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I discovered the following formula while working on the sequence A141697 from the OEIS. I have no idea whether it is something trivial or not. I would be very happy to know more about it.

$$ \textrm{Li}_\nu\left(z\right)= \frac { 6 (1+z)^{-\nu-1} + \displaystyle\sum_{k=0}^{-\nu-1} \displaystyle\left( -6 \displaystyle{{-\nu-1}\displaystyle\choose k}+7\sum_{j=0}^{k+1}(-1)^j (k-j+1)^{-\nu} {{-\nu+1}\ \choose j} \right) z^k } { 7 (1 - z)^{-\nu+1} }z $$ when $\nu$ is a negative integer.

In case I mistyped the formula, here are my codes in two different languages:

For Pari-GP:

mypolylog(n, x) = { ( 6*(x+1)^(-n-1)
   + sum(k=0,-n-1, (-6*binomial(-n-1,k)
       + 7*sum(j=0,k+1, (-1)^j * (k-j+1)^(-n) * binomial(-n+1,j)))*x^k) ) * x
    / (7*(1-x)^(-n+1) ) }

For Mathematica:

mypolylog[n_, x_] :=  (6*(x+1)^(-n-1)
    + Sum[(x^k*(-6*Binomial[-n-1, k]
        + 7*Sum[(-1)^j*(k-j+1)^(-n)*Binomial[-n+1, j], {j, 0, k+1}])),
    {k, 0, -n-1}]) / ( 7*(1 - x)^(-n+1) ) * x
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  • $\begingroup$ How do got the formula? Have you tried a reformulation of the rational functions using Stirling or Eulerian numbers given in en.wikipedia.org/wiki/Polylogarithm#Particular_values? $\endgroup$ – gammatester Jun 3 '18 at 13:59
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    $\begingroup$ @gammatester I empirically discovered this formula by using a tool I wrote for discovering identities: github.com/baruchel/oeis which detected A141697(n)=3*A168524(n)-2*A154337(n). The formula above comes from that. $\endgroup$ – Thomas Baruchel Jun 3 '18 at 14:09
  • $\begingroup$ Nice work. You can use $6=t$ and $7=t+1$ where $t\neq -1$ in your formula instead. $\endgroup$ – Somos Jun 3 '18 at 16:01
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    $\begingroup$ This could similarly apply to A141696 as well. $\endgroup$ – Leucippus Jun 4 '18 at 2:31
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To simplify formulas define $\, B(n,k) := {-n-1\choose k}, \,$ $\, A(n,k) := \sum_{j=0}^{k+1} (-1)^j (k-j+1)^{-n} {-n+1 \choose j}, \,$ and $\, u := 1-t \,$ where $\, t\neq 0. \,$ Your formula is the $\, t=7 \,$ case of the lightly simplified equation

$$ \textrm{Li}_n(z) = \frac{z}t (1-z)^{n-1} (-u (1-z)^{-n+1} + u \sum_{k=0}^{-n-1} z^k B(n,k) + t \sum_{k=0}^{-n-1} z^k A(n,k) ) $$ but $\, (1-z)^{-n+1} = \sum_{k=0}^{-n-1} z^k B(n,k) \,$ simplies it to $ \textrm{Li}_n(z) = z (1-z)^{n-1} \sum_{k=0}^{-n-1} z^k A(n,k). \,$

The numbers in $\,A(-n,k-1)\,$ are the triangular OEIS sequence A008292 of Eulerian numbers which entry has the information "O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x".

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