%I #10 Mar 31 2012 10:23:46
%S 0,0,0,1,1,1,2,4,4,4,5,5,5,6,8,11,11,12,12,12,14,14,14,16,18,20,22,24,
%T 26,29,29,29,30,31,32,35,35,35,37,38,40,43,43,45,46,50,51,52,55,55,57,
%U 57,59,61,63,65,69,69,74,74,74,76,77,78,81,82,82,86,89,91,93,96,99,100,104,105,106,107,108,112,113,115,115,117,121,122,122,124,124,125,126,131,133,134,137,139,141,146,148,150
%N Number of collinear triples in graph of preceding terms
%C a(n) is the number of 3element subsets (i<j<k) of (0,...,n1) such that both i,j,k and a(i),a(j),a(k) are arithmetic progressions (including the case a(i)=a(j)=a(k)). That is, kj=ji>0 and a(k)a(j)=a(j)a(i).
%C The sequence appears to grow faster than n but slower than n^(1+c) for any positive c.
%e For n=7, the triples (0,1,2),(0,3,6),(2,4,6),(3,4,5) satisfy the stated conditions, so a(7)=4
%K nonn,easy
%O 0,7
%A _Alex Abercrombie_, Jan 06 2011
