A302939
Number of signed trees with n nodes and p positive edges. Triangle T(n,p) read by rows, 0<=p
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, 23, 96, 233, 349, 349, 233, 96, 23, 47, 239, 679, 1187, 1439, 1187, 679, 239, 47, 106, 622, 1987, 4017, 5639, 5639, 4017, 1987, 622, 106, 235, 1607, 5784, 13216, 21263, 24758, 21263, 13216, 5784, 1607, 235
Offset: 1
Examples
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.
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.
T(5,0)=3: there are 3 trees on 5 nodes (4 edges) where all edges are negative.
The triangle starts
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;
23, 96, 233, 349, 349, 233, 96, 23;
47, 239, 679, 1187, 1439, 1187, 679, 239, 47;
106, 622,...
Links
Crossrefs
Programs
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PARI
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; } M(n)={my(B=x*Ser(R(n, y))); B - (1+y)*(B^2 - substvec(B, [x, y], [x^2, y^2]))/2} { my(A=Vec(M(10))); for(n=1, #A, print(Vecrev(A[n]))) } \\ Andrew Howroyd, May 13 2018
Formula
T(n,p) = T(n,n-p-1), flipping all edge signs.
Extensions
Completed row 10. - R. J. Mathar, Apr 29 2018
Terms a(58) and beyond from Andrew Howroyd, May 13 2018
Comments