User blog comment:Mh314159/Alpha numbers (and beyond)/@comment-35470197-20191007022858/@comment-35470197-20191010013834

I should write more details. The extensions given by superscripts in your notation is the iteration of the power, and hence contributes as "+2". The extensions given by subscripts in your notation is the diagonalisation.

The family \((\alpha[a_0,x],\alpha[a_1,x],\alpha[a_2,x],\ldots)\) contributes as "+2", "+4", "+6", and so on. Therefore its diagonalisation \(\alpha[a,x]\) contributes as "+2w", which coincides with "+w" as an ordinal.

The family \((\alpha[0,x],\alpha[1,x],\alpha[2,x],\ldots)\) corresponds to "+w", "+w×2", "+w×3", and so on. Therefore its diagonalisation \(\alpha_0[x]\) contributes as "+w^2", which is the greatest contribution in your strategy.

Since \(\alpha_{b+1}[x]\) is defined as the iteration of the power of \(\alpha_b[x]\), its contribution is just "+2". Namely, the family \((\alpha_0[x],\alpha_1[x],\alpha_2[x],\ldots)\). just contributes as "+w^2" (the contribution of \(\alpha_0[x]\)), "+w^2+2", "+w^2+4", and so on. That is why the contribution of its diagonalisation \(\alpha[x]\) is bounded by "+w^2+w".

Diagonalisation itself is strong, but in order to maximise the contribution, we need to care about what family we diagonalise.