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A053983
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a(n)=(2*n-1)*a(n-1)-a(n-2), a(0)=a(1)=1.
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2
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1, 1, 2, 9, 61, 540, 5879, 75887, 1132426, 19175355, 363199319, 7608010344, 174621038593, 4357917954481, 117489163732394, 3402827830284945, 105370173575100901, 3473812900148044788, 121478081331606466679
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Denominators of successive convergents to tan(1) using continued fraction 1/(1-1/(3-1/(5-1/(7-1/(9-1/(11-1/(13-1/15-...))))))).
Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Apr 20 2009: (Start)
Equals eigensequence of an infinite lower triangular matrix with (1, 3, 5,...)
as the main diagonal and (-1, -1, -1,...) as the subdiagonal. (End)
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FORMULA
| E.g.f. = (sin(1)*cos(sqrt(1-2*x))+cos(1)*sin(sqrt(1-2*x)))/sqrt(1-2*x). If a(0)=0, a(n)=0, 1, 1, 2, 9, 61, 540, 5879, 75887, 1132426... then E.g.f. = sin(1)*cos(sqrt(1-2*x))-cos(1)*sin(sqrt(1-2*x)). - Miklos Kristof (kristmikl(AT)freemail.hu), Jun 15 2005
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EXAMPLE
| a(10)=363199319 because 1/(1-1/(3-1/(5-1/(7-1/(9-1/(11-1/(13-1/(15-1/(17-1/19)))))))))=565649425/363199319
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MAPLE
| E(x):=sin(1)*cos(sqrt(1-2*x))-cos(1)*sin(sqrt(1-2*x)): f[0]:=E(x): for n from 1 to 30 do f[n]:=diff(f[n-1], x) od: x:=0: for n from 1 to 30 do f[n]:=simplify(f[n]/(sin(1)^2+cos(1)^2)) od: seq(f[n], n=1..30); (Kristof)
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MATHEMATICA
| RecurrenceTable[{a[0]==a[1]==1, a[n]==(2n-1)a[n-1]-a[n-2]}, a, {n, 20}] (* From Harvey P. Dale, Dec 21 2011 *)
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CROSSREFS
| A053984(n)=(-1)^n*A053983(-1-n). A053983(n)=-(-1)^n*A053984(-1-n).
Cf. A053984.
Sequence in context: A120014 A036774 A166882 * A192939 A107883 A088182
Adjacent sequences: A053980 A053981 A053982 * A053984 A053985 A053986
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KEYWORD
| easy,frac,nonn
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AUTHOR
| Vladeta Jovovic (vladeta(AT)eunet.rs), Apr 02 2000
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EXTENSIONS
| Additional comments from Michael Somos, Aug 23, 2000
More terms from Miklos Kristof (kristmikl(AT)freemail.hu), Jun 15 2005
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