I am not sure if you had a chance to look at my node Re^7: Algorithm for cancelling common factors between two lists of multiplicands
I guess my approach is same as Limbic~Region as I was using subtraction of lower factorial terms with higher ones. I copied the input list from one of your previous examples and I have 45700 instead of 4570 but aside from that the implementation should be easy to understand.
I guess if you sort numerator and denominator and subtract them you should be all set. I have not proved this formally but here is a stab at a simple proof that shows sorting and subtracting should work...
Let the fraction be
X! Y! Z!

a! b! c!
WLOG let's assume that X > Y > Z and a > b > c. Also let's assume that
+ b > Y , X > a (weaker assumption: Z > c)...<p>
NOTE: if we divide X!/a! we will have (Xa) elements <p>
To Prove: (Xa) + (Zc) + (bY) is the shortest list one can find
or in other words
(Xa) + (Zc) + (bY) <= (Xp) + (Zq) + (rY) for any p,q,r in permut
+ation(a,b,c)
Proof:
From the above equation since b > Y and Z > c, r should be equal to ei
+ther a or b. If r = b then the solution is trivial<p>
If r = a then we get
(Xa) + (Zc) + (bY) ?<= (Xb) + (Zc) + (aY)
canceling terms
a  c + b ?<= b c + a
a + b ?<= b + a ====> YES
since a > b we see that r = a is not the smallest list so r = b<p>
Similarly we can also show that (Xa) + (Zc) + (bY) <= (Xa) + (Yc)
+ + (bZ)
I don't think this is a rigourous proof this method but i sort of feel sorting and subtracting should give us what we need...
cheers
SK
PS: I think there will be 47448 elements and not 47444 as you suggested? as you need to count the first element too..

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