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#55 by N. J. A. Sloane at Sat Jun 05 16:36:48 EDT 2021
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#54 by Andrew Howroyd at Sat Jun 05 15:19:04 EDT 2021
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Discussion
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| Sat Jun 05
| 16:36
| N. J. A. Sloane: Andrew, the name suggests that a(0) should be 0. I suggest we leave the offset as 1, to avoid that difficulty!
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#53 by Andrew Howroyd at Sat Jun 05 15:18:03 EDT 2021
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(PARI) seq(n)={my(p=serreverse(2*x - x*exp(x + O(x^n)))/x); Vec(serlaplace(exp( intformal(p^2) )))} \\ Andrew Howroyd, Jun 05 2021
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proposed
editing
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Discussion
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| Sat Jun 05
| 15:19
| Andrew Howroyd: The formulas suggest a(0)=1.
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#52 by Michel Marcus at Sat Jun 05 14:40:18 EDT 2021
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#51 by Michel Marcus at Sat Jun 05 14:40:14 EDT 2021
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E.g.f. E(x) satisfies E'/E = y^2, where y=1+x+5*x^2/2+... is defined by y(*(2-exp(xyx*y))=1.
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proposed
editing
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#50 by Jon E. Schoenfield at Sat Jun 05 13:36:55 EDT 2021
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#49 by Jon E. Schoenfield at Sat Jun 05 13:36:25 EDT 2021
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E.g.f.: exp(int(RootOf(2*_Z-_Z*exp(x*_Z)-1)^2, x)).)) [in Maple notation].
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approved
editing
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Discussion
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| Sat Jun 05
| 13:36
| Jon E. Schoenfield: This is Maple notation, right? Is this edit a good way to indicate this to the user?
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#48 by Vaclav Kotesovec at Tue Mar 22 12:36:44 EDT 2016
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#47 by Vaclav Kotesovec at Tue Mar 22 12:36:08 EDT 2016
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a(n) ~ c * n^(n-2) / (r^n * exp(n)), where r = 2*(LambertW(2*exp(1))-1)^2 / LambertW(2*exp(1)) = 0.204378273928311464700648197201... and c = 1/((1 - 1/LambertW(2*exp(1))) * sqrt(2*exp(1)*(/2)*sqrt(2*(1 + 1/LambertW(2*exp(1))))) = 1.196923669815370203369255598062684... . - Vaclav Kotesovec, Mar 22 2016
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approved
editing
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#46 by Vaclav Kotesovec at Tue Mar 22 12:22:00 EDT 2016
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