%I #4 Apr 04 2014 19:41:34
%S 0,1,4,3,9,8,5,6,16,15,12,13,7,8,11,10,25,24,21,22,16,17,20,19,9,10,
%T 13,12,18,17,14,15,36,35,32,33,27,28,31,30,20,21,24,23,29,28,25,26,11,
%U 12,15,14,20,19,16,17,27,26,23,24,18
%N Area under the path specified by n-th composition.
%C Compositions are in the order specified in A066099.
%C Path has run lengths given by the specified composition followed by its reversal.
%C Can be considered a table with row lengths A011782 (2^(n-1)).
%C Maximum in row n is n^2, for [n]; minimum is 2*n-1 for [1,n-1].
%e Row 5 of A066099 is [2,1]. The path generated by [2,1,1,2] is:
%e __
%e |_
%e |
%e ____|
%e The area under the path is 8, so a(5) = 8.
%o (PARI) A066099row(n) = {local(v=vector(n),j=0,k=0);
%o while(n>0,k++; if(n%2==1,v[j++]=k;k=0);n\=2);
%o vector(j,i,v[j-i+1])} \\ row n of A066099.
%o patharea(v) = {local(m=#v,n=sum(k=1,m,v[k]),w,r=0);
%o w=vector(2*m,k,if(k<=m,v[k],v[2*m+1-k]));
%o forstep(k=1,2*m,2,r+=w[k]*n;n-=w[k+1]);
%o r}
%o a(n) = patharea(A066099row(n)) \\ _Franklin T. Adams-Watters_, Apr 02 2014
%Y Cf. A066099, A011782.
%K nonn,tabf
%O 0,3
%A _Franklin T. Adams-Watters_, Apr 02 2014