the labelled tree can be a linking edge, which corresponds to different Feynman diagram sets. Using factorization, as shown in Appendix \[appa\] , they together make up all Feynman diagrams on the RHS of Theorem \[identicalCc\] with coefficient $+1$, \[recursiongraph\] = + + +. While the two parts in figure \[factozation2\] themselves are also Feynman diagrams , whose corresponding labelled trees are $C^1,C^2$ in figure \[factozation\]. That means we can obtain Feynman diagrams from those of lower points. Let’s give the recursion explicitly \[recursion2\] T(C)= \_[ C\^1C\^2 - = C]{} { |c\^1T(C\^1),c\^2T(C\^2) }. Here we use the symbol $\bigsqcup$ as disjoint union.

A direct consequence is the recursion about the number of Feynman diagrams , \[recursion22\] |T(C)|= \_[ C\^1C\^2 - = C]{} |T(C\^1)||T(C\^2)|.

Let’s spell out some examples.

\[example4pt\] T( ) = { , }.

According to \[5ptafaf\] = + + , then \[example5pt\] T( ) = { , } { } { , }. 5 planar Feynman diagrams with ordering $1,2,3,4,5$ as expected.

Similarly for $C_n^{S}(1)$, according to = + + , then \[example5ptstar\] T( ) & =& { , } { , } && { , } , which are 6 multi-peripheral Feynman diagrams with the permutations of 