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Let $2\leq a_1<a_2<\cdots<a_n$ be pairwise coprime integers, let $a=a_1 \cdots a_n$, and let $b_i$ be the unique integer such that $b_i=(a/a_i)^{-1} \text{ mod } a_i$ and $0>b_i>-a_i$. Then let $b=-1/a + \sum_{i=1}^n (b_i/a_i) $. It is known that $b$ is a negative integer such that $-n<b\leq -1$.

Actually I am studying Saveliev's book Invariants for homology 3-spheres, and the integer $b$ arises here in the following way: The Seifert-fibered homology sphere $\Sigma(a_1,\dots,a_n)$ (defined in chapter 1.1.4) bounds a unique negative definite plumbing 4-manifold whose graph is a star-shaped tree, and the weight of the central node is exactly $b$.

Consider the case $n=4$, so in this case $b=-1, -2$ or $-3$. Is there a way to find a condition for $a_1,a_2,a_3,a_4$ so that $b=-1$? (Or at least an example.) I computed $b$ for some small $a_i$'s; for example if $(a_1,a_2,a_3,a_4)=(2,3,5,7)$ then $b=-2$, and if $(a_1,a_2,a_3,a_4)=(2,5,7,11)$ then $b=-2$, and I couldn't find $(a_1,\dots,a_4)$ with $b=-1$. I am wondering to find a criterion for $(a_1,\dots,a_4)$ to have $b=-1$ by using some number theory, etc.

blancket
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    Let $,x,$ be the solution of $,x\equiv 1\pmod{!a_i},, i=1,\cdots, n,$ that is given by the well-known CRT formula, so $,x = 1+a:!b,,$ for some $,b\in \Bbb Z.,$ Your $,b,$ is the "CRT quotient" $, b = (x!-!1)/a\ \ $ – Bill Dubuque Jul 25 '22 at 15:09
  • @BillDubuque Yes but $x$ is determined only modulo $\prod_i a_i$ – blancket Jul 28 '22 at 12:47

1 Answers1

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For $b=-1$:

$n\ \ (a_1,a_2,...,a_n)$

$3\ \ (2, 3, 7)\ (2, 5, 7)\ (3, 4, 5)\ (4, 5, 7)\ (5, 6, 7)\ ...$

$4\ \ (2, 5, 9, 13)\ (3, 4, 7, 11)\ (3, 5, 8, 11)\ (5, 8, 9, 11)\ ...$

$5\ \ (3, 4, 11, 13, 19)\ (5, 7, 13, 16, 19)\ (5, 11, 14, 17, 19)\ ...$

$6\ \ (5, 9, 11, 13, 16, 17)\ (4, 7, 13, 17, 19, 23)\ (7, 9, 10, 13, 17, 23)\ ...$

Jinyuan
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