%I #21 Jan 30 2020 22:02:23
%S 1,1,1,1,1,1,2,3,3,2,3,6,9,6,3,6,16,27,27,16,6,11,37,79,96,79,37,11,
%T 23,96,233,349,349,233,96,23,47,239,679,1187,1439,1187,679,239,47,106,
%U 622,1987,4017,5639,5639,4017,1987,622,106,235,1607,5784,13216,21263,24758,21263,13216,5784,1607,235
%N Number of signed trees with n nodes and p positive edges. Triangle T(n,p) read by rows, 0<=p<n.
%H Andrew Howroyd, <a href="/A302939/b302939.txt">Table of n, a(n) for n = 1..1275</a>
%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>
%F T(n,p) = T(n,n-p-1), flipping all edge signs.
%e T(2,0)=T(2,1)=1: the tree on 2 nodes (one edge) has one variant with no positive edge and one variant with one positive edge.
%e T(4,1)=3: the 2 trees on 4 nodes (three edges) have two variants from the linear tree with a positive edge (edge in the middle or at the end) and one variant from the star graph with one positive edge.
%e T(5,0)=3: there are 3 trees on 5 nodes (4 edges) where all edges are negative.
%e The triangle starts
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 2, 3, 3, 2;
%e 3, 6, 9, 6, 3;
%e 6, 16, 27, 27, 16, 6;
%e 11, 37, 79, 96, 79, 37, 11;
%e 23, 96, 233, 349, 349, 233, 96, 23;
%e 47, 239, 679, 1187, 1439, 1187, 679, 239, 47;
%e 106, 622,...
%o (PARI)
%o R(n, y)={my(v=vector(n)); v[1]=1; for(k=1, n-1, my(p=(1+y)*v[k]); my(q=Vec(prod(j=0, poldegree(p, y), (1/(1-x*y^j) + O(x*x^(n\k)))^polcoeff(p, j)))); v=vector(n, j, v[j] + sum(i=1, (j-1)\k, v[j-i*k] * q[i+1]))); v; }
%o M(n)={my(B=x*Ser(R(n, y))); B - (1+y)*(B^2 - substvec(B, [x, y], [x^2, y^2]))/2}
%o { my(A=Vec(M(10))); for(n=1, #A, print(Vecrev(A[n]))) } \\ _Andrew Howroyd_, May 13 2018
%Y Cf. A000060 (row sums), A000055 (diagonal and 1st column), A027852 (subdiagonal and 2nd column), A304489 (rooted), A331113 (central coefficients).
%K nonn,tabl
%O 1,7
%A _R. J. Mathar_, Apr 16 2018
%E Completed row 10. - _R. J. Mathar_, Apr 29 2018
%E Terms a(58) and beyond from _Andrew Howroyd_, May 13 2018