Abstract
A graph is called $K_{1,n}$-free if it contains no $K_{1,n}$ as an induced subgraph. Let $a$, $b$ $(0\leq a<b)$, and $n$ $(\geq 3)$ be integers. Let $G$ be a $K_{1,n}$-free graph. We prove that $G$ has an $[a,b]$-factor if its minimum degree is at least \[ \left(\frac{(a+1)(n-1)}{b}+1\right)\left\lceil\frac{a}{2}+\frac{b}{2(n-1)}\right\rceil-\frac{n-1}{b}\left(\left\lceil\frac{a}{2}+\frac{b}{2(n-1)}\right\rceil\right)^2-1. \] This degree condition is sharp for any integers $a$, $b$, and $n$ with $b\leq a(n-1)$. If $b\geq a(n-1)$, it exists if its minimum degree is at least $a$.
Citation
Taro TOKUDA. "A Degree Condition for the Existence of $[a,b]$-Factors in $K_{1,n}$-Free Graphs." Tokyo J. Math. 21 (2) 377 - 380, December 1998. https://doi.org/10.3836/tjm/1270041821
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