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#41 by Alois P. Heinz at Fri Nov 27 07:40:15 EST 2020
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#40 by Jean-François Alcover at Fri Nov 27 07:33:56 EST 2020
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#39 by Jean-François Alcover at Fri Nov 27 07:33:50 EST 2020
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| MATHEMATICA
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RecurrenceTable[{a[0] == 1, a[1] == 1, a[n] == n a[n-2] + (-1)^Floor[n/2]}, a, {n, 0, 32}] (* Jean-François Alcover, Nov 27 2020 *)
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approved
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#38 by N. J. A. Sloane at Sun Oct 25 22:44:58 EDT 2020
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#37 by Joerg Arndt at Sun Oct 25 10:39:16 EDT 2020
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#36 by Michel Marcus at Sun Oct 25 07:43:01 EDT 2020
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#35 by Michel Marcus at Sun Oct 25 07:42:18 EDT 2020
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Discussion
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| Sun Oct 25
| 07:43
| Michel Marcus: Pi rather than pi; formulas in formula section : block attribution for 2 formulas : see stylesheet
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#34 by Michel Marcus at Sun Oct 25 07:41:45 EDT 2020
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| COMMENTS
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a(2n)/A006882(2n) ~ 1/sqrt(e) = A092605. _From _Ryan Brooks_, Oct 25 2020: (Start)
a(2n)/A006882(2n) ~ 1/sqrt(e) = A092605.
a(2n+1)/A006882(2n+1) ~ sqrt(pi/(2*e))*erfi(1/sqrt(2)) = A306858. _Ryan Brooks_, Oct 25 2020. (End)
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proposed
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#33 by Ryan Brooks at Sun Oct 25 07:39:04 EDT 2020
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#32 by Ryan Brooks at Sun Oct 25 07:38:32 EDT 2020
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| COMMENTS
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a(2n)/A006882(2n) ~ 1/sqrt(e) = A092605.. _Ryan Brooks_, Oct 25 2020
a(2n+1)/A006882(2n+1) ~ sqrt(pi/(2*e))*erfi(1/sqrt(2)) = A306858.. _Ryan Brooks_, Oct 25 2020
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