|
|
A191515
|
|
Number of vertices of outdegree >=2 in the rooted tree having Matula-Goebel number n.
|
|
0
|
|
|
0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 0, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 3, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,14
|
|
COMMENTS
|
The Matula-Goebel number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the Matula-Goebel numbers of the m branches of T.
|
|
LINKS
|
|
|
FORMULA
|
Let g(n)=G(n,x) be the generating polynomial of the vertices of outdegree >=2 of the rooted tree having Matula-Goebel number n, with respect to level. Then g(1)=0; if n = p(t) (=the t-th prime), then g(n)=x*g(t); if n=rs (r,s>=2), then g(n)=1+g(r)+g(s)-G(r,0)-G(s,0). Clearly, a(n)=G(n,1).
|
|
EXAMPLE
|
a(5)=0 because the rooted tree with Matula-Goebel number 5 is the path-tree on 4 vertices.
a(7)=1 because the rooted tree with Matula-Goebel number 7 is the rooted tree Y.
|
|
PROG
|
(PARI) a(n) = my(f=factor(n)); (vecsum(f[, 2])>=2) + [self()(primepi(p))|p<-f[, 1]]*f[, 2]; \\ Kevin Ryde, Oct 30 2021
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|