A283438 The largest possible size of a set in F_3^n, free of three-term arithmetic progressions with the difference in {0,1}^n.
1, 2, 6, 14, 36
Offset: 0
Examples
a(0) = 1: {()}.
a(1) = 2: {(0), (1)}.
a(2) = 6: {(0,0), (0,2), (1,1), (1,2), (2,0), (2,1)}.
a(3) = 14: {(0,0,0), (0,0,2), (0,2,0), (0,2,2), (1,0,1), (1,0,2), (1,1,0), (1,1,1), (1,2,0), (1,2,2), (2,0,0), (2,0,1), (2,1,0), (2,1,1)}.
Links
- Robert Israel, Set of size 36 for n=4
- Robert Israel, Set of size 102 for n=5
Programs
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PARI
\\ This code is for demonstration only, it is far too slow for computation! ok(v)=vecmax(lift(v))<2 has(v)=my(t); for(i=1,#v, for(j=1,#v, if(i==j,next); for(k=1,#v, if(i==k||j==k,next); t=v[i]-v[j]; if(v[j]-v[k]==t && ok(t), return(0))))); 1 a(n)=my(N=3^n,v=vector(N,i,digits(N+i-1,3)[2..n+1]),u,r,t);for(i=0,2^N-1, u=Mod(vecextract(v,i),3); if(has(u), t=hammingweight(i); if(t>r, r=t))); r \\ Charles R Greathouse IV, Mar 07 2017
Extensions
a(4) from Robert Israel, Mar 07 2017
Comments