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A013989
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a(n) = (n+1)( a(n-1)/n + a(n-2) ), with a(0)=1, a(1)=2.
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5
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1, 2, 6, 16, 50, 156, 532, 1856, 6876, 26200, 104456, 428352, 1821976, 7959056, 35857200, 165592576, 785514512, 3812387616, 18948962656, 96194028800, 498931946016, 2638959243712, 14234346694976
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| a(n) is also the number of fixed points in all involutions (= self-inverse permutations) of {1,2,...,n+1}. Example: a(2)=6 because the involutions of {1,2,3} are 1'2'3', 1'32, 32'1, and 213', containing 6 fixed points (marked). [From Emeric Deutsch (deutsch(AT)duke.poly.edu), May 28 2009]
a(n) is also the number of adjacent transpositions in all involutions (= self inverse permutations) of {1,2,...,n+2}. Example: a(2)=6 because the involutions of {1,2,3,4} are 1234, 124*3, 13*24, 1432, 2*134, 2*14*3, 3214, 3412, 4231, and 43*21, containing 6 adjacent transpositions (marked with *). [From Emeric Deutsch (deutsch(AT)duke.poly.edu), Jun 08 2009]
It might be more natural to shift the index by 1 and prefix a(0)=0, then this would be exactly the first differences of A000085, and satisfy a(n)=n for n<3, a(n)/n = a(n-1)/(n-1)-a(n-2). - M. F. Hasler, Dec 25 2010
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REFERENCES
| rec.puzzles Dec 10 1995
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FORMULA
| E.g.f: x*exp(x*(x/2+1)). - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 28 2005
a(n) = A000085(n) * (n+1).
a(n) = A000085(n+2) - A000085(n+1). - M. F. Hasler, Dec 26 2010
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MAPLE
| A013989 := proc(n) option remember; if n <=1 then n+1; else (n+1)*(A013989(n-1)/n+A013989(n-2)); fi; end;
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CROSSREFS
| First differences of A000085 (except for a missing leading zero).
Sequence in context: A151445 A195645 A000136 * A002841 A136509 A100664
Adjacent sequences: A013986 A013987 A013988 * A013990 A013991 A013992
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Dan Hoey (Hoey(AT)AIC.NRL.Navy.Mil), 1996
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