User blog comment:Denis Maksudov/M-notation (new version)/@comment-10262436-20160620212735/@comment-28606698-20160622112145

Well, I will check now2^ω=ωf_0 (ω)=ω+1f_0 (f_0 (ω))=(ω+1)+1=ω+2f_1 (ω)=f_0^ω (ω)=ω+ω=ω2f_1 (f_1 (ω))=(ω2)2=ω4=ω2^2f_2 (ω)=f_1^ω (ω)=ω2^ω=ω*ω=ω^2f_2 (f_2 (ω))=(ω^2 )^2 =ω^4f_3 (ω)=f_2^ω (ω)=ω^(2^ω )=ω^ω=ω↑ωf_3 (f_3 (ω))=(ω^ω)^(ω^ω)>ω^(ω^ω )f_4 (ω)>ω↑↑ω=ε_0f_5 (ω)>ω↑↑↑ω=ζ_0Yes, it is mistake in блогI will correctBut further is right f_(ω+1) (ω)=φ(1,0,0)