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A072985
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Coefficient of the highest power of q in the expansion of nu(0)=1, nu(1)=b and for n >= 2, nu(n) = b*nu(n-1) + lambda*(n-1)_q*nu(n-2) with (b,lambda)=(2,3), where (n)_q = (1+q+...+q^(n-1)) and q is a root of unity.
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1
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1, 2, 7, 6, 21, 18, 63, 54, 189, 162, 567, 486, 1701, 1458, 5103, 4374, 15309, 13122, 45927, 39366, 137781, 118098, 413343, 354294, 1240029, 1062882, 3720087, 3188646, 11160261, 9565938, 33480783, 28697814, 100442349, 86093442
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OFFSET
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0,2
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COMMENTS
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Instead of listing the coefficients of the highest power of q in each nu(n), if we list the coefficients of the smallest power of q (i.e., constant terms), we get a sequence of weighted Fibonacci numbers described by f(0)=1, f(1)=1, for n >= 2, f(n) = 2*f(n-1) + 3*f(n-2).
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LINKS
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FORMULA
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For given b and lambda, the recurrence relation is given by; t(0)=1, t(1)=b, t(2) = b^2+lambda and for n >= 3, t(n) = lambda*t(n-2).
G.f.: (1 + 2*x + 4*x^2)/(1-3*x^2). - R. J. Mathar, Dec 05 2007
a(n) = (1/6)*(13 + (-1)^n)*3^floor(n/2) for n>0. - Ralf Stephan, Jul 19 2013
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EXAMPLE
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nu(0) = 1;
nu(1) = 2;
nu(2) = 7;
nu(3) = 20 + 6q;
nu(4) = 61 + 33q + 21q^2;
nu(5) = 182 + 144q + 120q^2 + 78q^3 + 18q^4;
nu(6) = 547 + 570q + 585q^2 + 501q^3 + 381q^4 + 162q^5 + 63q^6; ...
The coefficients of the highest power of q give this sequence.
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MATHEMATICA
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CoefficientList[Series[-(1 + 2 x + 4 x^2) / (-1 + 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
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PROG
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(Magma) [1] cat [(1/6)*(13+(-1)^n)*3^Floor(n/2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013
(PARI) x='x+O('x^30); Vec((1+2*x+4*x^2)/(1-3*x^2)) \\ G. C. Greubel, May 26 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
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EXTENSIONS
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STATUS
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approved
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