The number of bounded‐degree spanning trees

Abstract

For a graph G$$ G $$ , let ck(G)$$ {c}_k(G) $$ be the number of spanning trees of G$$ G $$ with maximum degree at most k$$ k $$ . For k≥3$$ k\ge 3 $$ , it is proved that every connected n$$ n $$ ‐vertex r$$ r $$ ‐regular graph G$$ G $$ with r≥nk+1$$ r\ge \frac{n}{k+1} $$ satisfies ck(G)1/n≥(1−on(1))r·zk,$$ {c}_k{(G)}^{1/n}\ge \left(1-{o}_n(1)\right)r\cdotp {z}_k, $$where zk>0$$ {z}_k>0 $$ approaches 1 extremely fast (e.g., z10=0.999971$$ {z}_{10}=0.999971 $$ ). The minimum degree requirement is essentially tight as for every k≥2$$ k\ge 2 $$ there are connected n$$ n $$ ‐vertex r$$ r $$ ‐regular graphs G$$ G $$ with r=⌊n/(k+1)⌋−2$$ r=\left\lfloor n/\left(k+1\right)\right\rfloor -2 $$ for which ck(G)=0$$ {c}_k(G)=0 $$ . Regularity may be relaxed, replacing r$$ r $$ with the geometric mean of the degree sequence and replacing zk$$ {z}_k $$ with zk∗>0$$ {z}_k^{\ast }>0 $$ that also approaches 1, as long as the maximum degree is at most n(1−(3+ok(1))lnk/k)$$ n\left(1-\left(3+{o}_k(1)\right)\sqrt{\ln k/k}\right) $$ . The same holds with no restriction on the maximum degree as long as the minimum degree is at least nk(1+ok(1))$$ \frac{n}{k}\left(1+{o}_k(1)\right) $$ .