Fibonacci Divisibility Graph

Let { F i } i = 1 \{F_i\}_{i=1}^\infty be the sequence of Fibonacci numbers ; that is, F 1 = F 2 = 1 F_1 = F_2 = 1 and F n + 2 = F n + 1 + F n F_{n+2} = F_{n+1} + F_n for all n 1 n \ge 1 .

Consider a graph with vertices labeled F 3 , F 4 , F 5 , , F N F_3, F_4, F_5, \ldots, F_N , and two (different) vertices x , y x,y are connected by an edge if and only if x y x|y or y x y|x . The above picture is the graph for N = 12 N = 12 .

Find the largest N N such that this graph is planar. If there is no largest N N , enter 0 0 .


The answer is 29.

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1 solution

Ivan Koswara
Dec 29, 2015

The answer is 29.

First, the Fibonacci numbers have the property that F a F b F_a | F_b if and only if a b a|b . Thus, we can construct a graph with 3 , 4 , 5 , , N 3, 4, 5, \ldots, N instead of F 3 , F 4 , F 5 , , F N F_3, F_4, F_5, \ldots, F_N , apply the same divisibility rule (an edge if a vertex is multiple of the other), and obtain an isomorphic graph.

With this, the following is a planar embedding for N = 29 N = 29 , with labels 3 , 4 , , N 3,4,\ldots,N instead of F 3 , F 4 , , F N F_3,F_4,\ldots,F_N :

Now, showing that N 30 N \ge 30 is impossible is actually simpler. By Kuratowski's theorem , a graph is not planar if and only if there exists a subgraph of it isomorphic to a subdivision of either K 5 K_5 or K 3 , 3 K_{3,3} . We claim there exists a subgraph isomorphic to a subdivision of K 3 , 3 K_{3,3} . This is best shown by a picture:

Thus no N 30 N \ge 30 will make it planar (because it will contain the above subgraph), so the largest N N is 29 \boxed{29} .

Moderator note:

Great question that combines knowledge in two different areas of mathematics.

That's a really interesting question!

Calvin Lin Staff - 5 years, 5 months ago

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