Extended xor to Tetrion space

[Xorter]

Thanks for answering my question. I have been busy, but now I have time to ask another question. I still don't understand why (1;2;3) is a triad or why (10;27;51) is a pentad. Let me explain my thinking so you can correct me:
(1;2;3) = (1;2;10) in base 3. 1 xor[3] 2 = 0. 0 does not equal 10.
(10;27;51) = (20;102;201) in base 5. 20 xor[5] 102 = 122. 122 does not equal 201.

Thanks,

11Keith22.

It is not enough to convert a triad (Bion) to another number system and call it another k-ad. It does not work. You have to calculate with totally other numbers with a similar method I coded. Okay, I admit I made some mistakes in my code. Here is the good code:

Code:

mod(x,y)=x-y*floor(x/y);
blength(x,b)=if(x==0,1,floor(real(log(x)/log(b)))+1);

h=50;
xorb(x,y,b)=sum(k=1,h+blength(max(x,y),b),b^(k-1)*mod(if(x==0,0,iferr(Vecrev(digits(floor(x*b^h),b))[k],err,0))+if(y==0,0,iferr(Vecrev(digits(floor(y*b^h),b))[k],err,0)),b))/b^h*1.0;

KSearch(k,N)={s=0;for(y=1,N,for(z=y+1,N,for(w=z+1,N,if(xorb(y,z,k)==w&&xorb(z,w,k)==y&&xorb(w,y,k)==z,s+=1;print(Str(s,".: ( ",y,"; ",z,"; ",w,")"))))));return(s)};

KSearch counts the k-ads in the interval to N. And I could not see Trions, but 2^n-ions. Why?