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%I #6 Mar 29 2020 13:08:54
%S 1,10,4,31,35,9,68,102,74,16,125,214,211,127,25,206,380,436,358,194,
%T 36,315,609,765,734,543,275,49,456,910,1214,1280,1108,766,370,64,633,
%U 1292,1799,2021,1925,1558,1027,479,81,850,1764,2536,2982,3030,2700,2084,1326
%N Square array read by antidiagonals: T(n,k) is the Wiener index of the firecracker graph F(n,k) (n>=1, k>=2). F(n,k) is the graph obtained by the concatenation of n k-stars by linking one leaf from each. The Wiener index of a connected graph is the sum of distances between all unordered pairs of vertices in the graph.
%C F(3,4)=Y_Y_Y (roughly).
%H B. E. Sagan, Y-N. Yeh and P. Zhang, <a href="http://users.math.msu.edu/users/sagan/Papers/Old/wpg-pub.pdf">The Wiener Polynomial of a Graph</a>, Internat. J. of Quantum Chem., 60, 1996, 959-969.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FirecrackerGraph.html">Firecracker Graph</a>.
%F T(n,k)=(1/6)n(6+6k-7k^2+n^2*k^2+12nk^2-18nk).
%F The Wiener polynomial of the graph F(n,k) is W(n,k,t)=n*w+t[n(1-t)-(1-t^n)]r^2/(1-t)^2, where w=(k-1)t+(1/2)(k-1)(k-2)t^2 and r=1+t+(k-2)t^2.
%e T(1,3)=4 because the firecracker graph F(1,3) reduces to a path on 3 nodes, where the distances are 1, 1, and 2.
%e Square array T(n,k) begins:
%e 1,4,9,16,25,36,49, ...
%e 10,35,74,127,194,275,370, ...
%e 31,102,211,358,543,766,1027, ...
%e 68,214,436,734,1108,1558,2084, ...
%e 125,380,765,1280,1925,2700,3605, ...
%p T := proc (n, k) options operator, arrow; (1/6)*n*(6+6*k-7*k^2+n^2*k^2+12*n*k^2-18*n*k) end proc: for n to 10 do seq(T(n+2-i, i), i = 2 .. n+1) end do; # yields sequence in triangular form
%K nonn,tabl
%O 1,2
%A _Emeric Deutsch_, Sep 29 2010
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