|
%I #14 Jan 09 2021 22:10:07
%S 1,1,0,0,1,0,0,1,1,0,0,0,2,1,0,0,0,2,3,1,0,0,0,0,6,4,1,0,0,0,0,5,12,5,
%T 1,0,0,0,0,0,20,20,6,1,0,0,0,0,0,15,52,31,7,1,0,0,0,0,0,0,76,107,43,8,
%U 1,0,0,0,0,0,0,49,242,192,58,9,1,0,0,0,0,0,0,0,313,589,313,75,10,1,0
%N Triangle read by rows: T(n,k) is the number of unlabeled trees on n vertices with independence number k.
%C For n > 1, a star graph on n nodes has independence number n-1 and a path graph has independence number ceiling(n/2) which is the least possible in a tree.
%C A maximum independent set can be found in a tree using a greedy algorithm. First choose any node to be the root and perform a depth first search from the root. Include all leaves in the independent set (except possibly the root) and also greedily include any other node if no children are in the set. This method can be converted into an algorithm to compute the number of trees by independence number. See the PARI program for technical details.
%H Andrew Howroyd, <a href="/A339829/b339829.txt">Table of n, a(n) for n = 1..1275</a> (rows 1..50)
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/IndependenceNumber.html">Independence Number</a>
%e Triangle begins:
%e 1;
%e 1, 0;
%e 0, 1, 0;
%e 0, 1, 1, 0;
%e 0, 0, 2, 1, 0;
%e 0, 0, 2, 3, 1, 0;
%e 0, 0, 0, 6, 4, 1, 0;
%e 0, 0, 0, 5, 12, 5, 1, 0;
%e 0, 0, 0, 0, 20, 20, 6, 1, 0;
%e 0, 0, 0, 0, 15, 52, 31, 7, 1, 0;
%e ...
%e There are 3 trees with 5 nodes:
%e x x
%e | |
%e o---x---o x---o---x---o---x x---o---x
%e | |
%e x x
%e The first 2 of these have a maximum independent set of 3 nodes and the last has a maximum independent set of 4 nodes, so T(5,3)=2 and T(5,4)=1.
%o (PARI)
%o EulerMT(u)={my(n=#u, p=x*Ser(u), vars=variables(p)); Vec(exp( sum(i=1, n, substvec(p + O(x*x^(n\i)), vars, apply(v->v^i,vars))/i ))-1, -n)}
%o \\ In the following, u,v count rooted trees weighted by independence number: u is root in the set and v is root not in the set.
%o T(n)={my(u=[y], v=[0]); for(n=2, n, my(t=EulerMT(v)); v=concat([0], EulerMT(u+v)-t); u=y*concat([1], t)); my(g=x*Ser(u+v), gu=x*Ser(u), r=Vec(g + (substvec(g, [x,y], [x^2,y^2]) - (1-1/y)*substvec(gu, [x,y], [x^2,y^2]) - g^2 + (1-1/y)*gu^2 )/2)); vector(#r, n, Vecrev(r[n]/y, n))}
%o { my(A=T(10)); for(n=1, #A, print(A[n])) }
%Y Row sums are A000055.
%Y Cf. A294490 (connected graphs), A339830, A339831, A339833 (domination number).
%K nonn,tabl
%O 1,13
%A _Andrew Howroyd_, Dec 18 2020
|
