User blog comment:Edwin Shade/The Challenge Of Minimum Rings/@comment-1605058-20171102114301

Clearly we are going to need 1 yellow and 3 green rings.

The largest number of red rings will appear after the last blue ring is taken off (I'm not going to bother justifying this, though I could). Say this happens when we have the score \(N\), so we get \(N\) red rings and we end up with the score \(2N=f_3(3)\). Hence we need \(\frac{1}{2}f_3(3)\) red rings.

Similarly, the most blue rings will appear after the last green ring is taken off. Since in a few steps after the beginning we have three green rings, the number of blue rings we will end up with is going to be the same as the number of red rings we get if we started with (from the bottom) a blue, green and green ring, with the starting score of three. Like above, we find that this number is \(\frac{1}{2}f_1(f_2(f_2(3)))\).

Therefore, if I'm not mistaken, the total number of rings is \(4+\frac{1}{2}f_3(3)+\frac{1}{2}f_1(f_2(f_2(3)))\).