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Search: id:A099496
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| 1, -2, 5, -13, 34, -89, 233, -610, 1597, -4181, 10946, -28657, 75025, -196418, 514229, -1346269, 3524578, -9227465, 24157817, -63245986, 165580141, -433494437, 1134903170, -2971215073, 7778742049, -20365011074, 53316291173, -139583862445, 365435296162, -956722026041
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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With interpolated zeros, a Chebyshev transform of A056594, which has g.f. 1/(1+x^2). The image of G(x) under the Chebyshev transform is (1/(1+x^2))G(x/(1+x^2)).
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LINKS
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Index entries for sequences related to linear recurrences with constant coefficients
Tanya Khovanova, Recursive Sequences
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FORMULA
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G.f.: (1+x)/(1+3x+x^2);(with interpolated zeros) (1+x^2)/(1+3x^2+x^4); a(n)=sum{k=0..floor(n/2), binomial(n-k, k)(-1)^k*cos((n-2k)*pi/2)} (with interpolated zeros); a(n)=F(n+1)(-1)^(n/2)(1+(-1)^n)/2 (with interpolated zeros).
a(n)=[(-1)^n]*[Sum{k=0..n+1}(binomial(n+k,n-k)], with n>=0 - Paolo P. Lava (ppl(AT)spl.at), Apr 13 2007
a(n)=-3*a(n-1)-a(n-2),a(0)=1, a(1)=-2. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 03 2008]
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MATHEMATICA
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lst={}; Do[AppendTo[lst, (-1)^n*Fibonacci[2*n+1]], {n, 5!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Jan 18 2009]
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CROSSREFS
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Adjacent sequences: A099493 A099494 A099495 this_sequence A099497 A099498 A099499
Sequence in context: A122367 A001519 A048575 this_sequence A114299 A112842 A097417
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KEYWORD
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easy,sign
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AUTHOR
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Paul Barry (pbarry(AT)wit.ie), Oct 19 2004
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