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%I #6 Mar 30 2020 04:18:23
%S 1,4,8,9,25,25,16,52,72,56,25,89,145,154,105,36,136,244,304,280,176,
%T 49,193,369,506,545,459,273,64,260,520,760,900,884,700,400,81,337,697,
%U 1066,1345,1451,1337,1012,561,100,424,900,1424,1880,2160,2184,1920,1404,760
%N Square array read by antidiagonals: T(m,n) is the Wiener index of the stacked book graph B(m,n) (m>=1, n>=1). B(m,n) is defined as the graph Cartesian product S(m+1) x P(n), where S(m+1) is the star graph on m+1 nodes and P(n) is the path graph on n nodes. The Wiener index of a connected graph is the sum of distances between all unordered pairs of nodes in the graph.
%C T(1,n) = A131423(n).
%C T(2,n) = A180569(n).
%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/StackedBookGraph.html">Stacked Book Graph</a>.
%F T(m,n) = (1/6)n[n^2-(m+1)^2+mn(mn+6m+2n)].
%F The Wiener polynomial p[n](t) of the graph B(m,n) satisfies the recurrence relation p[n] = p[n-1]+mt+(1/2)m(m-1)t^2+[t+mt+2mt^2+m(m-1)t^3]*sum(t^j,j=0..n-2).
%e T(2,1)=4 because B(2,1) reduces to the path graph P(3) which has 2 pairs of nodes at distance 1 and 1 pair at distance 2.
%e Square array T(m,n) begins:
%e 1, 8, 25, 56, 105, ...
%e 4, 25, 72, 154, 280, ...
%e 9, 52, 145, 304, 545, ...
%e 16, 89, 244, 506, 900, ...
%p T := proc (m, n) options operator, arrow: (1/6)*n*(n^2-(m+1)^2+m*n*(m*n+2*n+6*m)) end proc: for n to 10 do seq(T(n+1-j, j), j = 1 .. n) end do; # yields sequence in triangular form
%Y Cf. A131423, A180569.
%K nonn,tabl
%O 1,2
%A _Emeric Deutsch_, Sep 28 2010