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Yeah, it seems important that the formula support unbounded numbers, because the numbers can grow inside the sequence. Naively, it seems like any starting valu
by leereeves 4y ago
Yeah, it seems important that the formula support unbounded numbers, because the numbers can grow inside the sequence.
Naively, it seems like any starting value (except those for which you already know the full sequence) might lead to values that exceed any arbitrary bound.
- pclmulqdq 4y agoFrom a theoretical standpoint, I think you could also attempt to approach this by thinking about a field where the "remove trailing zeros" step is built in. It may be a very weird field to do math on, and we may be unprepared for the math that is required, but it's a way to remove branching from the operations required for a proof. There's nothing inherently special about the branchiness of "remove trailing zeros" as an operation if you drill down into it: addition, for example, also needs conditional execution (based on the carry in to each section of a number). Otherwise, if you're searching for a counterexample, bound n by some ridiculous number of bits (whatever your supercomputer supports) and run your solver with a simple closed form.