|
Search: id:A105384
|
|
|
| A105384 |
|
Expansion of x/(1+x+x^2+x^3+x^4). |
|
+0
3
|
|
| 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
(list; graph; listen)
|
|
|
OFFSET
|
0,1
|
|
|
COMMENT
|
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 (mathar(AT)strw.leidenuniv.nl), Apr 07 2008
|
|
FORMULA
|
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 (ppl(AT)spl.at), Nov 21 2006
a(n)=A010891(n-1). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 07 2008
|
|
CROSSREFS
|
Sequence in context: A030301 A071981 A093692 this_sequence A057212 A023959 A076182
Adjacent sequences: A105381 A105382 A105383 this_sequence A105385 A105386 A105387
|
|
KEYWORD
|
easy,sign
|
|
AUTHOR
|
Paul Barry (pbarry(AT)wit.ie), Apr 02 2005
|
|
EXTENSIONS
|
Corrected by njas, Nov 05 2005
|
|
|
Search completed in 0.002 seconds
|
