"All in" anti-Pascal triangle; (This could be a legendary weapon)

Ladies and Gentlemen, here it comes, from the deepest pit of hell, the LINE that divides mortals from Demigods;

A problem so hardcore, it's reputation better be unknown, so people don't quit before trying. Let me be clear, if you can solve this, you won't be a mortal anymore, you would be a DemiGod or a Goddess! Your soul would be near ultimate perfectness.

An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it.

For example, the following array is an anti-Pascal triangle with four rows which contains every integer from 1 to 10.

                                                                           4
                                                                        2    6
                                                                    5     7     1
                                                                 8     3     10    9

Does there exist an anti-Pascal triangle with 2018 rows which contains every integer from 1 to 1 + 2 + · · · + 2018?

Side note: Note that the exact amount of int variables of the triangle is equal to the summation of every integer from 1 to n = amount of rows.

Note by Xizlon Tho
2 weeks, 4 days ago

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If you change all the even numbers to \(0\) and all the odd numbers to \(1\), you can use a computer program to at least eliminate some larger triangles. (Two evens or two odds make an even, and one even and one odd make an odd, and we would require the whole triangle to have round(\(\frac{n(n + 1}{4})\) odd numbers for \(n\) rows.)

There are no possible even-odd bottom row combinations for \(n = 6\) or \(n = 14\) that will create triangle with round(\(\frac{n(n + 1}{4})\) odd numbers, so there are no anti-Pascal triangles with \(6\) or \(14\) rows.

Unfortunately, this method did not eliminate \(n = 22\) like I was hoping. I was able to test \(n \leq 22\) before there were too many possibilities for my computer program, and \(n = 6\) and \(n = 14\) were the only \(n\)-values that I could eliminate.

David Vreken - 1 week, 3 days ago

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I'm starting to recover; Mr. David Vreken.

Thanks to you, now I know there might be no solution for n=2018 rows, but, how can that be proven?

""A problem so hardcore, it's reputation better be unknown"" R/ Why?? Because this IS the hardest problem ever in the International Mathematical Olympiad, it´s from this year(2018) Olympiad; It was proposed by one of the fathers of the Cryptographic field, Martin Gardner.

That is why, I was being literal.

Xizlon Tho - 1 week, 1 day ago

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WOW, oh boy, If I were not with the flu, I would be deeply analyzing your approach; It looks very smart :D But, I do have the flu right now, :v

Xizlon Tho - 1 week, 1 day ago

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I used a computer program, and:

for \(n = 3\) there are \(4\) possible bottom rows: \((1, 3), (2, 3), (3, 1), (3, 2)\)

for \(n = 4\) there are \(8\) possible bottom rows: \((6, 1, 10, 8), (6, 10, 1, 8), (8, 1, 10, 6), (8, 3, 10, 9), (8, 10, 1, 6), (8, 10, 3, 9), (9, 3, 10, 8), (9, 10, 3, 8)\)

for \(n = 5\) there are \(2\) possible bottom rows: \((6, 14, 15, 3, 13), (13, 3, 15, 14, 6)\)

Unfortunately, there were too many possibilities to test \(n > 5\).

David Vreken - 1 week, 3 days ago

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