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This is a purely recreational question, just curious if anything out there exists about these.

Let $x^\triangle$ denote the triangular number $x(x+1)/2$, then by a triangular equation, I mean an equation of the form $$ax^\triangle+bx+c=0,$$ where $a\neq 0$.

The analogue of completing the square for these equations is $x^\triangle + bx = (x+b)^\triangle - b^\triangle$, so solving the monic equation in general gives $$x^\triangle + bx+c=0\iff(x+b)^\triangle-b^\triangle+c = 0\iff x=-b+\sqrt[\triangle]{b^\triangle-c}.$$ On the other hand, some ugly facts about the "triangle" operation are that it doesn't distribute nicely over products. In fact, we have

  • $(a+b)^\triangle = a^\triangle + ab + b^\triangle$,
  • $(ab)^\triangle = 2a^\triangle b^\triangle - a^\triangle b - b^\triangle a + ab$,
  • $(1/b)^\triangle = b^\triangle / (b+2b^\triangle (1-b)) = b^\triangle/(b+6(b^{\triangle_3}-b^\triangle))$, where $x^{\triangle_3} = x(x+1)(x+2)/6$ is the $x$th tetrahedral number.
J. W. Tanner
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Luke Collins
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    Have you seen the falling / rising factorial? It's similar to what you're considering, just a multiple times of it. – Calvin Lin Nov 30 '20 at 22:16
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    Not quite what you are asking, but Gauss was quite pleased when he proved that all natural numbers are the sum of three triangular numbers "In 1796, German mathematician and scientist Carl Friedrich Gauss discovered that every positive integer is representable as a sum of three triangular numbers (possibly including T0 = 0), writing in his diary his famous words, "ΕΥΡΗΚΑ! num = Δ + Δ + Δ". This theorem does not imply that the triangular numbers are different (as in the case of 20 = 10 + 10 + 0), nor that a solution with exactly three nonzero triangular numbers must exist. " – Will Jagy Nov 30 '20 at 22:19

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