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Revision History for A056040 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
Swinging factorial, a(n) = 2^(n-(n mod 2))*Product_{k=1..n} k^((-1)^(k+1)).
(history; published version)
#115 by Peter Luschny at Wed Dec 07 03:01:15 EST 2022
STATUS

editing

approved

#114 by Peter Munn at Tue Dec 06 16:18:31 EST 2022
LINKS

Didier Guillet, <a href="/A056040/a056040.pdf">On swinging factorials and the lonely runner conjecture</a> (Text in French).

Discussion
Wed Dec 07
03:01
Peter Luschny: Thank you!
#113 by Peter Munn at Tue Dec 06 15:27:05 EST 2022
FORMULA

a(n) = n!/floor(n/2)!^2. [Essentially the original name.]

a(n) = n!/floor(n/2)!^2.

Discussion
Tue Dec 06
15:37
Peter Munn: This was the only formula in A056040 when this entry was created.  I see no reason that it should not preceed the more complex neighboring formula.
#112 by Peter Munn at Tue Dec 06 15:06:17 EST 2022
COMMENTS

a(n) is the number of vertices of the polytope resulting from the intersection of a an n-hypercube with the hyperplane perpendicular to and bisecting one of its long diagonals at its midpoint. - Didier Guillet, Jun 11 2018 [Edited by Peter Munn, Dec 06 2022]

LINKS

Didier Guillet, <a href="/A056040/a056040.pdf">On swinging factorials and the lonely runner conjecture</a>.

Didier Guillet, <a href="/A056040/a056040.pdf">On swinging factorials and the lonely runner conjecture</a>.

Discussion
Tue Dec 06
15:23
Peter Munn: I have been corresponding about this with Didier Guillet, who very happily gives permission to upload and store the document, and hadn't realised this could be done.
#111 by Peter Munn at Tue Dec 06 14:57:23 EST 2022
COMMENTS

a(n) is the number of vertices of the polytope resulting of from the intersection of a n-hypercube with the hyperplane perpendicular to one of its diagonals at its diagonalmidpoint. - Didier Guillet, Jun 11 2018 [Edited by _Peter Munn_, Dec 06 2022]

LINKS

Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/Orbitals">Orbitals</a>.

Didier Guillet, <a href="/A056040/a056040.pdf">On swinging factorials and the lonely runner conjecture</a>.

STATUS

approved

editing

#110 by Joerg Arndt at Thu Mar 10 03:17:11 EST 2022
STATUS

reviewed

approved

#109 by Michel Marcus at Thu Mar 10 03:13:30 EST 2022
STATUS

proposed

reviewed

#108 by Amiram Eldar at Thu Mar 10 03:11:08 EST 2022
STATUS

editing

proposed

#107 by Amiram Eldar at Thu Mar 10 03:10:47 EST 2022
COMMENTS

Sum_{n>=0} 1/a(n) = 4/3 + 8*Pi/(9*sqrt(3)). - Alexander R. Povolotsky, Aug 18 2012

FORMULA

Sum_{n>=0} 1/a(n) = 4/3 + 8*Pi/(9*sqrt(3)). - Alexander R. Povolotsky, Aug 18 2012

From _Sum_{n>=0} (-1)^n/a(n) = 4/3 - 4*Pi/(9*sqrt(3)). - _Amiram Eldar_, Mar 10 2022: (Start)

Sum_{n>=0} 1/a(n) = 8*Pi/(9*sqrt(3)) + 4/3.

Sum_{n>=0} (-1)^n/a(n) = 4/3 - 4*Pi/(9*sqrt(3)). (End)

STATUS

proposed

editing

Discussion
Thu Mar 10
03:11
Amiram Eldar: Yes. Fixed. Thanks.
#106 by Amiram Eldar at Thu Mar 10 03:02:07 EST 2022
STATUS

editing

proposed

Discussion
Thu Mar 10
03:09
Michel Marcus: your 1st sum is also Alexander R. Povolotsky, Aug 18 2012