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A131743
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Period 4: repeat 0, 1, 0, 2.
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4
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0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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LINKS
| Index to sequences with linear recurrences with constant coefficients, signature (0,0,0,1).
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FORMULA
| a(n)=(1/8)*{5*(n mod 4)-3*[(n+1) mod 4]+3*[(n+2) mod 4]-[(n+3) mod 4]}, with n>=0 - Paolo P. Lava (paoloplava(AT)gmail.com), Oct 02 2007
G.f.: -x*(1+2*x^2)/ ((x-1)*(x+1)*(x^2+1)). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 15 2007
a(n) = 3/4-1/2*sin(1/2*Pi*n)+3/4*(-1)^(1+n). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 15 2007
a(n)= (1-(-1)^n)*(3+i^(n+1))/4, with i=sqrt(-1) - Paolo P. Lava (paoloplava(AT)gmail.com), Jul 17 2008, Bruno Berselli, Mar 14 2011
a(n)=Fibonacci(2*n) mod 3. [From Gary Detlefs (gdetlefs(AT)aol.com) Feb 13 2011]
a(n)==A006368(n) mod 3. - From DELEHAM Philippe, Oct 24 2011.
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CROSSREFS
| Sequence in context: A194591 A070105 A111397 * A147648 A113686 A192033
Adjacent sequences: A131740 A131741 A131742 * A131744 A131745 A131746
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KEYWORD
| nonn,easy
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AUTHOR
| Paul Curtz (bpcrtz(AT)free.fr), Sep 20 2007
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