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Let $ [n] $ be the set $ \{1,2,\ldots n\}$.

A summoid is a subset $ A \subset [n] $ of the form $ \{a,b,a+b\} $ (you can choose a better name, if it doesn't exist already).

Now, I developed by accident this simple result:

There is no partition of $ [9] $ into (disjoint) summoids.

I want to ask the following questions:

  1. Is it true for general $ [3k] $ when $ k > 1 $? According to a computer program I found that it is true for $ [12] $, but it seems my method for $ [9]$ can't be applied to the general case (maybe my program for $[12]$ isn't trustworthy). According to the comment below by R. van Dobben de Bruyn, there is no such partition when $ k \equiv 2,3 \pmod 4$. According to the comment below by Gerhard Paseman, there is a counterexample for $ k = 5 $ that extends also to $ n = 3 \cdot 4^k $ and to $ n = 3 \cdot (1+4^k) $.

  2. Is it useful for something?

  3. Is there any study of such results?
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    $\begingroup$ Just a comment: $[3]$ is itself a summoid, so you should at least assume $k>1$. $\endgroup$ – Daniel Litt Aug 10 '17 at 15:25
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    $\begingroup$ There is a local obstruction at $2$: the sum of the elements of $\{a,b,a+b\}$ is even, hence $\frac{3}{2}k(3k+1)$ (the sum of the elements of $\{1,\ldots,3k\}$) should be even as well. This already proves that there is no such partition when $k \equiv 2, 3 \pmod{4}$. But this argument doesn't tell us anything about the other two cases $k \equiv 0, 1 \pmod{4}$. $\endgroup$ – R. van Dobben de Bruyn Aug 10 '17 at 15:42
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    $\begingroup$ I get 5,7,12-3,8,11-1,9,10-2,4,6, and a similar decomposition for n=15. Perhaps this pattern can be extended for n=24 and 27? Gerhard "Working From The Inside Out" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 10 '17 at 16:08
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    $\begingroup$ I can extend this to n =3*4^k and n+1, but n=3*2^k is a challenge for k odd Perhaps someone can prove it impossible? Gerhard "Hard To Leave Construction Mode" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 10 '17 at 16:20
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    $\begingroup$ Shorthand with error. I can do it for 3*4^k and 3*(1+4^k) for positive k. I prefer to keep symbols to a minimum, so I define a variable and reuse it. Gerhard "Sort Of Like Macro Programming" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 10 '17 at 16:39
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I have the following results:

  • N = 12: (1, 11, 12) (3, 7, 10) (4, 5, 9) (2, 6, 8)

  • N = 15: (1, 14, 15) (3, 10, 13) (4, 8, 12) (5, 6, 11) (2, 7, 9)

  • N = 24: (1, 23, 24) (2, 20, 22) (5, 16, 21) (6, 13, 19) (7, 11, 18) (8, 9, 17) (3, 12, 15)

  • N = 27: (1, 26, 27) (2, 23, 25) (4, 20, 24) (7, 15, 22) (8, 13, 21) (9, 10, 19) (6, 12, 18) (3, 14, 17) (5, 11, 16)

and so on.

My guess would be that there are always (a lot of) solutions for any $k \equiv 0, 1\mod 4$. In my opinion, this is some kind of Goldbach-ish problem. It is likely to get some probabilistic heuristic, but may be difficult to prove.


For reference, here I add the number of different solutions for each $k$.

  • N = 12: 8 solutions;
  • N = 15: 21 solutions;
  • N = 24: 3040 solutions;
  • N = 27: 20505 solutions;

and so on.


EDIT:

Thanks to OEIS, we now have some reference.

In this paper: http://oeis.org/A104429/a104429.pdf, the existence of solutions is discussed in detail in section I.3 (starting from page 22).

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  • $\begingroup$ Check out a284737 at oeis.org. Does this match up? Gerhard "It May Be Langford Disguised" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 11 '17 at 3:21
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    $\begingroup$ A108235 is better. Gerhard "It Has Some More References" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 11 '17 at 3:24
  • $\begingroup$ I think a104429 is about arithmetic progressions, not about summoids. Gerhard "We May Need Different Terminology" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 11 '17 at 3:53
  • $\begingroup$ Ah, I see it handles both. Gerhard "Thanks For Posting The Link" Paseman, 2017.08.10. $\endgroup$ – Gerhard Paseman Aug 11 '17 at 3:56

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