, we order it like so:

Let $S_{17} = [\ldots]$ be our sequence that orders the elements from $\{1, 2, 3,\ldots, 17\}$ in a certain way as mentioned in the foregoing then, $$S_{17} := \big[17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16\big].$$ Here, $17 + 8 = 5^2$, $8 + 1 = 3^2$, $1 + 15 = 4^2,\ldots$

The sequence in the sandbox above is a special case where it has $17$ elements and begins with $17$. Take the sequence $S_{16}$ then this sequence ends with $16$. In fact, it is exactly the same as $S_{17}$ except it does not start from $17$, but $8$ instead. However, the sequence $S_{18}$ does not exist, and there are many sequences with this property that do not exist. The conjecture is interesting because if true, it will prove that there are only finitely many sequences $S_n$ that do not exist, also proving the contrapo-sitive.

I did some research and it seems like this is true for $89\leqslant n \leqslant 300$ thus far, but I do not know the name of this conjecture. Does it even have a name? I also haven’t stumbled across any attempts of proving this conjecture. Can it be done? I guess that is two questions, then.

Thank you in advance.

Mark S.
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Mr Pie
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  • Thanks @GTonyJacobs for the edit :) – Mr Pie Feb 05 '18 at 20:10
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    Related but different question: https://math.stackexchange.com/questions/1168983/arranging-numbers-from-1-to-n-such-that-the-sum-of-every-two-adjacent-number – Patrick Stevens Feb 05 '18 at 21:17
  • @PatrickStevens thank you very much for that :) $\dots$ However do you know the name of this conjecture by any chance? – Mr Pie Feb 07 '18 at 06:16
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    The meaning of the title is impossible to decipher. – Did Mar 05 '18 at 11:58
  • @Did there exists infinitely many of the special sequences that orders elements from $1$ to $n$, for all $n\geqslant 89$. These special sequences, denoted as $S_n$, are defined above. – Mr Pie Mar 05 '18 at 22:31
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    If you say so. Anyway, the title does not say that (actually it is difficult to know what the title is really saying). Please modify. – Did Mar 06 '18 at 06:55
  • @Did ok, I will. Thank u for telling me, so I know that maybe other users may not understand the title :) – Mr Pie Mar 06 '18 at 09:42
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    I'm with @Did. However, if the title is meant as a negative advertisement, it worked on me :) – yo' Mar 06 '18 at 10:08
  • @yo' so you understand the title? I just don't want to make it too wordy, is all. But I guess that if nobody understands it, what's the point anyway? – Mr Pie Mar 06 '18 at 10:31
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    @user477343 If you really want to know my frank opinion: The title is stupid. It uses non-standard notation ($\mathop{\exists}\limits^\infty$), it has the operators in the wrong order (I suppose you wanna say "for all $n\geq89$ there exist infinitely many") and it contains undefined symbols ($S_n$) that have many meanings in many contexts, and also, it's a title that uses math notation in a place where none is appropriate. Actually, why do you write $\forall n\in \mathbb{N}_{\geq89}$ for something as simple as $\forall n\geq89$? – yo' Mar 06 '18 at 10:46
  • "there are many sequences with this property that do not exist." Sorry, but that's a self-contradicting sentence. What do you really mean? – Gerry Myerson Mar 06 '18 at 11:48
  • @GerryMyerson, I figured from context that they meant "there are many numbers $n$ such that it is not possible to arrange the numbers from $1$ to $n$ in a list with pairs of adjacent numbers summing to a square". (In fact, it seems there are 20 such numbers.) – Mark S. Mar 06 '18 at 11:54
  • @Mark, you may be right, but I'd like to see OP edit the body of the question into something that makes sense. – Gerry Myerson Mar 06 '18 at 12:18
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    https://math.stackexchange.com/questions/1676413/for-which-n-can-1-2-n-be-rearranged-so-that-the-sum-of-each-two-adja is very closely related, and a number of other questions were asked on MSE a while back. – Mark S. Mar 06 '18 at 14:16
  • @MarkS. Thank you for the edit :) – Mr Pie Mar 07 '18 at 03:09

1 Answers1

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On the OEIS sequence for counterexamples, R.K.Guy claims that "the problem originated (for n = 15) with Bernardo Recamán Santos of Colombia.", but does not give a name for the problem.

On a Numberphile episode, Matt Parker himself has given this problem the name "The Square-Sum Problem".

I have not evaluated the proof myself, but there is what amounts to a program and description (a preprint?) purporting to prove this conjecture at this mersenneforum post (and the preceding posts help for context) by Robert Gerbicz.

For the same problem but requiring the numbers to make a loop (so that the first and last numbers sum to a square as well), the numbers of solutions for small $n$ are documented in OEIS A071984. Guy attributed that version of the problem to Joe Kisenwether. In fact, the above post by Robert Gerbicz handles the corresponding conjecture for loops (loops exist for all $n\ge32$) as well.

There is a related MSE question and corresponding MO question about perfect powers with or without loops in more generality.

Mark S.
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