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**Unformatted text preview: **has one more element than earlier, 4m + 1, so has
m + 1 elements. The extra vertex will connect the m vertices in [2], so all the
vertices in [2] have degree m + 1, while the vertices in [0] ∪ [1] ∪ [3] have degree
m. So
ΦV (G4m+1 ) (x) = (3m + 1)xm + mxm+1 .
2 If n = 4m + 2, the classes [1] and [2] have now m + 1 elements (we have added
4m + 2 to [2]) while the classes [0] and [3] still have m elements. The vertices in
[1] ∪ [2] have degree m + 1 while the vertices in [3] ∪ [0] have degree m. Hence
ΦV (G4m+2 ) (x) = 2mxm + (2m + 2)xm+1 .
Lastly, if n = 4m + 3, the classes [1], [2] and [3] now all have m + 1 elements while
[0] still has m elements. The vertices in [0], [1] and [2] therefore have degree m + 1
while the vertices in [3] have degree m. Hence
ΦV (G4m+3 ) (x) = (m + 1)xm + (3m + 2)xm+1 .
4. Let Gn be the set of all ﬁnite graphs with vertices {1, . . . , n}, and let G = n≥0 Gn . Let
the weight function w : G → N≥0 be given by w(G) = |V (G)|. Find ΦG (x)....

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