We know the cyclic quotient singularity of type $\frac{1}{dn^2}(1,dna-1)$, where $n,a$ are coprime, has $Q-$Gorenstein smoothing. The second Betti number $b_2 = d-1$ on its smoothing. I'm wondering if there is a formula (or inequality) for $b_2$ of the minimal resolution of $\frac{1}{dn^2}(1,dna-1)$?
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$\begingroup$ Wouldn't this number be the length of the negative continued fraction expansion of $dn^2/(dna-1)$? $\endgroup$– Marco GollaCommented Sep 27, 2017 at 22:53
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$\begingroup$ Yes it is. But is there any effective formula to compute it, or maybe give a general upper, or lower bound of b_2? $\endgroup$– jhanCommented Sep 27, 2017 at 23:11
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1 Answer
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By Proposition 3.11 (i) and (ii) of Kollár and Shepherd-Barron's paper, if $N_{n,a,d}$ is the length of the continued fraction expansion of $\frac{dn^2}{dna-1}$, then we get the formulae $N_{2,1,d}=d$ and $N_{n+a,a,d}=N_{2n-a,n-a,d}=N_{n,a,d}+1$.
By the inductive process in part (iii) of the same Proposition, we can reduce $\frac{dn^2}{dna-1}$ to $\frac{4d}{2d-1}$ in $M_{n,a}$ steps (say), independent of $d$. This gives $N_{n,a,d}=M_{n,a}+d$, where $M_{n,a}$ is something like the number of steps in the Euclidean division algorithm for $\frac{n}{a}$. I don't think any nice closed formula exists in terms of $n,a,d$, but I guess this gives the lower bound $M_{n,a} \geq \lfloor\frac{n}{a}\rfloor - 2$ (adjusted so $M_{2,1}=0$).
