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Search: id:A057682
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| A057682 |
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Sum((-1)^j*binomial(n,3*j+1),j=0..floor(n/3)). |
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+0
4
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| 0, 1, 2, 3, 3, 0, -9, -27, -54, -81, -81, 0, 243, 729, 1458, 2187, 2187, 0, -6561, -19683, -39366, -59049, -59049, 0, 177147, 531441, 1062882, 1594323, 1594323, 0, -4782969, -14348907, -28697814, -43046721, -43046721, 0, 129140163, 387420489, 774840978
(list; graph; listen)
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OFFSET
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0,3
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FORMULA
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Starting at 1, the binomial transform of A000484. - Paul Barry (pbarry(AT)wit.ie), Jul 21 2003
It appears that Abs[a(n)]=Floor[Abs[A000748(n)]/3]. - John W. Layman (layman(AT)math.vt.edu), Sep 05 2003
a(n)={(3+i*sqrt(3))/2}^(n-2)+{(3-i*sqrt(3))/2}^(n-2) - Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 27 2003
G.f.:(x-x^2)/(1-3*x+3*x^2). a(n)=3*a(n-1)-3*a(n-2), if n>1.
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PROGRAM
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(PARI) a(n)=sum(j=0, n\3, (-1)^j*binomial(n, 3*j+1)) /* Michael Somos May 26 2004 */
(PARI) a(n)=if(n<2, n>0, n-=2; polsym(x^2-3*x+3, n)[n+1]) /* Michael Somos May 26 2004 */
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CROSSREFS
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Alternating row sums of triangle A030523.
Sequence in context: A106242 A121474 A138003 this_sequence A124841 A085355 A103120
Adjacent sequences: A057679 A057680 A057681 this_sequence A057683 A057684 A057685
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KEYWORD
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sign
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Oct 20 2000
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