克拉夫楚克多项式

克拉夫楚克多项式以超几何函数定义如下

${\displaystyle K_{n}(x;p,N)={N \choose n}^{-1}*}$${\displaystyle \sum _{k=0}^{n}{N-x \choose n-k}*{x \choose k}*(1-1/p)^{k}}$=${\displaystyle _{2}F_{1}(-n,-x;N;1/p)}$

克拉夫楚克多项式的头几项是

${\displaystyle K_{0}(x,p,N)=1}$

${\displaystyle K_{1}(x,p,N)=1-{\frac {x}{p*N}}}$

${\displaystyle K_{2}(x,p,N)=1+({\frac {-2}{p*N}}+{\frac {1}{(p^{2}*N*(-N+1)}})*x-{\frac {x^{2}}{(p^{2}*N*(-N+1))}}}$

${\displaystyle K_{3}(x,p,N)=1+({\frac {-3}{(p*N)}}+{\frac {3}{(p^{2}*N*(-N+1))}}-{\frac {2}{(p^{3}*N*(-N+1)*(-N+2))}})*x+(-3/(p^{2}*N*(-N+1))+3/(p^{3}*N*(-N+1)*(-N+2)))*x^{2}-x^{3}/(p^{3}*N*(-N+1)*(-N+2))}$

量子Q克拉夫楚克多项式→ 克拉夫楚克多项式：

${\displaystyle \lim _{a\to 1}=K_{n}^{qtm}(q^{-}{x};p,N;q)=K_{n}(x;p^{-1},N)}$

令双Q克拉夫楚克多项式${\displaystyle c=1=p^{-1}}$,并令q→1,即得克拉夫楚克多项式

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