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\begin{gather*} \frac{1}{4} \\ \frac{1}{4} \quad \frac{1}{4} \\ \frac{11}{48} \quad \frac{1}{4} \quad \frac{11}{48} \\ \frac{5}{24} \quad \frac{11}{48} \quad \frac{11}{48} \quad \frac{5}{24} \\ \frac{137}{720} \quad \frac{5}{24} \quad \frac{121}{576} \quad \frac{5}{24} \quad \frac{137}{720} \\\frac{7}{40} \quad \frac{49}{240} \quad \frac{55}{288} \quad \frac{55}{288} \quad \frac{49}{240} \quad \frac{7}{40} \\ \ldots \end{gather*} Does anyone know what's the generate term expression for this number sequence。

This sequence comes the coefficients of $a_0^2a_{1/x}^2$ in A5 \begin{eqnarray} A5&=&(-1+\mathbf{u})^{3}\left(-\frac{a_0^{4}}{24}-\frac{a_0^{3} a_{1/x} x}{6(-1+x)}-\frac{a_0^{2} a_{1/x}^{2} x^{2}}{4(-1+x)^{2}}-\frac{a_0 a_{1/x}^{3} x^{3}}{6(-1+x)^{3}}-\frac{a_{1/x}^{4} x^{4}}{24(-1+x)^{4}}\right)+\notag \\ &&(-1+\mathbf{u})^{4}\left(\frac{a_0^{4}}{12}+\frac{a_0^{3} a_{1/x} x}{4(-1+x)}+\frac{a_0 a_{1/x}\left(a_0^{2}+3 a_0 a_{1/x}\right) x^{2}}{12(-1+x)^{2}}-\right.\notag \\&&\qquad \left.\frac{a_0(-3 a_0-a_{1/x}) a_{1/x}^{2} x^{3}}{12(-1+x)^{3}}+\frac{a_0 a_{1/x}^{3} x^{4}}{4(-1+x)^{4}}+\frac{a_{1/x}^{4} x^{5}}{12(-1+x)^{5}}\right)\notag \\ && (-1+u)^{5}\left(-\frac{17 a_0^{4}}{144}-\frac{7 a_0^{3} a_{1/x} x}{24(-1+x)}-\frac{a_0 a_{1/x}\left(6 a_0^{2}+11 a_0 a_{1/x}\right) x^{2}}{48(-1+x)^{2}}-\frac{a_0 a_{1/x}\left(2 a_0^{2}+9 a_0 a_{1/x}+2 a_{1/x}^{2}\right) x^{3}}{36(-1+x)^{3}}+\right.\notag \\&&\qquad \left.\frac{a0(-11 a_0-6 a_{1/x}) a_{1/x}^{2} x^{4}}{48(-1+x)^{4}}-\frac{7 a 0 a_{1/x}^{3} x^{5}}{24(-1+x)^{5}}-\frac{17 a_{1/x}^{4} x^{6}}{144(-1+x)^{6}}\right)+... \end{eqnarray} which is the from the $ \epsilon^4$ of expension of $(1-u)^{\text{a1} \epsilon -1} \left(u^{\text{a0} \epsilon } (1-u x)^{\text{a1x} \epsilon }-(1-x)^{\text{a1x} \epsilon }\right)$ around $u=1$

I will also note that this is a hypergeometrical integration giving stuff proportional to ${}_2F_1$ with singular term at $u=1$ being substracted.

Here is the mathematica code:

H1 = Normal[Series[-(1 - u)^(-1 +  a1 \[Epsilon]) (u^(a0 \[Epsilon]) (1 - u x)^(  a1x \[Epsilon]) - (1 - x)^(a1x \[Epsilon])), {u, 1, 5}]];

H1e2 = SeriesCoefficient[H1, {\[Epsilon], 0, 4}] /. Log[1 - x] -> 0
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  • $\begingroup$ multiplying the first line by 4, the second by 8, the third and fourth by 48, the fifth by $2880$ yields the triangle \begin{gather*} 1 \\ 2 \quad 2 \\ 11 \quad 12 \quad 11 \\ 10 \quad 11 \quad 11 \quad 10 \\ 548 \quad 600 \quad 605 \quad 600 \quad 548 \end{gather*} Maybe this helps? $\endgroup$
    – ViktorStein
    Commented Dec 11, 2020 at 18:19
  • 3
    $\begingroup$ The first number in row $k$ is $\frac{H_k}{2(k+1)}$, where $H_k$ is the $k$-th harmonic number. If you multiply the entries in row $k$ by $2(k+1)!$, the first numbers in each row are the unsigned Stirling numbers of the first kind. $\endgroup$
    – Brian M. Scott
    Commented Dec 11, 2020 at 19:54
  • 1
    $\begingroup$ @BrianM.Scott Thank you for your answer. I am sorry I don't quite clearly see how this works. Would you give me more details? $\endgroup$
    – YU MU
    Commented Dec 12, 2020 at 9:14
  • 1
    $\begingroup$ @YUMU: I really don’t know what’s going on here, since I don’t yet see where the numbers in the interior of the triangle are coming from. My comment was simply an empirical observation. $\endgroup$
    – Brian M. Scott
    Commented Dec 12, 2020 at 19:45
  • 1
    $\begingroup$ @BrianM.Scott Okok thank you $\endgroup$
    – YU MU
    Commented Dec 14, 2020 at 6:58

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