User blog comment:Simplicityaboveall/Insanely Fast-Growing Functions/@comment-25912386-20171103024156/@comment-30754445-20171103180826

BTW, the following statement of yours in 100% correct:

"The ordinal growth of the H(n) function grows by one as n increases by 1"

And this is exactly why H(n) is no stronger than ω×2 in the standard FGH. Because ω×2 is the limit of:

ω+1, ω+2, ω+3, ...

Note that in the standard FGH we are not allowed to replace ω+10 with ω×2. So we just continue adding 1 and get:

... ω+9, ω+10, ω+11, ω+12, ...

So H(1000000) would correspond to the ordinal ω+999,990.

And we call the limit of this process ω×2.

Now, I fully understand that you're using a different system. I get that in your system, ω+999990 is actually equal to ω6. But that's just notation. You can call this level of recursion by whatever name you wish, but this wouldn't change the actual size if the number H(1000000). It doesn't change the fact that the calculation of H(1000000) involves precisely 1 diagonalization and 999990 recursions, making it comparable to f ω×2 (990,000) in the standard system.

And what most of us would call "f ω6(10)" is so much bigger than this, that it isn't even funny.

So why the huge difference? Because you're limiting the possible value of all your ω's to 10, and this robs the entire concept of ordinals of its power.