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A105384
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Expansion of x/(1+x+x^2+x^3+x^4).
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3
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0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0, 0, 1, -1, 0, 0
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OFFSET
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0,1
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COMMENTS
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Inverse binomial transform of A103311. A transform of the Fibonacci numbers: apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x),-x/(1-x)) followed by the inverse binomial transform (1/(1+x), x/(1+x)) (expressed as Riordan arrays) to the -F(n); equivalently, apply (1/(1+x^2),-x/(1+x^2)) to -F(n). Periodic {0,1,-1,0,0}.
Essentially the same as A010891. - R. J. Mathar, Apr 07 2008
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LINKS
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Table of n, a(n) for n=0..104.
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FORMULA
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Euler transform of length 5 sequence [ -1, 0, 0, 0, 1].
G.f.: x(1-x)/(1-x^5); a(n)=-sqrt(1/5+2sqrt(5)/25)cos(4*pi*n/5+pi/10)+sqrt(5)sin(4*pi*n/5+pi/10)/5+ sqrt(1/5-2sqrt(5)/25)cos(2*pi*n/5+3*pi/10)+sqrt(5)sin(2*pi*n/5+3*pi/10)/5
a(n)=-1/5*{[(n+2) mod 5]-2*[(n+3) mod 5]+[(n+4) mod 5]} with n>=0 - Paolo P. Lava, Nov 21 2006
a(n)=A010891(n-1). - R. J. Mathar, Apr 07 2008
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CROSSREFS
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Sequence in context: A030301 A071981 A093692 * A057212 A023959 A076182
Adjacent sequences: A105381 A105382 A105383 * A105385 A105386 A105387
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KEYWORD
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easy,sign
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AUTHOR
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Paul Barry, Apr 02 2005
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EXTENSIONS
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Corrected by N. J. A. Sloane, Nov 05 2005
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STATUS
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approved
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