What is the relation between Surreal birth order and its numeric value I found an equation that matches surreal numbers to their birth order. The surreal numbers are born in a sequence, but their numeric values (labels) come in a different order. This equation matches the two orderings.
$$ \large \frac{2 n + 1}{2^{d}}-2 = \frac{s(f-2)}{2^{w+1}} \: \text{mod} \: {2}$$
Where n & v are given...
d = $\lfloor log _{2} \: {n} \rfloor$
w = $\lfloor \left| {v} \right| \rfloor$
f = $\left|v\right| - w$
s = $\large \frac {\left| {v} \right|}{v}$
Symbol meaning:
n = given surreal appearance order or "birth" as a positive integer ($n \in {Z \gt {0}}$)
d = digits in binary representation of order number n
v = given numeric value or "label" ($\text{v} \in \text{dyadic rational or} s(w+f)$)
w = whole value part of absolute v
f = fractional part of absolute v
s = sign of the v
This equation came from a circle for me. I expect that this was already discovered by others. I want to know more about finite dyatic surreals and their creation.
Can someone direct me to the source of information that would provide me with these sorts of equations?
I found the equation like this: https://github.com/peawormsworth/tools/blob/master/surreal_order_value/Birth%20Order%20and%20Label%20Value%20-%20Surreal%20Equality.ipynb
I choose mod 2 because it fits around the unit circle when multiplied by pi i and is raised to e:



Update: I found an answer to my question which allows the value to be directly determined from the order and vice-versa. I  will post an answer when I can transfer the formulas from paper. In the meantime, this is the python code I used to verify the logic:
from math import floor, log2, copysign
from fractions import Fraction

def ov(n):
   """given natural n, return the value of the nth surreal number"""
   if n == 0:
      return None
   l = floor(log2(n+1/2))
   r = (2*n+1)/2**l-2
   a = abs(r-1)
   s = copysign(1,r-1)
   w = floor(-log2(1-a))
   f = (a-1)*2**(w+1)+2
   v = Fraction(s*(w+f))
   return v

def vo(v):
   """given dyadic v, return the ordinal of the surreal having value v"""
   if v is None:
      return 0
   u = abs(v)
   s = copysign(1,v)
   w = floor(u)
   f = u-w
   r = Fraction(s*((f/2-1)/2**w+1)+1)
   a = r.numerator
   b = r.denominator
   o = int((a-1)/2+b)
   return o

# testing the first 128 surreal numbers:
for n in range(128):
   print(n,ov(n),vo(ov(n)))

 A: Given surreal order number $n$, find surreal value $v$:
$$ l = \lfloor log_{2} (n+ {1 \over 2} ) \rfloor $$
$$ r = {2n+1 \over 2^l} -3  $$
$$ m = | r | $$
$$ s = {r \over m} $$
$$ w = \lfloor -log_{2} (1-m) \rfloor  $$
$$ f = 2^{ w+1 }( m-1 ) $$
$$ v=s(w+f) $$
... value $v$ has been found from order $n$
Given surreal value $v$ find surreal order $n$:
$$ u = |v|  $$
$$ s = {v \over u} $$
$$ w = \lfloor u \rfloor  $$
$$ f=u-w $$
$$ r = s \left ( {{f \over 2} - 1 \over 2^{w}} + 1 \right ) $$
let..
$$ {a \over b} = r + 1  $$
$$ n = {a-1 \over 2} + b  $$
... order $n$ has been found from value $v$
Finding a and b in the $2^{nd}$ to last equation should not be difficult. $r + 1$ is a dyadic rational between 0 and 2 so $a$ will be a positive odd, $b$ will be a power of two and $a < 2b$.
$n$ is the $n^{th}$ surreal number born where $ n \in \mathbb{N} $, a natural number (0, 1, 2, 3, ...).
$v$ is the value of the surreal number $ v \in base2 $
$r$ is a fraction between -1 and +1. Specifically $ r \in dyadic, |r| < 1 $. It can be plotted around the unit circle as $x$ using $ e^{x \cdot i \cdot \pi} $. Allowing the base2 numbers and the surreal order to be observed over top of one another, like the diagrams posted in the question. However, the circle points here will be rotated 180 degrees from the diagrams above.
In all formulas, $w$ is the absolute whole amount of the value of the surreal number referenced. $f$ is the remaining absolute fractional portion of the value. $s$ is the sign of the value of the surreal number.
I believe this is a correct solution because python code using these formulas correctly converts between surreal number order and value.
