We'll use ordinal notation for this:

Let α,β,≤⌈₀ be ordinals, with β being a limit ordinal.

Let n,k be positive integers.

Define a function Λ : (⌈₀ ∪ {⌈₀})×ℕ → ℕ as:

Λ(α+1,1) = 10
Λ(1,n) = En = 10n
Λ(α+1,n+1) = Λ(α,Λ(α+1,n))
Λ(β,n) = Λ(β[n],10)

Where β[n] is the n-th member of the fundamental sequence of β, given as follows:

For β=β₁+β₂ (with β₂≤β₁) we have β[n] = β₁+(β₂[n])
For β=ω we have β[n] = n
For β=ωα+1 we have β[n] = ωα×n
For β=ωβ₁ (with β₁<β a limit ordinal) we have β[n] = ωβ₁
For β=φ(α+1,0) we have β[1]=φ(α,0), β[n+1]=φ(α,β[n])
For β=φ(α₁+1,α₂+1) we have β[1]=φ(α₁+1,α₂)+1, β[n+1]=φ(α₁,β[n])
For β=φ(β₁,0) (with β₁<β a limit ordinal) we have β[n+1]=φ(β₁[n],0)
For β=φ(β₁,α+1) (with β₁<β a limit ordinal) we have β[n+1]=φ(β₁[n],φ(β₁,α)+1)
For β=φ(α,β₁) (with β₁<β a limit ordinal) we have β[n+1]=φ(α,β₁[n])
For β=⌈₀ we have ⌈₀[1]=1, ⌈₀[n+1]=φ(⌈₀[n],0)

And using these definitions, we can now define:

Q1 = R1 = 10
Qn = Λ(ε₀,n) (for n>1)
Rn = Λ(⌈₀,n) (for n>1)
nQm = Λ(ω↑↑m,n) for n+m>2

And retroactively, for n>1:

En = Λ(1,n)
Fn = Λ(2,n)
Gn = Λ(3,n)
Hn = Λ(4,n)
Jn = Λ(ω,n)
Kn = Λ(ω+1,n)
Ln = Λ(ω+2,n)
Mn = Λ(ω×2,n)
Nn = Λ(ω2,n)
Pn = Λ(ωω,n)

If I've made any mistake (which is pretty likely) please let me know.