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%I #10 Jul 21 2017 10:46:13
%S 1,3,4,2,5,5,6,6,3,7,7,7,8,8,8,4,9,9,9,9,10,10,10,10,5,11,11,11,11,11,
%T 12,12,12,12,12,6,13,13,13,13,13,13,14,14,14,14,14,14,7,15,15,15,15,
%U 15,15,15,16,16,16,16,16,16,16,8,17,17,17,17,17,17,17,17,18,18,18,18,18,18
%N Triangle read by rows: T(n,k) is the number of unordered pairs of vertices at distance k in the cycle C_n (1 <= k <= floor(n/2)).
%C Row n contains floor(n/2) entries.
%C The entries in row n are the coefficients of the Wiener polynomial of the cycle C_n.
%C Sum of entries in row n = n(n-1)/2 = A000217(n-1).
%C Sum_{k=1..floor(n/2)} k*T(n,k) = the Wiener index of the cycle C_n = A034828(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.
%F T(2n+1,k) = 2n+1 (1<=k<=n); T(2n,k)=2n (1<=k<=n-1); T(2n,n)=n.
%F G.f. = G(q,z) = qz^2/(1+z-z^2-qz^3)/((1-qz^2)^2*(1-z)^2).
%e T(4,2)=2 because in C_4 (a square) there are 2 distances equal to 2.
%e Triangle starts:
%e 1;
%e 3;
%e 4, 2;
%e 5, 5;
%e 6, 6, 3;
%e 7, 7, 7;
%p P:=proc(n) if `mod`(n, 2)=0 then n*(sum(q^j,j=1..(1/2)*n-1))+(1/2)*n*q^((1/2)*n) else n*(sum(q^j,j=1..(1/2)*n-1/2)) end if end proc: for n from 2 to 18 do p[n]:=P(n) end do: for n from 2 to 18 do seq(coeff(p[n],q,j),j=1..floor((1/2)*n)) end do; # yields sequence in triangular form
%Y Cf. A000217, A034828.
%K nonn,tabf
%O 2,2
%A _Emeric Deutsch_, Sep 06 2008
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