Does there exist a bijection $f:\mathbb{N}\to\mathbb{N}$ such that $\vert f(n) - f(n+1)\vert = n\ \forall\ n\in\mathbb{N}?$
(If this was $f:\mathbb{N}\to\mathbb{Z}$ then this would be trivial: $0,1,-1,2,-2,3,-3,\ldots .$)
The greedy algorithm starting at $1$ doesn't work. Curiously, we get:
$$1,2,4,7,3,8,14,21,13,22,12,23,11,24,10,25,9,26,44,63,43,64,42,19,$$
but then we cannot continue.
Maybe there is a strategy to answer this question, but I do not see anything straight away.
Some ideas here are possibly useful, in particular the idea of blocks.
