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A007980
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Expansion of (1+x^2)/((1-x)^2*(1-x^3)).
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5
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1, 2, 4, 7, 10, 14, 19, 24, 30, 37, 44, 52, 61, 70, 80, 91, 102, 114, 127, 140, 154, 169, 184, 200, 217, 234, 252, 271, 290, 310, 331, 352, 374, 397, 420, 444, 469, 494, 520, 547, 574, 602, 631, 660, 690, 721, 752, 784, 817, 850, 884, 919, 954, 990, 1027, 1064
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Molien series for ternary self-dual codes over GF(3) of length 12n containing 11...1.
(1+x)*(1+x^2) / ((1-x)*(1-x^2)*(1-x^3)) is the Poincare series (or Molien series) for H^*(O_3(q); F_2).
a(n) is the position of the n-th triangular number in the running sum of the (pseudo-Orloj) sequence 1,2,1,2,1,2,1...., cf. A028355. - Wouter Meeussen (wouter.meeussen(AT)pandora.be), Mar 10 2002
a(n) = [a(n-1) + (number of even terms so far in the sequence)]. Example: 14 is [10 + 4 even terms so far in the sequence (they are 0,2,4,10)]. See A096777 for the same construction with odd integers. - Eric Angelini (eric.angelini(AT)kntv.be), Aug 05 2007
a(n)=A002378(n+1)/3 if 3 divides A002378(n+1), a(n)=(A002378(n)+1)/3 otherwise. [From Bruno Berselli (berselli.bruno(AT)yahoo.it), Oct 22 2010]
a(n) = A192736(n+1) / (n+1). [Reinhard Zumkeller, Jul 08 2011]
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REFERENCES
| A. Adem and R. J. Milgram, Cohomology of Finite Groups, Springer-Verlag, 2nd. ed., 2004; p. 233.
C. L. Mallows and N. J. A. Sloane, Weight enumerators of self-orthogonal codes over GF(3), SIAM J. Alg. Discrete Methods, 2 (1981), 452-460.
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LINKS
| Index entries for two-way infinite sequences
A. R. Calderbank and N. J. A. Sloane, Double circulant codes over Z_4, J. Algeb. Combin., 6 (1997) 119-131 (Abstract, pdf, ps).
Index entries for Molien series
Index to sequences with linear recurrences with constant coefficients, signature (2,-1,1,-2,1)
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FORMULA
| G.f.: (1+x^2)/((1-x)^2*(1-x^3)). a(n)=a(n-1)+a(n-3)-a(n-4)+2=a(-3-n).
a(n)=ceil((n+1)*(n+2)/3). - Paul Boddington (psb(AT)maths.warwick.ac.uk), Jan 26 2004
Contribution from Bruno Berselli (berselli.bruno(AT)yahoo.it), Oct 22 2010: (Start)
a(n) = ((n+1)*(n+2)+(2*cos(2*pi*n/3)+1)/3)/3 = sum[i=1..n+1] A004396(i).
a(n)-2*a(n-1)+a(n-2)-a(n-3)+2*a(n-4)-a(n-5) = 0 for n>4. (End)
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MAPLE
| with (combinat):seq(count(Partition((2*n+1)), size=3), n=1..56); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 28 2008
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MATHEMATICA
| Table[ Ceiling[ n(n+1)/3], {n, 56}]
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PROG
| (PARI) a(n)=if(n<-1, a(-3-n), polcoeff((1+x^2)/((1-x)^2*(1-x^3))+x*O(x^n), n))
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CROSSREFS
| Sequence in context: A055607 A024512 A047808 * A022339 A025711 A117634
Adjacent sequences: A007977 A007978 A007979 * A007981 A007982 A007983
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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