David Hilbert  -  double recursion by φn(a,b)

Two pages from the article in Mathematische Annalen 95:
David Hilbert "Über das Unendliche", 1925.
Hilbert's 1925 lecture (online) stops short of the part on double recursive functions.
For a full English translation of the article, with an introduction, see:
Jean van Heijenoort "From Frege to Gödel", 1967, pp 367-392 "On the infinite".

D. Hilbert - Über das Unendliche - Seite 185 D. Hilbert - Über das Unendliche - Seite 186

So Hilbert's index under φ functions as the second iterator of a double recursion, suited as the third parameter of a superexponential function H:

H(a,1,n1) = φn+1(a,1) 
       = ι(φn,a,1) = a
       
H(a,b1,n1) = φn+1(a,b+1) 
         = ι(φn,a,b+1)
       = φn(a,ι(φn,a,b))
     = φn(a,φn+1(a,b))
   = H(a,H(a,b,n1),n) 
  == H(a,..a..,n) :b:  for n>0

From the initial φ1(a,b) = a+b follows that ι(φ0,a,1) = a+1 adds one.
The primitive function φ0 that Hilbert held behind is a choice successor function.

φ1(a,b+1) = a+b+1
        = ι(φ0,a,b+1)
      = φ0(a,ι(φ0,a,b))
    = φ0(a,φ1(a,b))
  = φ0(a,a+b)  then
φ0(a,b) = φ(b) = b+1

Hilbert's function φn+2(a,b) expresses exactly the superpowers of Knuth's up-arrows a↑{n}b where n counts the number of arrows in the operator.
From here Ackermann went on to prove that the function A(n) = H(n,n,n1) is not primitive recursively definable and strictly faster than any such φ functions.