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Aug 8, 2020 at 15:14 answer added Alexandre Eremenko timeline score: 3
Aug 8, 2020 at 8:29 history edited user163402 CC BY-SA 4.0
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Aug 7, 2020 at 15:58 comment added მამუკა ჯიბლაძე Should be derivable easily from the recurrence$$nE_{n+1}(z)+zE_n(z)=e^{-z}$$
S Aug 7, 2020 at 11:49 history suggested GohPiHan CC BY-SA 4.0
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Aug 7, 2020 at 11:19 review Suggested edits
S Aug 7, 2020 at 11:49
Aug 7, 2020 at 6:36 comment added მამუკა ჯიბლაძე Mathematica says that Sum[k! x^k, {k, 0, n-1}] is$$(-1/x)e^{-1/x}(E_1(-1/x)-n!x^nE_{n+1}(-1/x))$$
Aug 7, 2020 at 6:19 comment added Gerry Myerson Letting $x=1$, we get the sequence tabulated at oeis.org/A007489 and since there isn't any useful formula given there it's highly unlikely there is one for general $x$.
Aug 7, 2020 at 6:16 history edited YCor CC BY-SA 4.0
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Aug 7, 2020 at 5:37 history edited user163402 CC BY-SA 4.0
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Aug 7, 2020 at 4:07 review First posts
Aug 7, 2020 at 6:11
Aug 7, 2020 at 4:01 history asked user163402 CC BY-SA 4.0