User blog:Mh314159/new YIP notation

My YIP notation was not using the most powerful form of recursion. I was wasting my time trying to fix it or extend it. Going back to the beginning, this is another attempt to use advice.

f0(a) = a + 1

fa(1) = fa-1(a+1)

fa(b) = fa-1fa(b-1)(b)

Example: f1(2) = f03(2) because f1(1) = fa-1(a+1) = f0(2) = 3, so f1(2) = 5

Single bracketed number:

[0] = 1

[a] = f[a-1][a-1](a)

[1] = f11(1) = f1(1) = f0(2) = 3

[2] = f33(2) where f3(2) = f2f3(1)(2) = f2f2(4)(2) etc.

Two bracketed numbers:

[a,0] = [a]

given g0(a) = f[a][a](a) [a,1] = g[a](a)

given ga(1) = ga-1(a+1)

given ga(b) = ga-1ga(b-1)(b)

[a,1] = g[a][a](a)

given h0(a) = g[a,1][a,1](a)

given ha(1) = ha-1(a+1)

given ha(b) = ha-1ha(b-1)(b)

[a,2] = h[a,1][a,1](a)

etc., increase the function level by one each time: f,g,h, etc., assuming no limit on the number of available function symbols.

Next step: change f,g,h, etc to {1}, {2}, etc? and then [n,n] would use the {n} function which would recurse {n-1} exactly as h recurses g. Then, a three entry array would use a new series of functions? Starting with {1,1} which would associate with [n,n,1] and would recurse {n} the same way that g recurses f and working up to Perhaps {1,n} for an array of type [n,n,n]?