User blog:Mh314159/YIP notation

In a recent post I lamented that I am a dog, which unlike cats, cannot climb TREEs. Actually, given how small my growth rates have been, I am a puppy. And puppies go YIP. So here is YIP notation. I have tried to break my habit of thinking linearly. So far, my notation only goes to [a,b] where (b+1) is the dimensionality of the recursion structure. I'm not sure yet what [a,b,c] would mean, but probably something like recursing the dimensionality using dimensional numbers. Anyway, before I go much farther, I'd love to know if what I'm doing is better than what I've done in the past or if it's just more of the same in a different format. Thanks!!

↓x/indicates beginning and end of subscript

↑x\indicates beginning and end of superscript (i.e. functional recursion)

f0(a) = a + 1 = f↓0/(a)

fa(1) = fa-1(a+1) = f↓(a-1)/(a+1)

fa(b) = f↓(a-1)/↑f↓(a-1)/...↑f↓(a-1)/(b)\...(b)\(b)

conceptually, this is a pyramid with all of the steps on the left being f(a-1) and all of the steps on the right being (b) and the number of levels = fa(b-1)

Example: f1(2) = f0f0 f0(2)(2) (2) with 3 levels because f1(1) = fa-1(a+1) = f0(2) = 3, so f1(2) = f0f0 3(2) (2) = f05(2) = 7

The values of f1(n) starting from 1 are 3, 7, 22, 89, 2676, indeed the formula is f1(n) = 1 + (n)(f1(n-1)) so it is factorial-like. Given that f1(6)= 2676 and 6! = 720 (ratio 3.716...) and that f1(7)= 18,734 and 7! = 5,040 (ratio 3.7170634920635) it appears that f1(x)undefined> x!

Single bracketed number:

[0] = 1

[a] = fa(a)

Two bracketed numbers represent a higher dimensional structure, with the second number setting the dimensionality of the recursion and both numbers setting the edge length.

[a,b] is a recursive construction of dimensionality (b+1) with edge length [a, b-1].

[a,0] is a linear recursion of the form f↓[f↓...[f↓a/(a)].../(a)]/(a) with [a]undefinedinstances of f.

[a,1] is constructed as follows

Start with a linear expression of the following type: f↓[f↓...[f↓a/(a)].../(a)]/(a) with [a]undefinedinstances of f. This is the base expression, or X0.

Use this expression to determine the number of f's in a similarly constructed expression, which we will call expression X1. The value of Xn determines the number of f's in expression Xn+1.The value of [a,1] is the output of row [a].

Think of this as a page of rows of recursed functions, starting with the shortest row top, we build successive rows representing larger expression with the bottom and longest one being the output.

Example: [1,1] starts, given that [1] = 3, with f↓[f↓[f↓1/(1)]/(1)]/(1) the value of which is X0 = f↓[f↓3/(1)]/(1) and f↓3/(1) = f↓2/(4), already large because it is a recursion pyramid of f1 fuctions f↓1/(4) = 89 levels high where f1as discussed is more powerful than the factorial. X1 = f↓...[f↓[f↓1/(1)]/(1)].../(1) with X0 instances of f, X2has X1 instances, and the value of [1,1] = X3which has X2 instances of f.

[a,2] is three dimensional, and constructed as follows

Start with the 2D expression [a,1]. This is the base expression and the edge length, or X0.

Use this expression to determine the edge length in a similarly constructed 2D expression, which we will call expression X1. In other words, if X0 = [a,1], then Xn+1 = [Xn,1] and [a,2] = X[a,1].

Think of this as a cube of pages of recursed rows, starting with the 2D expression [a,1], we recurse the edge length of each successive page using the value of the page before it, the bottom and largest page being the output.