posts from @belarius tagged #Mertens conjecture

also:

hthrflwrs
@hthrflwrs

the two kinds of unsolved math problem are "we think it might be infinite but we've solved it definitively up to 12,478,329,709,179,875,728,943,982,427,349,234,293,487" and "we've spent the past thirty years trying to find out if 2 is a valid answer"


cgranade
@cgranade

Graham's number is one of the most ridiculously large numbers to appear in any kind of mathematical argument, requiring several new forms of notation to even define. Essentially, let π‘Ž ↑ 𝑏 mean "π‘Ž raised to the power 𝑏," then let π‘Ž ↑↑ 𝑏 mean "π‘Ž ↑ π‘Ž ↑ π‘Ž … ↑ π‘Ž, where ↑ appears 𝑏 βˆ’ 1 times." For example, 2 ↑↑ 4 = 2 ↑ 2 ↑ 2 ↑ 2 = 2^(2^(2^(2))) = 2^16 = 65,536. Using this notation, let 𝑔₁ = 3 ↑↑↑↑ 3, already a ridiculously large number. Then, let 𝑔ₙ = 3 ↑…↑ 3, where ↑ appears 𝑔ₙ₋₁ times.

Graham's number is what you get when you do this ridiculously large recursion 64 times, 𝑔₆₄. Why would anyone care? Because at one point, this was the best upper bound known for a not too ugly graph theory problem on hypercubes:

Connect each pair of geometric vertices of an 𝑛-dimensional hypercube to obtain a complete graph on 2𝑛 vertices. Color each of the edges of this graph either red or blue. What is the smallest value of 𝑛 for which every such coloring contains at least one single-colored complete subgraph on four coplanar vertices?

The best lower bound at the time?

6.


belarius
@belarius

The current known values of the bounds (huge but not remotely on the level of Graham's number) are provided in the Wikipedia article, but as with many fun numbers, there's also a Numberphile video about it:


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