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#60 by Michael De Vlieger at Tue Jul 16 16:15:05 EDT 2024
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#59 by Peter Luschny at Tue Jul 16 16:04:37 EDT 2024
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#58 by Alois P. Heinz at Tue Jul 16 15:52:15 EDT 2024
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#57 by Alois P. Heinz at Tue Jul 16 15:32:08 EDT 2024
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a(n) = (1/e)*Sum_{k >= >=1} (k^n - 1)/((k - 1)()*(k - 1)!). - Joseph Wheat, Jul 16 2024
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Discussion
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| Tue Jul 16
| 15:33
| Alois P. Heinz: added "*" ... now it works ...
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#56 by Alois P. Heinz at Tue Jul 16 15:31:28 EDT 2024
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#55 by Joseph Wheat at Tue Jul 16 15:16:34 EDT 2024
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Discussion
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| Tue Jul 16
| 15:31
| Alois P. Heinz: I cannot verify the new formula ...
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#54 by Joseph Wheat at Tue Jul 16 15:14:55 EDT 2024
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| FORMULA
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a(n) = (1/e)*Sum_{k >= 1} (1 + k + k^2 + ... + k^(n - 1)/((k - 1))/()(k - 1)!. - _)!). - _Joseph Wheat_, Jul 16 2024
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proposed
editing
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Discussion
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| Tue Jul 16
| 15:16
| Joseph Wheat: That looks much nicer, thank you Stefano.
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#53 by Joseph Wheat at Tue Jul 16 13:02:29 EDT 2024
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Discussion
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| Tue Jul 16
| 14:45
| Stefano Spezia: 1 + k + k^2 + ... + k^(n - 1) = (k^n - 1)/(k - 1). It is better to simplify rather to keep a double summation
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#52 by Joseph Wheat at Tue Jul 16 13:02:22 EDT 2024
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| FORMULA
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a(n) = (1/e)*Sum_{k >= 1} (1 + k + k^2 + ... + k^(n- - 1))/(k - 1)!. - Joseph Wheat, Jul 16 2024
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proposed
editing
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#51 by Joseph Wheat at Tue Jul 16 12:29:12 EDT 2024
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